Quadratic Equations Class 10 Previous Year Questions: 1 Mark (Chapter 4)

1-mark questions on Quadratic Equations

This page has 29 1-mark questions on Quadratic Equations from CBSE Class 10 board papers of 2023, 2024, 2025 and 2026. Of these, 12 appeared only in Standard papers and 17 appeared only in Basic papers. Question formats: 29 MCQs. 16 of them were asked in two or more paper sets.

0 of 29 practised

All 29 Quadratic Equations 1-mark questions

Q1. Which of the following equations is a quadratic equation ?

Mark practised

(A) \(x^2 = (x + 1)^2\)
(B) \((x - 1)(x + 2) = 2x + 1\)
(C) \((x + 2)^3 = 2x(x^2 - 1)\)
(D) \(\sqrt{x} = x^2\)

Marks: 1
Level: Basic
Asked in: 6 paper sets
Years: 2025, 2026
Topic: 4.2 Quadratic Equations
Source: (CBSE 2025 Mathematics Basic, Set 430/2/1, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) (x − 1)(x + 2) = 2x + 1, because it simplifies to x² − x − 3, a quadratic polynomial.
Why? A quadratic equation can be written with a non-zero x² term and no higher power of x. Option B simplifies to x² − x − 3, which has degree 2, so it is quadratic.

The full 1 mark is for the correct option.

Answer:

(B) (x − 1)(x + 2) = 2x + 1

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/2/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing A because both sides contain x², without simplifying; x² = (x + 1)² becomes 2x + 1 = 0, which is linear.
Choosing C because it contains (x + 2)³, without simplifying; on simplification it becomes x³ − 6x² − 14x − 8 = 0, which is cubic.
Choosing D because √x = x² looks like it has an x² term; √x is not a polynomial term, so the equation is not of the form ax² + bx + c = 0.

Why this works

To decide, simplify each equation and compare with ax² + bx + c = 0, a ≠ 0. A: x² = (x + 1)² = x² + 2x + 1, so 2x + 1 = 0; this is linear. B: (x − 1)(x + 2) = x² + x − 2, so x² + x − 2 = 2x + 1, giving x² − x − 3 = 0; this is quadratic. C: (x + 2)³ = x³ + 6x² + 12x + 8 and 2x(x² − 1) = 2x³ − 2x, so the equation becomes x³ − 6x² − 14x − 8 = 0, which is cubic. D: √x = x² has √x, a term with power 1/2, so it is not a polynomial equation of degree 2. Therefore B is the quadratic equation. Another correct method also earns full marks.

Key idea to remember

Standard form of a quadratic equation: see Key ideas used on this page.

Explanation

A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. To test an equation, simplify it completely and check that the highest power of x is 2 with non-zero coefficient.

Important formulas

(x + y)² = x² + 2xy + y² · Used to expand option A.
(x + y)³ = x³ + 3x²y + 3xy² + y³ · Used to expand option C.
(x − 1)(x + 2) = x² + x − 2 · Used to simplify option B.

Q2. The discriminant of the quadratic equation \(x^2 - 3x - 2 = 0\) is :

Mark practised

(A) 1
(B) 17
(C) \(\sqrt{17}\)
(D) \(-\sqrt{17}\)

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2025, 2026
Topic: 4.4 Nature of Roots
Source: (CBSE 2025 Mathematics Basic, Set 430/1/1, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 17, because b² − 4ac = (−3)² − 4(1)(−2) = 9 + 8 = 17.
Why? The discriminant of the quadratic equation ax² + bx + c is b² − 4ac; here a is 1, b is −3 and c is −2.

The full 1 mark is for the correct option.

Answer:

(B) 17

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/1/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking c as 2 instead of −2, which gives 1, option A.
Writing b² as −9 because b is −3; the square of −3 is 9.
Subtracting the product instead of adding it: −4(1)(−2) is +8, so the discriminant is 17, not 1.

Why this works

The given quadratic equation is already in standard form, so we read off the coefficients directly: a is 1, b is −3, c is −2. The discriminant is b² − 4ac. Put the numbers in carefully: (−3)² gives 9, and −4(1)(−2) gives +8, so the discriminant is 17. A positive discriminant means the equation has two distinct real roots, but this question only asks for the discriminant itself. Quick check: (−3)² gives 9 and 4 × 1 × 2 gives 8, so 9 plus 8 gives 17.

Key idea to remember

Discriminant: see Key ideas used on this page.

Explanation

For a quadratic equation ax² + bx + c, the expression b² − 4ac is called the discriminant. It tells whether the roots are real and distinct, equal, or not real. Here we only evaluate it, not decide the nature of the roots.

Important formulas

b² − 4ac · The discriminant of the quadratic equation ax² + bx + c.

Q3. The discriminant of the quadratic equation \(2x^2 - 3x - 5 = 0\) is :

Mark practised

(A) \(-31\)
(B) \(49\)
(C) \(7\)
(D) \(\sqrt{-31}\)

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2025, 2026
Topic: 4.4 Nature of Roots
Source: (CBSE 2025 Mathematics Basic, Set 430/1/3, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 49, because b² − 4ac = (−3)² − 4(2)(−5) = 9 + 40 = 49.
Why? For a quadratic equation ax² + bx + c = 0, the discriminant is b² − 4ac; its value decides whether the roots are real and distinct, equal, or not real.

The full 1 mark is for the correct option.

Answer:

(B) 49

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/1/3: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking c = 5 instead of c = −5, which gives 9 − 40 and the wrong option −31.
Choosing 7, which is √49, instead of the discriminant 49 itself.
Using a = 2, b = 3, c = 5 and ignoring the minus signs in b and c.

Why this works

For 2x² − 3x − 5 = 0, the coefficient of x² is 2, the coefficient of x is −3, and the constant term is −5. So D = (−3)² − 4(2)(−5) = 9 + 40 = 49. Since 49 > 0, the roots are two distinct real roots. Option C is 7, which is √49, not the discriminant. Option A is −31, which comes from the sign error 9 − 40.

Key idea to remember

Discriminant of a quadratic equation: see Key ideas used on this page.

Explanation

For ax² + bx + c = 0, the expression b² − 4ac is called the discriminant. Its sign tells the nature of the roots: positive means two distinct real roots, zero means two equal real roots, negative means no real roots. Here D = 49 > 0, so the roots are real and distinct.

Important formulas

D = b² − 4ac · Discriminant of a quadratic equation in standard form.

Q4. If \(x = \sqrt{x}\) , ( \(x \neq 0\) ) is expressed as a quadratic equation in the form \(ax^2 + bx + c = 0\) , then the value of \(a + b + c\) is :

Mark practised

(A) 0
(B) 1
(C) 2
(D) 3

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2025, 2026
Topic: 4.2 Quadratic Equations
Source: (CBSE 2025 Mathematics Basic, Set 430/3/1, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 0, because squaring x = √x gives x² − x + 0 = 0, so a = 1, b = −1, c = 0 and a + b + c = 1 − 1 + 0 = 0.
Why? Squaring both sides removes the square root, and the simplified equation is compared with the standard form ax² + bx + c = 0 to read the coefficients.

The full 1 mark is for the correct option.

Answer:

(A) 0

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/3/1: option (A).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Writing x² + x = 0 after squaring, because the sign of x must change when it is brought to the left side; this gives b = 1 instead of b = −1.
Choosing 1 as the answer because x ≠ 0 makes x = 1 the root, but the question asks for a + b + c, not the root.

Why this works

The equation contains a square root, so square both sides. Since (√x)² = x, we get x² = x. Bring x to the left side: x² − x = 0. This is the standard form ax² + bx + c = 0 with a = 1, b = −1 and c = 0. Therefore a + b + c = 1 − 1 + 0 = 0, so option A is correct. Quick check: x = 1 satisfies the original equation, since 1 = √1, and the quadratic gives 1² − 1 = 0.

