Quadratic Equations — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Quadratic Equations" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Quadratic Equations" — 9 important questions with detailed answers for CBSE board…
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NCERT Solutions for Class 10 Mathematics — Quadratic Equations. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Quadratic Equations — All 9 Questions Solved
Q1. Define a quadratic equation and state the general form.
Quadratic Equation:
A quadratic equation in one variable is a polynomial equation of degree 2.
General form: ax² + bx + c = 0
where a, b, and c are real numbers and a ≠ 0.
The condition a ≠ 0 is essential because if a = 0, it becomes a linear equation, not quadratic.
Examples of quadratic equations:
- x² + 5x + 6 = 0
- 2x² - 7x + 3 = 0
- x² - 9 = 0
- 3x² + 2x = 0
Characteristics:
- The highest power of the variable is 2
- A quadratic equation has at most 2 real roots
- The graph of y = ax² + bx + c is a parabola
Q2. Prove that the discriminant determines the nature of roots of a quadratic equation.
For the quadratic equation ax² + bx + c = 0, the discriminant is Δ = b² - 4ac.
The nature of roots is determined as follows:
1) If Δ > 0 (discriminant is positive):
- The equation has two distinct real roots
- Roots are given by x = (-b ± √Δ)/(2a)
- Example: x² + 5x + 6 = 0 has Δ = 25 - 24 = 1 > 0, so two distinct real roots
2) If Δ = 0 (discriminant is zero):
- The equation has two equal (repeated) real roots
- Both roots are x = -b/(2a)
- Example: x² - 4x + 4 = 0 has Δ = 16 - 16 = 0, so equal roots
3) If Δ < 0 (discriminant is negative):
- The equation has no real roots
- Roots are complex conjugates
- Example: x² + x + 1 = 0 has Δ = 1 - 4 = -3 < 0, so no real roots
This proves that the discriminant completely determines the nature of roots.
Q3. Solve x² + 5x + 6 = 0 by factorisation.
Given equation: x² + 5x + 6 = 0
Step 1: Identify factors
We need two numbers that multiply to 6 and add to 5.
Factors of 6 are: 1×6, 2×3
Checking: 2 + 3 = 5 ✓ and 2 × 3 = 6 ✓
Step 2: Factorise
x² + 5x + 6 = (x + 2)(x + 3) = 0
Step 3: Apply zero product property
Either (x + 2) = 0 or (x + 3) = 0
Therefore: x = -2 or x = -3
Step 4: Verification
For x = -2: (-2)² + 5(-2) + 6 = 4 - 10 + 6 = 0 ✓
For x = -3: (-3)² + 5(-3) + 6 = 9 - 15 + 6 = 0 ✓
Therefore, the roots are x = -2 and x = -3
Q4. Solve 2x² - 7x + 3 = 0 using the quadratic formula.
Given equation: 2x² - 7x + 3 = 0
Identify coefficients:
a = 2, b = -7, c = 3
Step 1: Calculate the discriminant
Δ = b² - 4ac
= (-7)² - 4(2)(3)
= 49 - 24
= 25
Since Δ = 25 > 0, the equation has two distinct real roots.
Step 2: Apply the quadratic formula
x = (-b ± √Δ)/(2a)
x = (-(-7) ± √25)/(2 × 2)
x = (7 ± 5)/4
Step 3: Calculate the roots
Root 1: x = (7 + 5)/4 = 12/4 = 3
Root 2: x = (7 - 5)/4 = 2/4 = 1/2
Step 4: Verification
For x = 3: 2(3)² - 7(3) + 3 = 18 - 21 + 3 = 0 ✓
For x = 1/2: 2(1/2)² - 7(1/2) + 3 = 2(1/4) - 7/2 + 3 = 1/2 - 7/2 + 3 = 0 ✓
Therefore, the roots are x = 3 and x = 1/2
Q5. For what values of k does the equation 2x² - 3x + k = 0 have equal roots?
For a quadratic equation to have equal roots, the discriminant must be zero: Δ = 0
Given equation: 2x² - 3x + k = 0
Coefficients: a = 2, b = -3, c = k
Step 1: Write the discriminant condition
Δ = b² - 4ac = 0
(-3)² - 4(2)(k) = 0
9 - 8k = 0
Step 2: Solve for k
8k = 9
k = 9/8
Step 3: Verification
With k = 9/8, the equation becomes:
2x² - 3x + 9/8 = 0
Multiplying by 8: 16x² - 24x + 9 = 0
Discriminant: (-24)² - 4(16)(9) = 576 - 576 = 0 ✓
The equal root is: x = -b/(2a) = -(-3)/(2 × 2) = 3/4
Verification: 2(3/4)² - 3(3/4) + 9/8 = 2(9/16) - 9/4 + 9/8 = 9/8 - 18/8 + 9/8 = 0 ✓
Therefore, k = 9/8
Q6. A rectangle has length 2 m more than its width. Its area is 48 m². Find the dimensions.