Key idea to remember

Standard form of a quadratic equation: see Key ideas used on this page.

Explanation

A quadratic equation in the variable x is of the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. The given equation is first simplified and written in this form; then a, b and c are the coefficients of x², x and the constant term.

Important formulas

ax² + bx + c = 0, a ≠ 0 · Standard form of a quadratic equation in the variable x.
(√x)² = x · Squaring removes the square root sign.

Q5. If \((\sqrt{x} + 1)^2 = x^2 + 2\sqrt{x}\) is expressed as a quadratic equation in the form of \(ax^2 + bx + c = 0\), then the value of \(a - b + c\) is :

Mark practised

(A) \(-1\)
(B) \(0\)
(C) \(1\)
(D) \(2\)

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2025, 2026
Topic: 4.2 Quadratic Equations
Source: (CBSE 2025 Mathematics Basic, Set 430/3/2, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 1, because (√x + 1)² = x + 2√x + 1, so the equation becomes x² − x − 1 = 0, and a − b + c = 1 − (−1) + (−1) = 1.
Why? In the standard form ax² + bx + c = 0, a is the coefficient of x², b is the coefficient of x, and c is the constant term.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/3/2: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Thinking (√x + 1)² expands to x² + 2√x + 1, so the x term and the x² term are not distinguished.
Taking b as 1 instead of −1 in the simplified quadratic, and then computing 1 − 1 − 1, which gives option A (−1).
Taking c as zero because the 2√x terms cancel, and then adding 1 and 1, which leads to option D (2).

Why this works

Expand the left side: (√x + 1)² = x + 2√x + 1. So the given equation becomes x + 2√x + 1 = x² + 2√x. The term 2√x is on both sides, so it cancels, leaving x + 1 = x². Rearranging, and keeping the x² coefficient positive, gives x² − x − 1 = 0, the standard form, with coefficients 1, −1, −1. Therefore a − b + c = 1 − (−1) + (−1) = 1. NCERT Example 2 shows that an equation must be simplified before deciding whether it is quadratic. The usual standard form keeps the x² coefficient positive, so the form −x² + x + 1 is not used.

Key idea to remember

Standard form of a quadratic equation: see Key ideas used on this page.

Explanation

A quadratic equation in x is ax² + bx + c = 0, a ≠ 0, and this is called the standard form. The given equation contains √x, so first simplify by expanding and cancelling equal terms. Comparing the simplified equation with the standard form gives the coefficients, and then a − b + c is evaluated.

Important formulas

(m + n)² = m² + 2mn + n² · Used to expand (√x + 1)².
ax² + bx + c = 0, a ≠ 0 · Standard form of a quadratic equation; after simplification the coefficients are read off.

Q6. The value(s) of k for which the roots of quadratic equation \(x^2 + 4x + k = 0\) are real, is :

Mark practised

(A) \(k \geq 4\)
(B) \(k \leq 4\)
(C) \(k \geq -4\)
(D) \(k \leq -4\)

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2023
Topic: 4.4 Nature of Roots
Source: (CBSE 2023 Mathematics Basic, Set 430/5/1, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) k ≤ 4, because the roots are real when b² − 4ac = 4² − 4(1)(k) = 16 − 4k ≥ 0, i.e. k ≤ 4.
Why? For a quadratic equation ax² + bx + c = 0, the roots are real exactly when the discriminant b² − 4ac is not negative.

The full 1 mark is for the correct option.

Answer:

(B) k ≤ 4

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/5/1: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing k ≥ 4 because the inequality is reversed while solving 16 − 4k ≥ 0.
Dropping k = 4 and writing k < 4, forgetting that equal roots are also real roots.
Using b² − 4ac < 0 as the condition for real roots instead of b² − 4ac ≥ 0.

Why this works

Compare x² + 4x + k = 0 with ax² + bx + c = 0. Here a = 1, b = 4 and c = k. The discriminant is D = b² − 4ac = 16 − 4k. Roots are real exactly when D ≥ 0. So 16 − 4k ≥ 0, which gives k ≤ 4. Quick check: at k = 4, D = 16 − 16 = 0, so the roots are equal and real; at k = 3, D = 16 − 12 = 4 > 0, so the roots are distinct and real; at k = 5, D = 16 − 20 = −4 < 0, so there are no real roots. Therefore the boundary is k = 4, and every value below it works.

Key idea to remember

Discriminant: see Key ideas used on this page.

Explanation

For a quadratic equation ax² + bx + c = 0, the expression b² − 4ac is called the discriminant. It decides the nature of the roots: positive means two distinct real roots, zero means two equal real roots, and negative means no real roots. This question uses the condition b² − 4ac ≥ 0 for real roots.

Important formulas

b² − 4ac · The discriminant of ax² + bx + c = 0; roots are real when b² − 4ac ≥ 0.
x = (−b ± √(b² − 4ac))/(2a) · The quadratic formula, which gives real roots when the discriminant is non-negative.

Q7. If one root of the quadratic equation \(x^2 - 4x + 3 = 0\) is 1, then the other root is :

Mark practised

(A) 4
(B) -4
(C) 3
(D) -3

Marks: 1
Level: Basic
Asked in: 3 paper sets
Compartment: 2024
Topic: 4.3 Solution of a Quadratic Equation by Factorisation
Source: (CBSE 2024 Compartment Mathematics Basic, Set 430/S/1, Q13, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 3, because x² − 4x + 3 = (x − 1)(x − 3), so the roots are 1 and 3.
Why? The factorisation method states that a quadratic written as a product of two linear factors has its roots where each factor equals zero. Since one root is already known, the other root comes from the other factor.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Compartment, Paper 430/S/1: option (C).

Open CBSE’s official marking scheme for the 2024 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 4 because the coefficient of x in x² − 4x + 3 is 4, without using the constant term 3.
Choosing −3 because the constant term is 3, without factorising x² − 4x + 3.
Writing the other root as −3, confusing the factor (x − 3) with x = −3.

Why this works

The middle term −4x splits as −x − 3x because (−1) + (−3) = −4 and (−1)(−3) = 3. So x² − 4x + 3 = x² − x − 3x + 3 = x(x − 1) − 3(x − 1) = (x − 1)(x − 3). Equating each factor to zero gives x = 1 and x = 3. Since one root is given as 1, the other root is 3. Quick check: 3² − 4(3) + 3 = 0 and 1² − 4(1) + 3 = 0. Another correct method also earns full marks.

Key idea to remember

Solution of a Quadratic Equation by Factorisation: see Key ideas used on this page.

Explanation

The factorisation method rewrites the quadratic polynomial as a product of two linear factors. Equating each factor to zero gives the roots of the equation. For x² − 4x + 3, the factors are (x − 1) and (x − 3), so the roots are 1 and 3; the root already given is 1, so the other root is 3.

Important formulas

x = (−b ± √(b² − 4ac))/(2a) · Quadratic formula, used here to verify the roots.

Q8. Which of the following is not a quadratic equation ?

Mark practised

(A) \((x - 2)^2 + 1 = 2x - 3\)
(B) \((2x - 1)(x - 3) = (x + 5)(x - 1)\)
(C) \(x(x + 1) + 8 = (x + 2)(x - 2)\)
(D) \(2x + \frac{3}{x} = 5\)

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2024
Topic: 4.2 Quadratic Equations
Source: (CBSE 2024 Mathematics Basic, Set 430/4/2, Q15, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) x(x + 1) + 8 = (x + 2)(x − 2), because after simplification it becomes x + 12, which is linear, not quadratic.
Why? A quadratic equation must be of the form ax² + bx + c = 0 with a ≠ 0; when the x² terms cancel, the equation is linear, not quadratic.

The full 1 mark is for the correct option.

Answer:

(C)

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/4/2: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Marking option C as quadratic because x² appears on the left side; simplifying cancels the x² terms and leaves x + 12.
Marking option B as not quadratic because x² appears on both sides; simplifying gives x² − 11x + 8 = 0, which is quadratic.
Marking option D as the answer because of 3/x; for x ≠ 0, multiplying by x gives 2x² − 5x + 3 = 0, which is quadratic.

Why this works

Simplify each option and see whether the x² term survives. For option C, x(x + 1) + 8 = x² + x + 8 and (x + 2)(x − 2) = x² − 4, so x² + x + 8 = x² − 4 gives x + 12 = 0. The x² terms cancel, so the degree is 1, not 2. This is exactly the equation in NCERT Example 2 (ii), which is not quadratic. Check the others: option A gives x² − 6x + 8 = 0, option B gives x² − 11x + 8 = 0, and option D, for x ≠ 0, gives 2x² − 5x + 3 = 0. Only option C is not of the form ax² + bx + c = 0 with a ≠ 0.

Key idea to remember

Standard form of a quadratic equation: see Key ideas used on this page.

Explanation

An equation in x is a quadratic equation only when it can be written as ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. So we must simplify each equation before deciding. If the x² terms cancel, the equation is linear, not quadratic.

Important formulas

ax² + bx + c = 0, a ≠ 0 · Standard form of a quadratic equation.
(x + a)(x − a) = x² − a² · Used to simplify option C.
(x − a)² = x² − 2ax + a² · Used to simplify option A.

Q9. If the quadratic equation \(ax^2 + ax + c = 0\) ( \(a \neq 0\) ) has real and equal roots, then :

Mark practised

(A) \(a = 4c\)
(B) \(4a = c\)
(C) \(a = -4c\)
(D) \(c = -4a\)

Marks: 1
Level: Basic
Asked in: 3 paper sets
Compartment: 2025
Topic: 4.4 Nature of Roots
Source: (CBSE 2025 Compartment Mathematics Basic, Set 430/S/1, Q15, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) a = 4c, because equal real roots make the discriminant \(b^2 - 4ac\) zero, and here \(b = a\), so \(a = 4c\).
Why? For a quadratic equation, the discriminant must be zero when the roots are real and equal. Substituting \(b = a\) into the discriminant gives \(a^2 - 4ac = a(a - 4c)\); the condition \(a \neq 0\) leaves \(a = 4c\).

The full 1 mark is for the correct option.

Answer:

(A) a = 4c

✓ Answer matches the CBSE answer key, 2025 Compartment, Paper 430/S/1: option (A).

Open CBSE’s official marking scheme for the 2025 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing \(4a = c\): from \(a^2 = 4ac\), dividing by \(a\) gives \(a = 4c\), not \(4a = c\).
Choosing \(a = -4c\): the discriminant is \(b^2 - 4ac\), not \(b^2 + 4ac\), so the sign of \(4ac\) stays negative.
Forgetting the condition \(a \neq 0\) and keeping \(a = 0\) as a second possibility; only \(a = 4c\) remains.

Why this works

Compare \(ax^2 + ax + c = 0\) with the standard form \(ax^2 + bx + c = 0\). The coefficient of \(x\) is \(a\), so \(b = a\). Two equal real roots mean the discriminant is zero: \(b^2 - 4ac = 0\). Put \(b = a\) to get \(a^2 - 4ac = 0\), i.e. \(a(a - 4c) = 0\). Since \(a \neq 0\), the factor \(a - 4c\) must be zero, so \(a = 4c\). Quick check: when \(a = 4c\), the discriminant is \((4c)^2 - 4(4c)c = 0\), so the condition is satisfied.

Key idea to remember

Discriminant: see Key ideas used on this page.

Explanation

NCERT states that a quadratic equation \(ax^2 + bx + c = 0\) has two equal real roots when its discriminant \(b^2 - 4ac\) is zero. Here the coefficient of \(x\) is \(a\), so \(b = a\), and the condition becomes \(a(a - 4c) = 0\). Since \(a \neq 0\), the required relation is \(a = 4c\).

Important formulas

\(D = b^2 - 4ac\) · Discriminant of the quadratic equation \(ax^2 + bx + c = 0\).
For equal real roots, \(D = 0\) · Condition given in NCERT Section 4.4 for two equal real roots.

Q10. The value of 'a' for which \(ax^2 + x + a = 0\) has equal and positive roots is :

Mark practised

(A) 2
(B) -2
(C) \(\frac{1}{2}\)
(D) \(-\frac{1}{2}\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 4.4 Nature of Roots
Source: (CBSE 2025 Mathematics Standard, Set 30/4/1, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) −1/2, because 1² − 4(a)(a) = 0 gives a = ±1/2, and the equal root −1/(2a) is positive only for a = −1/2.
Why? The discriminant b² − 4ac decides the nature of roots: equal roots occur exactly when it is 0. The repeated root is −b/(2a), so its sign decides which value of a fits "positive".

The full 1 mark is for the correct option.

Answer:

(D) −1/2

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/4/1: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 1/2 because it makes the discriminant 1 − 4a² equal to 0, without checking that the equal root −1/(2a) is negative.
Writing the discriminant as 1 − 4a instead of 1 − 4a², because the constant term of ax² + x + a is a.
Taking 1/2 or −1/2 as the answer and ignoring the word "positive" in the question.

Why this works

For equal roots, the discriminant b² − 4ac must be 0. Here b = 1 and c = a, so the condition is 1 − 4a² = 0, which gives a = 1/2 or a = −1/2. When the roots are equal, the quadratic formula gives the repeated root x = −b/(2a) = −1/(2a). For a = 1/2 this root is −1, so it is not positive. For a = −1/2 this root is 1, which is positive. Check: a = −1/2 gives the equation −(1/2)x² + x − 1/2 = 0, or x² − 2x + 1 = 0, whose roots are 1 and 1.

Key idea to remember

Discriminant of a quadratic equation: see Key ideas used on this page.

Explanation

For ax² + bx + c = 0, the discriminant is b² − 4ac. If it is 0, the equation has two equal real roots, and each root is −b/(2a). This solution sets the discriminant to 0 to find the possible values of a, then uses the repeated root to choose the value that gives a positive root.

Important formulas

D = b² − 4ac · The discriminant; D = 0 gives two equal real roots.
x = (−b ± √(b² − 4ac))/(2a) · Quadratic formula; when D = 0, both roots are −b/(2a).

Q11. If \(x^2 + bx + b = 0\) has two real and distinct roots, then the value of \(b\) can be

Mark practised

(A) 0
(B) 4
(C) 3
(D) -3

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 4.4 Nature of Roots
Source: (CBSE 2025 Mathematics Standard, Set 30/5/1, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) −3, because for this value the discriminant b² − 4b > 0.
Why? A quadratic equation has two distinct real roots exactly when its discriminant is positive, so the value of b that makes b² − 4b positive is correct.

The full 1 mark is for the correct option.

Answer:

(D) −3

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/5/1: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 0 or 4, because the discriminant becomes 0, which means two equal real roots, not two distinct real roots.
Choosing 3, because 3² − 4×3 is negative, so the equation has no real roots.
Forgetting that the constant term is also b, so the discriminant is b² − 4b, not b² − b.

Why this works

The discriminant of x² + bx + b = 0 is D = b² − 4b. For two distinct real roots, D must be positive. Test the options: for 0, D = 0² − 4×0 = 0; for 4, D = 4² − 4×4 = 0; for 3, D = 3² − 4×3 = −3; for −3, D = (−3)² − 4×(−3) = 9 + 12 = 21. Only 21 is positive, so option −3 gives two distinct real roots.

Key idea to remember

Discriminant: see Key ideas used on this page.

Explanation

For a quadratic equation ax² + bx + c = 0, the value b² − 4ac is called the discriminant. If the discriminant is positive, the equation has two distinct real roots; if it is zero, two equal real roots; if it is negative, no real roots. This solution uses the discriminant to test each option and picks the one that makes it positive.

Important formulas

D = b² − 4ac · Discriminant of the quadratic equation ax² + bx + c = 0.
D > 0 ⇒ two distinct real roots · NCERT condition used to choose the value of b.

Q12. The discriminant of the quadratic equation \(2x^2 - 5x - 3 = 0\) is

Mark practised

(A) 1
(B) 49
(C) 7
(D) 19

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2023
Topic: 4.4 Nature of Roots
Source: (CBSE 2023 Mathematics Basic, Set 430/1/1, Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 49, because b² − 4ac = (−5)² − 4 × 2 × (−3) = 25 + 24 = 49.
Why? The discriminant of a quadratic equation is b² − 4ac; substituting the coefficients of the given equation gives its value.

The full 1 mark is for the correct option.

Answer:

(B) 49

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/1/1: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking the constant term with the wrong sign, so the last term of the discriminant changes sign.
Choosing 7, the square root of 49, instead of the discriminant itself.
Forgetting that the square of a negative number is positive.

Why this works

Substitute the coefficients of the given equation into the discriminant formula. The coefficient of x is negative, so its square is positive; the product of the coefficient of x² and the constant term is negative, so subtracting that product adds. That gives (−5)² = 25 and 4 × 2 × (−3) = −24, so b² − 4ac = 25 − (−24) = 25 + 24 = 49. Since 49 is positive, the roots are real and distinct, but only the value of the discriminant is asked. Quick check: 49 = 7².

Key idea to remember

Discriminant: see Key ideas used on this page.

Explanation

The discriminant b² − 4ac of a quadratic equation decides whether the roots are two distinct real roots, two equal real roots, or no real roots. Here it is positive, so the correct option is 49.

Important formulas

b² − 4ac · Discriminant of a quadratic equation in standard form.

Q13. A quadratic equation whose roots are \((2 + \sqrt{3})\) and \((2 - \sqrt{3})\) is :

Mark practised

(A) \(x^2 - 4x + 1 = 0\)
(B) \(x^2 + 4x + 1 = 0\)
(C) \(4x^2 - 3 = 0\)
(D) \(x^2 - 1 = 0\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2023
Topic: 4.3 Solution of a Quadratic Equation by Factorisation
Source: (CBSE 2023 Mathematics Standard, Set 30/5/1, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) x² − 4x + 1 = 0, because the sum of the roots is (2 + √3) + (2 − √3) = 4 and the product is (2 + √3)(2 − √3) = 1.
Why? When the roots are known, the quadratic is the product (x − α)(x − β); expanding it gives the coefficient of x as the negative of the sum of the roots and the constant term as their product.

The full 1 mark is for the correct option.

Answer:

(A) x² − 4x + 1 = 0

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/5/1: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option B because the roots contain +√3 and −√3, without adding them: the sum is 4, so the middle term must be −4x.
Thinking (2 + √3)(2 − √3) is 2² + (√3)², instead of using (a + b)(a − b) = a² − b², which gives 1.
Writing the constant term as −1 because one root has a minus sign, although the product of the two roots is +1.

Why this works

For roots α and β, the quadratic expression is (x − α)(x − β). Expanding gives x² − (α + β)x + αβ. With the roots 2 + √3 and 2 − √3, adding the roots gives 4 and multiplying them gives 1, so the expression is x² − 4x + 1. Setting this expression to zero gives the required equation. Option B would need the sum of the roots to be −4, but the sum is 4. Another correct method also earns full marks.

Key idea to remember

Roots of a Quadratic Equation: see Key ideas used on this page.

Explanation

A real number α is a root of a quadratic equation if substituting it makes the quadratic expression zero. When α and β are the roots, the factors are (x − α) and (x − β). Expanding their product gives x² − (α + β)x + αβ, so the middle coefficient is the negative of the sum of the roots and the constant term is their product.

Important formulas

x² − (α + β)x + αβ = 0 · Use this when the two roots α and β are given; the middle coefficient is minus the sum of the roots and the constant term is their product.

Q14. Which of the following equations has 2 as a root ?

Mark practised

(A) \(x^2 - 4x + 5 = 0\)
(B) \(x^2 + 3x - 12 = 0\)
(C) \(2x^2 - 7x + 6 = 0\)
(D) \(3x^2 - 6x - 2 = 0\)

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2024
Topic: 4.3 Solution of a Quadratic Equation by Factorisation
Source: (CBSE 2024 Mathematics Basic, Set 430/2/1, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 2x² − 7x + 6 = 0, because substituting 2 for x gives 2(2)² − 7(2) + 6 = 0.
Why? A number is a root of a quadratic equation if substituting it for x makes the left-hand side zero. Only option C gives zero when 2 is substituted.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/2/1: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

For option D, forgetting the constant term −2: 3(2)² − 6(2) alone is 0, but the whole left side is −2.
For option A, thinking 4 − 8 + 5 = 1 is close enough to 0; the value must be exactly 0.
For option B, seeing 4 + 6 = 10 and not subtracting 12; the whole left side is −2.

Q15. The roots of the quadratic equation \(x^2 - 4 = 0\) is/are :

Mark practised

(A) 2 only
(B) -2, 2
(C) 4 only
(D) -4, 4

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2024
Topic: 4.3 Solution of a Quadratic Equation by Factorisation
Source: (CBSE 2024 Mathematics Basic, Set 430/4/1, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) -2, 2, because x² − 4 = (x − 2)(x + 2) = 0, so x = 2 or x = −2.
Why? By the zero product rule, if (x − 2)(x + 2) = 0, then x − 2 = 0 or x + 2 = 0, so both roots must be listed.

The full 1 mark is for the correct option.

Answer:

(B) -2, 2

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/4/1: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing '2 only' because √4 is taken as 2 alone, forgetting that (−2)² = 4 also.
Choosing '4 only' or '−4, 4' by taking the square root of 4 as 4 itself.
Forgetting that x² − 4 = 0 is the same as (x − 2)(x + 2) = 0 and that both factors give a root.

Why this works

The NCERT factorisation method says: write the quadratic as a product of two linear factors and equate each factor to zero. Here x² − 4 = x² − 2² = (x − 2)(x + 2). So x − 2 = 0 or x + 2 = 0, giving x = 2 or x = −2. A quick check: 2² − 4 = 0 and (−2)² − 4 = 0, so both are roots. Option A gives only 2 and misses −2. Options C and D give 4 and −4, which do not satisfy the equation because 4² − 4 = 12 and (−4)² − 4 = 12. Another correct method also earns full marks.

Key idea to remember

Root of a quadratic equation: see Key ideas used on this page.

Explanation

A real number α is a root of ax² + bx + c = 0 if aα² + bα + c = 0. A quadratic equation can have at most two roots. Here x² − 4 is factored as (x − 2)(x + 2) = 0, and equating each factor to zero gives the two roots 2 and −2.

Important formulas

x² − a² = (x − a)(x + a) · Difference of squares identity used to factorise x² − 4.
If (x − a)(x − b) = 0, then x = a or x = b. · Equating each linear factor to zero gives the roots.

Q16. If the discriminant of the quadratic equation \(3x^2 - 2x + c = 0\) is 16, then the value of c is :

Mark practised

(A) 1
(B) 0
(C) -1
(D) \(\sqrt{2}\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 4.4 Nature of Roots
Source: (CBSE 2024 Mathematics Standard, Set 30/4/1, Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) −1, because D = (−2)² − 4(3)(c) = 4 − 12c = 16, so c = −1.
Why? For a quadratic equation of the form ax² + bx + c, the discriminant is b² − 4ac; equating it to the given value gives a linear equation in c.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/4/1: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Using 4 + 12c instead of 4 − 12c while forming the discriminant, which gives a positive value for c and leads to option A.
Taking the given value 16 as the constant term c and writing b² − 4a(16), which is not what the question says.
Choosing √2 without checking whether the discriminant actually becomes 16 for it.

Why this works

Compare 3x² − 2x + c with the standard form ax² + bx + c: here the coefficient of x² is 3, the coefficient of x is −2, and the constant term is c. The discriminant is D = b² − 4ac = (−2)² − 4(3)(c) = 4 − 12c. The question gives D = 16, so 4 − 12c = 16, which gives c = (4 − 16)/12 = −1. Quick check: with c = −1, D = 4 − 12(−1) = 16. Another correct method, substituting each option into the discriminant, is also acceptable.

Key idea to remember

Discriminant of a quadratic equation: see Key ideas used on this page.

Explanation

For a quadratic equation ax² + bx + c, a ≠ 0, the expression b² − 4ac is called its discriminant. It tells whether the roots are real, equal or not real. Here the discriminant is given, so we substitute the coefficients and solve for c.

Important formulas

b² − 4ac · Discriminant of the quadratic equation ax² + bx + c, a ≠ 0.
D = (−2)² − 4(3)(c) · Substituting the coefficient of x² as 3 and the coefficient of x as −2.

Q17. The discriminant of the quadratic equation \(2x^2 + x - 1 = 0\) is :

Mark practised

(A) -9
(B) -7
(C) 9
(D) 7

Marks: 1
Level: Basic
Years: 2023
Topic: 4.4 Nature of Roots
Source: (CBSE 2023 Mathematics Basic, Set 430/1/2, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 9, because \(b^2 - 4ac = 1^2 - 4(2)(-1) = 1 + 8 = 9\).
Why? The discriminant of a quadratic \(ax^2 + bx + c\) is \(b^2 - 4ac\).

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/1/2: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking \(c\) as \(+1\) instead of \(-1\), so \(4ac\) is treated as \(+8\) and the answer comes out as \(-7\) (option B).
Forgetting that subtracting a negative number means adding, so the answer is positive, not \(-7\).

Q18. If \(x = 0.3\) , is a root of the equation \(x^2 - 0.9k = 0\) , then \(k\) is equal to :

Mark practised

(A) 1
(B) 10
(C) 0.1
(D) 100

Marks: 1
Level: Standard
Years: 2023
Topic: 4.3 Solution of a Quadratic Equation by Factorisation
Source: (CBSE 2023 Mathematics Standard, Set 30/2/2, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 0.1, because substituting x = 0.3 gives (0.3)² − 0.9k = 0, so k = 0.09 ÷ 0.9 = 0.1.
Why? A root of a quadratic equation satisfies it, so the left side becomes zero and the equation turns into a linear equation in k.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/2/2: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Substituting 3 in place of 0.3 and getting 10 for k.
Multiplying 0.09 by 0.9 and getting 0.081, which is not an option.
Taking k as just x², that is 0.09, and forgetting the factor 0.9 that multiplies k.

Q19. A quadratic equation whose roots are \((3 - \sqrt{2})\) and \((3 + \sqrt{2})\) is :

Mark practised

(A) \(x^2 - 6x + 7 = 0\)
(B) \(x^2 + 6x + 7 = 0\)
(C) \(9x^2 - 2 = 0\)
(D) \(x^2 - 7 = 0\)

Marks: 1
Level: Standard
Years: 2023
Topic: 4.3 Solution of a Quadratic Equation by Factorisation
Source: (CBSE 2023 Mathematics Standard, Set 30/5/3, Q10, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) \(x^2 - 6x + 7 = 0\), because \((x - (3 - \sqrt{2}))(x - (3 + \sqrt{2})) = x^2 - 6x + 7\)
Why? If a quadratic equation has roots \(\alpha\) and \(\beta\), then the corresponding linear factors are \(x - \alpha\) and \(x - \beta\); equating their product to zero gives the equation.

The full 1 mark is for the correct option.

Answer:

(A) x² − 6x + 7 = 0

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/5/3: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing B, \(x^2 + 6x + 7 = 0\), because the sum of the roots is 6 and the x-term is written with + instead of −.
Writing \((3-\sqrt{2})(3+\sqrt{2})\) as \(9 - \sqrt{2}\); the correct product is \(9 - 2 = 7\).
Choosing D, \(x^2 - 7 = 0\), because its roots are \(\pm \sqrt{7}\), not \(3 \pm \sqrt{2}\).

Why this works

Each root gives one linear factor. For the root \(3 - \sqrt{2}\), the factor is \(x - (3 - \sqrt{2})\); for the root \(3 + \sqrt{2}\), the factor is \(x - (3 + \sqrt{2})\). The product of these two factors is a quadratic equation with exactly those roots. Rewrite the product as \((x - 3 + \sqrt{2})(x - 3 - \sqrt{2})\). With \(y = x - 3\), this is \((y + \sqrt{2})(y - \sqrt{2}) = y^2 - 2\), which becomes \((x - 3)^2 - 2 = x^2 - 6x + 9 - 2 = x^2 - 6x + 7\). So option A is the equation. Quick check: put \(x = 3 - \sqrt{2}\) into \(x^2 - 6x + 7\); the value is 0. The same holds for \(x = 3 + \sqrt{2}\).

Key idea to remember

Roots of a quadratic equation: see Key ideas used on this page.

Explanation

NCERT says a real number \(\alpha\) is a root of \(ax^2 + bx + c = 0\) if \(a\alpha^2 + b\alpha + c = 0\), and roots are found by factorising the quadratic into linear factors and equating each factor to zero. This solution works backwards: since the roots are \(3 - \sqrt{2}\) and \(3 + \sqrt{2}\), the linear factors are \(x - (3 - \sqrt{2})\) and \(x - (3 + \sqrt{2})\), so their product gives the required equation.

Important formulas

\((x - \alpha)(x - \beta) = 0\) · A quadratic equation with roots \(\alpha\) and \(\beta\).
\((a + b)(a - b) = a^2 - b^2\) · Used to multiply \((y + \sqrt{2})(y - \sqrt{2})\).
\((x - 3)^2 = x^2 - 6x + 9\) · Expansion used to rewrite the product.

Q20. If 'p' is a root of the quadratic equation x² - (p + q) x + k = 0, then the value of 'k' is

Mark practised

(A) p
(B) q
(C) p + q
(D) pq

Marks: 1
Level: Standard
Years: 2023
Topic: 4.3 Solution of a Quadratic Equation by Factorisation
Source: (CBSE 2023 Mathematics Standard, Set 30/6/3, Q16, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) pq, because putting x = p in the equation gives p² − (p + q)p + k = 0, so k = pq.
Why? A root of a quadratic equation satisfies the equation, so substituting x = p makes the left side zero; solving this gives k = pq.

The full 1 mark is for the correct option.

Answer:

(D) pq

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/3: option (D).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing p because p is the root, without realising that the root and the constant k need not be equal.
Choosing p + q because (p + q) is the coefficient of x, without substituting x = p.
Dropping the negative sign: p² − (p + q)p is p² − p² − pq, so k = pq, not −pq.

Q21. The equation \(3x^2\) + 12 = 0 has :

Mark practised

(A) real and distinct roots
(B) no real roots
(C) real and equal roots
(D) roots which are reciprocal of each other

Marks: 1
Level: Basic
Compartment: 2024
Topic: 4.4 Nature of Roots
Source: (CBSE 2024 Compartment Mathematics Basic, Set 430/S/1, Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) no real roots, because b² − 4ac = 0² − 4×3×12 = −144, which is less than 0.
Why? By the discriminant rule, a quadratic equation has no real roots when b² − 4ac < 0.

The full 1 mark is for the correct option.

Answer:

(B) no real roots

✓ Answer matches the CBSE answer key, 2024 Compartment, Paper 430/S/1: option (B).

Open CBSE’s official marking scheme for the 2024 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing A or C because 3x² and 12 both look positive, without checking b² − 4ac.
Thinking that x squared can be −4 for a real x, and then writing the roots as ±2; no real number has a negative square.
Taking b as 0 and c as 12 but computing the discriminant as 0² + 4×3×12, which gives 144 instead of −144.

Why this works

The discriminant b² − 4ac decides the nature of the roots before solving. Here the coefficient of x (the number multiplying x) is 0, so b² − 4ac = 0² − 4×3×12 = −144, which is less than 0. Therefore there is no real number whose square equals b² − 4ac, so the equation has no real roots. A quick check: 3x² + 12 rearranges to x² = −12/3 = −4, and no real number has a negative square.

Key idea to remember

Discriminant: see Key ideas used on this page.

Explanation

The discriminant b² − 4ac decides the nature of roots. If b² − 4ac < 0, there is no real number whose square equals b² − 4ac, so the quadratic equation has no real roots. Here b is 0, so b² − 4ac = 0² − 4×3×12 = −144, which is less than 0.

Important formulas

b² − 4ac · Discriminant of a quadratic equation in the form ax² + bx + c. If it is negative, the equation has no real roots.

Q22. The equation \(x^2 + x + 1 = 0\) has :

Mark practised

(A) real and distinct roots
(B) no real roots
(C) real and equal roots
(D) both negative roots

Marks: 1
Level: Basic
Compartment: 2024
Topic: 4.4 Nature of Roots
Source: (CBSE 2024 Compartment Mathematics Basic, Set 430/S/2, Q13, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) no real roots, because the discriminant b² − 4ac = 1² − 4(1)(1) = −3 < 0.
Why? The discriminant b² − 4ac decides the nature of roots of ax² + bx + c; a negative discriminant means the equation has no real roots.

The full 1 mark is for the correct option.

Answer:

(B) no real roots

✓ Answer matches the CBSE answer key, 2024 Compartment, Paper 430/S/2: option (B).

Open CBSE’s official marking scheme for the 2024 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 'real and distinct roots' without first checking the discriminant of x² + x + 1.
Forgetting the negative sign when simplifying 1² − 4(1)(1), so taking the discriminant as positive and choosing real and distinct roots.
Assuming every quadratic equation has real roots, so missing that a negative discriminant means no real roots.

Why this works

Compare x² + x + 1 with ax² + bx + c, so the coefficient of x² is 1, the coefficient of x is 1 and the constant term is 1. The discriminant is b² − 4ac = 1² − 4(1)(1), which is −3. Since −3 < 0, there is no real number whose square is −3, so the quadratic formula x = (−b ± √(b² − 4ac))/(2a) cannot give real values. Therefore the equation has no real roots. Option B is correct.

Key idea to remember

Discriminant: see Key ideas used on this page.

Explanation

The discriminant is b² − 4ac for the quadratic equation ax² + bx + c. Its sign tells whether the roots are real: positive gives two distinct real roots, zero gives two equal real roots, and negative gives no real roots. Here the discriminant is negative, so the equation has no real roots.

Important formulas

b² − 4ac · Discriminant of the quadratic equation ax² + bx + c.
x = (−b ± √(b² − 4ac))/(2a) · Quadratic formula, used only when b² − 4ac ≥ 0.

Q23. The equation \(4x^2 - 25 = 0\) has :

Mark practised

(A) no real roots
(B) real and equal roots
(C) real and distinct roots
(D) real roots of same sign

Marks: 1
Level: Basic
Compartment: 2024
Topic: 4.4 Nature of Roots
Source: (CBSE 2024 Compartment Mathematics Basic, Set 430/S/3, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) real and distinct roots, because the discriminant b² − 4ac = 0² − 4 × 4 × (−25) = 400 > 0.
Why? For a quadratic equation, a positive discriminant gives two distinct real roots.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Compartment, Paper 430/S/3: option (C).

Open CBSE’s official marking scheme for the 2024 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing A because the middle term is zero, without computing the discriminant; the discriminant is positive here, so real roots exist.
Taking the constant term with the wrong sign gives a negative discriminant, which wrongly suggests no real roots.
Choosing D because the roots are 2.5 and −2.5, but these two real roots have opposite signs, so they are not of the same sign.

Q24. Which of the following quadratic equations has \(-1\) as a root?

Mark practised

(A) \(x^2 - 4x - 5 = 0\)
(B) \(-x^2 - 4x + 5 = 0\)
(C) \(x^2 + 3x + 4 = 0\)
(D) \(x^2 - 5x + 6 = 0\)

Marks: 1
Level: Basic
Years: 2024
Topic: 4.3 Solution of a Quadratic Equation by Factorisation
Source: (CBSE 2024 Mathematics Basic, Set 430/2/2, Q16, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) x² − 4x − 5 = 0, because (−1)² − 4(−1) − 5 = 1 + 4 − 5 = 0.
Why? By the definition of a root, x = −1 is a root exactly when substituting x = −1 makes the left-hand side of the equation equal to 0.

The full 1 mark is for the correct option.

Answer:

(A) x² − 4x − 5 = 0

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/2/2: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Substituting x = 1 instead of x = −1, which wrongly makes option (B) true.
Writing (−1)² as −1, so the test for option (A) is done with the wrong sign.
Choosing an option by looking at its constant term instead of substituting x = −1.

Why this works

Substitute x = −1 into each equation and simplify. For (A): (−1)² − 4(−1) − 5 = 1 + 4 − 5 = 0, so the equation is satisfied. For (B): −(−1)² − 4(−1) + 5 = −1 + 4 + 5 = 8 ≠ 0. For (C): (−1)² + 3(−1) + 4 = 1 − 3 + 4 = 2 ≠ 0. For (D): (−1)² − 5(−1) + 6 = 1 + 5 + 6 = 12 ≠ 0. Only (A) gives 0. A root is a solution, so (A) is the correct option.

Key idea to remember

Root of a quadratic equation: see Key ideas used on this page.

Explanation

A root of ax² + bx + c = 0 is a number α for which aα² + bα + c = 0. The zeroes of the quadratic polynomial ax² + bx + c are the same as the roots of the equation. So test each option by substituting x = −1 and seeing which left-hand side becomes 0.

Important formulas

aα² + bα + c = 0 · This is the test for a real number α to be a root of ax² + bx + c = 0.

Q25. If the quadratic equation \(ax^2 + bx + c = 0\) has real and equal roots, then the value of \(c\) is :

Mark practised

(A) \(\frac{b}{2a}\)
(B) \(-\frac{b}{2a}\)
(C) \(\frac{b^2}{4a}\)
(D) \(-\frac{b^2}{4a}\)

Marks: 1
Level: Standard
Years: 2024
Topic: 4.4 Nature of Roots
Source: (CBSE 2024 Mathematics Standard, Set 30/5/2, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) \(\frac{b^2}{4a}\), because real and equal roots mean the discriminant is zero, so \(c = \frac{b^2}{4a}\).
Why? The discriminant determines the nature of the roots: zero means both roots are \(-\frac{b}{2a}\), so the equation \(b^2 - 4ac = 0\) is the condition to solve.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/5/2: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option D, \(-\frac{b^2}{4a}\), by making the sign negative when moving \(4ac\) to the other side.
Choosing option B, \(-\frac{b}{2a}\), which is the equal root itself, not the value of \(c\).
Writing \(c = b^2\) or \(c = \frac{b^2}{4}\), forgetting to divide by \(a\) as well.

Why this works

The discriminant \(b^2 - 4ac\) decides whether the roots are real. For real and equal roots, the discriminant is zero, so \(b^2 - 4ac = 0\). Then \(4ac = b^2\), and because \(a \neq 0\), \(c = \frac{b^2}{4a}\). This is option C. Quick check: take \(x^2 + 2x + 1 = 0\), where \(a=1\), \(b=2\), \(c=1\). Then \(\frac{b^2}{4a} = \frac{2^2}{4 \times 1} = \frac{4}{4} = 1 = c\), and \(2^2 - 4 \times 1 \times 1 = 0\), so the roots are equal.

Key idea to remember

Discriminant: see Key ideas used on this page.

Explanation

The discriminant \(b^2 - 4ac\) tells the nature of the roots of \(ax^2 + bx + c = 0\). When it is zero, the quadratic formula gives both roots as \(-\frac{b}{2a}\), so the roots are real and equal. This solution sets the discriminant equal to zero and solves for \(c\).

Important formulas

\(b^2 - 4ac = 0\) · Condition for real and equal roots.
\(c = \frac{b^2}{4a}\) · Value obtained by solving the condition, since \(a \neq 0\).

Q26. Which of the following equations is a quadratic equation ?

Mark practised

(A) \(x^3 = (x - 1)^3 + 3x^2\)
(B) \(x^3 = (x + 1)^3\)
(C) \(x^2 = \sqrt{x}\)
(D) \(x^2 + 1 = \frac{1}{x}\)

Marks: 1
Level: Standard
Years: 2025
Topic: 4.2 Quadratic Equations
Source: (CBSE 2025 Mathematics Standard, Set 30/5/2, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) x³ = (x + 1)³, because expanding the cube leaves the quadratic polynomial 3x² + 3x + 1.
Why? A quadratic equation is an equation that can be written as a polynomial of degree 2; only option B simplifies to a polynomial of degree 2.

The full 1 mark is for the correct option.

Answer:

(B) x³ = (x + 1)³

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/5/2: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing A because x³ appears on both sides; after expanding and cancelling, the remaining expression 3x − 1 is linear, not quadratic.
Choosing C because x² is visible, but √x is not a polynomial term, so the equation is not a quadratic equation.
Choosing D because x² is present, but 1/x is not a polynomial term, and after clearing the denominator the degree becomes 3.

Why this works

To decide, simplify each option and see whether the final expression is a polynomial of degree 2. In option A, (x − 1)³ expands to x³ − 3x² + 3x − 1; adding 3x² gives x³ + 3x − 1, so the x³ terms cancel and only the linear expression 3x − 1 remains. Option C has √x, which is not a polynomial term. Option D has 1/x, which is not a polynomial term; multiplying by x gives a cubic expression. In option B, (x + 1)³ expands to x³ + 3x² + 3x + 1, so the x³ terms cancel and the remaining polynomial 3x² + 3x + 1 has degree 2. Hence option B is the quadratic equation.

Key idea to remember

Standard form of a quadratic equation: see Key ideas used on this page.

Explanation

A quadratic equation in the variable x is an equation whose standard form is a polynomial of degree 2, namely ax² + bx + c with a, b, c real numbers and a not zero, set equal to zero. This solution simplifies each option and checks whether the remaining expression is a polynomial of degree 2.

Important formulas

(x + a)³ = x³ + 3x²a + 3xa² + a³ · Used to expand option B.
(x − a)³ = x³ − 3x²a + 3xa² − a³ · Used to expand option A.

Q27. Which of the following quadratic equations has real and equal roots?

Mark practised

(A) \((x+1)^2 = 2x + 1\)
(B) \(x^2 + x = 0\)
(C) \(x^2 - 4 = 0\)
(D) \(x^2 + x + 1 = 0\)

Marks: 1
Level: Standard
Years: 2025
Topic: 4.4 Nature of Roots
Source: (CBSE 2025 Mathematics Standard, Set 30/6/1, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) \((x+1)^2 = 2x+1\), because it simplifies to \(x^2 = 0\), whose discriminant is zero.
Why? By Section 4.4 of the NCERT text, a quadratic equation has two equal real roots exactly when its discriminant is zero.

The full 1 mark is for the correct option.

Answer:

(A)

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/6/1: option (A).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option C because it has two real roots, without noticing that equal roots must be the same number.
For option D, forgetting the minus sign in the term \(4ac\) of the discriminant, and wrongly concluding that the roots are real.
Testing option A without first simplifying the left side as a square, so the coefficients are read from the wrong standard form.

Why this works

Write each option in the standard form \(ax^2 + bx + c = 0\) and find its discriminant \(b^2 - 4ac\). For option A, \((x+1)^2 = 2x+1\) simplifies to \(x^2 = 0\), and the discriminant is \(0^2 - 4(1)(0) = 0\), so the roots are real and equal. For option B, \(x^2 + x = 0\) gives \(1^2 - 4(1)(0) = 1\), which is positive, so the roots are distinct. For option C, \(x^2 - 4 = 0\) gives \(0^2 - 4(1)(-4) = 16\), which is positive, so the roots are distinct. For option D, \(x^2 + x + 1 = 0\) gives \(1^2 - 4(1)(1) = -3\), which is negative, so there are no real roots. Therefore option A is correct.

Key idea to remember

Discriminant: see Key ideas used on this page.

Explanation

For the quadratic equation \(ax^2 + bx + c = 0\), the expression \(b^2 - 4ac\) is called the discriminant. When the discriminant is zero, the quadratic formula gives both roots as \(\frac{-b}{2a}\), so the roots are real and equal. This solution puts each option in standard form and compares its discriminant with zero.

Important formulas

\(b^2 - 4ac\) · Discriminant of \(ax^2 + bx + c = 0\); the roots are real and equal when it is zero.
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) · Quadratic formula; when the discriminant is zero, both roots equal \(\frac{-b}{2a}\).

Q28. Which of the following quadratic equations has real and distinct roots?

Mark practised

(A) \(x^2 + 2x = 0\)
(B) \(x^2 + x + 1 = 0\)
(C) \((x - 1)^2 = 1 - 2x\)
(D) \(2x^2 + x + 1 = 0\)

Marks: 1
Level: Standard
Years: 2025
Topic: 4.4 Nature of Roots
Source: (CBSE 2025 Mathematics Standard, Set 30/6/2, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) x² + 2x = 0, because its discriminant is 2² − 4(1)(0) = 4 > 0, so it has two distinct real roots.
Why? For ax² + bx + c = 0, the discriminant b² − 4ac decides the nature of the roots: a positive discriminant gives two distinct real roots.

The full 1 mark is for the correct option.

Answer:

(A) x² + 2x = 0

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/6/2: option (A).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing (C) because (x − 1)² looks like a perfect square; after simplifying it gives x² = 0, whose discriminant is zero, so the roots are equal, not distinct.
For (B) or (D), thinking that a negative discriminant still gives real roots; the discriminants are −3 and −7, both less than zero.
For option C, reading off a, b, c before simplifying; the equation must first be simplified to standard form before using the discriminant.

Why this works

The roots of ax² + bx + c = 0 are given by the quadratic formula x = (−b ± √(b² − 4ac))/(2a). The expression b² − 4ac is the discriminant. If it is positive, the ± gives two different real numbers, so the roots are real and distinct. If it is zero, both roots are −b/(2a), so equal. If it is negative, no real number has that square, so there are no real roots.

Check each option in the form ax² + bx + c = 0: A: for x² + 2x, b² − 4ac = 2² − 4 × 1 × 0 = 4 > 0. B: for x² + x + 1, b² − 4ac = 1² − 4 × 1 × 1 = −3 < 0. C: (x − 1)² = 1 − 2x simplifies to x² − 2x + 1 = 1 − 2x, so x² = 0; its discriminant is 0² − 4 × 1 × 0 = 0, so the roots are equal. D: for 2x² + x + 1, b² − 4ac = 1² − 4 × 2 × 1 = −7 < 0. Only A gives a positive discriminant. Quick check: x² + 2x = x(x + 2), so the roots are 0 and −2, two distinct real numbers.

Key idea to remember

Nature of Roots: see Key ideas used on this page.

Explanation

The discriminant b² − 4ac of ax² + bx + c = 0 tells the nature of the roots: if b² − 4ac > 0, there are two distinct real roots; if b² − 4ac = 0, there are two equal real roots; if b² − 4ac < 0, there are no real roots. This solution writes each option in standard form, computes its discriminant, and selects the one whose discriminant is positive.

Important formulas

b² − 4ac · Discriminant of ax² + bx + c = 0; positive means two distinct real roots.
x = (−b ± √(b² − 4ac))/(2a) · Quadratic formula for the roots when b² − 4ac ≥ 0.

Q29. Which of the following equations does not have a real root ?

Mark practised

(A) \(x^2 = 0\)
(B) \(2x - 1 = 3\)
(C) \(x^2 + 1 = 0\)
(D) \(x^3 + x^2 = 0\)

Marks: 1
Level: Standard
Years: 2025
Topic: 4.4 Nature of Roots
Source: (CBSE 2025 Mathematics Standard, Set 30/6/3, Q12, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) x² + 1 = 0, because its discriminant b² − 4ac = 0² − 4 × 1 × 1 = −4 < 0.
Why? For a quadratic equation ax² + bx + c = 0, a negative discriminant b² − 4ac means the equation has no real roots.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/6/3: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Picking x² = 0 because zero looks like no root, but x = 0 is a real root.
Treating 2x − 1 = 3 as a quadratic and applying the discriminant; it is linear and its real root is x = 2.
Choosing x³ + x² = 0 just because it is cubic, without trying x = 0.

Why this works

Check the options one by one. Option (A), x² = 0, has the real solution x = 0 because 0² = 0. Option (B), 2x − 1 = 3, gives 2x = 4, so x = 2; putting x = 2 gives 2 × 2 − 1 = 3. Option (D), x³ + x² = 0, has the real solution x = 0 because 0³ + 0² = 0. Option (C), x² + 1 = 0, has coefficients 1, 0 and 1, so its discriminant is 0² − 4 × 1 × 1 = −4, which is negative. A negative discriminant means no real roots, so the required option is (C). Quick check: if x² + 1 = 0, then x² would have to be negative, but the square of a real number is never negative.

Key idea to remember

Discriminant: see Key ideas used on this page.

Explanation

For a quadratic equation ax² + bx + c = 0, the value b² − 4ac is called the discriminant. According to NCERT Section 4.4, if b² − 4ac < 0, the equation has no real roots. For x² + 1 = 0, the discriminant is 0² − 4 × 1 × 1 = −4, so it has no real root.

Important formulas

b² − 4ac · Discriminant of the quadratic equation ax² + bx + c = 0.
b² − 4ac < 0 ⇒ no real roots · Nature-of-roots rule from NCERT Section 4.4.

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Key ideas used on this page

Discriminant
Used in questions 2 6 9 11 12 17 21 22 25 27 29
Standard form of a quadratic equation
Used in questions 1 4 5 8 26
Root of a quadratic equation
Used in questions 14 15 18 20 24
Discriminant of a quadratic equation
Used in questions 3 10 16 23
Roots of a Quadratic Equation
Used in questions 13 19
Solution of a Quadratic Equation by Factorisation
Used in question 7
Nature of Roots
Used in question 28

How to use these questions

Try each question on paper first. Then read the steps: the mark beside each step shows what the examiner looks for. If a step is new to you, revise Section 4.2, Section 4.3 and Section 4.4 of the NCERT chapter.

To revise the lessons behind these questions, read the NCERT chapter, exactly as NCERT publishes it on its own website.

Official papers behind this page

Every question on this page comes from an official CBSE Class 10 Mathematics board paper. The board publishes the question papers on cbse.gov.in and the marking schemes on cbse.gov.in, where you can see how marks are given.

Year	Exam	Course	Question paper	Marking scheme
2025	Main	Standard (041)	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 54.5 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Main	Basic (241)	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 49.9 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Compartment	Basic (241)	Open CBSE's official question paper for the 2025 compartment exam (one ZIP file of about 2.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 compartment exam (one ZIP file of about 18.7 MB, Class 10 Maths Basic (241)) to see how marks are given.
2024	Main	Standard (041)	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 35.3 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2024	Main	Basic (241)	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 43.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.8 MB, Class 10 Maths Basic (241)) to see how marks are given.
2024	Compartment	Basic (241)	Open CBSE's official question paper for the 2024 compartment exam (one ZIP file of about 2.2 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2024 compartment exam (one ZIP file of about 5.6 MB, Class 10 Maths Basic (241)) to see how marks are given.
2023	Main	Standard (041)	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 44 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 17.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2023	Main	Basic (241)	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 49.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 26 MB, Class 10 Maths Basic (241)) to see how marks are given.

Sources and session

Each question is reproduced as printed in its CBSE paper, with its year, paper code and question number; only the layout is tidied. The same question printed in several paper sets counts once, and the card says how often it was printed in 2023–2026. These counts describe past papers; they are not predictions.

Every solution is original and follows NCERT Mathematics Class X (reprint 2026-27). The mark beside each step follows the official CBSE marking scheme where the scheme shows it; a mark labelled “(suggested)” follows the board’s usual pattern for that type of question.

Reference: CBSE Class 10 Mathematics board question papers via cbse.gov.in; all solutions are original. learncbse.net is an independent website and is not affiliated with CBSE or NCERT.

1-mark questions on Quadratic Equations

All 29 Quadratic Equations 1-mark questions

Q1. Which of the following equations is a quadratic equation ?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q2. The discriminant of the quadratic equation \(x^2 - 3x - 2 = 0\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q3. The discriminant of the quadratic equation \(2x^2 - 3x - 5 = 0\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q4. If \(x = \sqrt{x}\) , ( \(x \neq 0\) ) is expressed as a quadratic equation in the form \(ax^2 + bx + c = 0\) , then the value of \(a + b + c\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q5. If \((\sqrt{x} + 1)^2 = x^2 + 2\sqrt{x}\) is expressed as a quadratic equation in the form of \(ax^2 + bx + c = 0\), then the value of \(a - b + c\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q6. The value(s) of k for which the roots of quadratic equation \(x^2 + 4x + k = 0\) are real, is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q7. If one root of the quadratic equation \(x^2 - 4x + 3 = 0\) is 1, then the other root is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q8. Which of the following is not a quadratic equation ?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q9. If the quadratic equation \(ax^2 + ax + c = 0\) ( \(a \neq 0\) ) has real and equal roots, then :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q10. The value of 'a' for which \(ax^2 + x + a = 0\) has equal and positive roots is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q11. If \(x^2 + bx + b = 0\) has two real and distinct roots, then the value of \(b\) can be

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q12. The discriminant of the quadratic equation \(2x^2 - 5x - 3 = 0\) is

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q13. A quadratic equation whose roots are \((2 + \sqrt{3})\) and \((2 - \sqrt{3})\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q14. Which of the following equations has 2 as a root ?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q15. The roots of the quadratic equation \(x^2 - 4 = 0\) is/are :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q16. If the discriminant of the quadratic equation \(3x^2 - 2x + c = 0\) is 16, then the value of c is :

Step-by-step solution

Step 1: Choose the option