Let the width of the rectangle be x m.
Then the length is (x + 2) m.
Step 1: Form the equation using area formula
Area = length × width
48 = (x + 2) × x
48 = x² + 2x
x² + 2x - 48 = 0
Step 2: Solve using factorisation
Find two numbers that multiply to -48 and add to 2.
Factors: 8 and -6 (since 8 × (-6) = -48 and 8 + (-6) = 2)
x² + 2x - 48 = (x + 8)(x - 6) = 0
Step 3: Find the roots
Either x + 8 = 0 or x - 6 = 0
Therefore: x = -8 or x = 6
Since width cannot be negative, x = 6 m.
Step 4: Find dimensions
Width = 6 m
Length = x + 2 = 6 + 2 = 8 m
Step 5: Verification
Area = 8 × 6 = 48 m² ✓
Length is 2 m more than width: 8 - 6 = 2 m ✓
Therefore, the dimensions are length = 8 m and width = 6 m
Q7. Show that x² + 2x + 5 = 0 has no real roots.
Given equation: x² + 2x + 5 = 0
Coefficients: a = 1, b = 2, c = 5
Step 1: Calculate the discriminant
Δ = b² - 4ac
= (2)² - 4(1)(5)
= 4 - 20
= -16
Step 2: Interpret the result
Since Δ = -16 < 0 (discriminant is negative), the equation has no real roots.
This means:
- The parabola y = x² + 2x + 5 does not intersect the x-axis
- The equation has two complex conjugate roots (not real)
Alternative verification using completing the square:
x² + 2x + 5 = 0
(x² + 2x + 1) + 5 - 1 = 0
(x + 1)² + 4 = 0
(x + 1)² = -4
Since (x + 1)² is always ≥ 0 for real x, it cannot equal -4.
Therefore, there are no real solutions.
Conclusion: The equation x² + 2x + 5 = 0 has no real roots
Q8. State Vieta's formulas for a quadratic equation ax² + bx + c = 0 with roots p and q.
Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots.
For a quadratic equation ax² + bx + c = 0 with roots p and q:
1) Sum of roots:
p + q = -b/a
2) Product of roots:
p × q = c/a
Proof:
The quadratic equation with roots p and q can be written as:
a(x - p)(x - q) = 0
Expanding:
a(x² - qx - px + pq) = 0
a(x² - (p + q)x + pq) = 0
ax² - a(p + q)x + apq = 0
Comparing with ax² + bx + c = 0:
Coefficient of x: b = -a(p + q) → p + q = -b/a
Constant term: c = apq → pq = c/a
Example: For 2x² - 5x + 3 = 0 with roots p and q:
p + q = -(-5)/2 = 5/2
pq = 3/2
We can verify: the roots are 1 and 3/2
1 + 3/2 = 5/2 ✓
1 × 3/2 = 3/2 ✓
Applications:
- Constructing polynomials when roots are known
- Finding relationships between roots without solving explicitly
- Checking if roots satisfy required conditions
Q9. If one root of the equation 3x² + kx + 15 = 0 is 5/3, find the other root and the value of k.
Given equation: 3x² + kx + 15 = 0
One root: p = 5/3
Find: other root q and the value of k
Step 1: Use Vieta's formulas
Coefficients: a = 3, b = k, c = 15
Product of roots: pq = c/a
(5/3) × q = 15/3
(5/3) × q = 5
q = 5 ÷ (5/3)
q = 5 × (3/5)
q = 3
Step 2: Find k using sum of roots
Sum of roots: p + q = -b/a
(5/3) + 3 = -k/3
5/3 + 9/3 = -k/3
14/3 = -k/3
k = -14
Step 3: Verification
With k = -14, the equation is: 3x² - 14x + 15 = 0
Check p = 5/3:
3(5/3)² - 14(5/3) + 15 = 3(25/9) - 70/3 + 15
= 25/3 - 70/3 + 45/3
= (25 - 70 + 45)/3
= 0/3 = 0 ✓
Check q = 3:
3(3)² - 14(3) + 15 = 27 - 42 + 15 = 0 ✓
Therefore, the other root is 3 and k = -14
Quadratic Equations — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following is a quadratic equation?
- A. A) x³ + 2x + 1 = 0
- B. B) x(x + 2) + 3 = 0
- C. C) (x + 1)³ = x³ + x
- D. D) x² + 1/x = 0
Answer: B. B) x(x + 2) + 3 = 0 — Option B simplifies to x² + 2x + 3 = 0, which is in the form ax² + bx + c = 0 where a ≠ 0. Options A and C are cubic after simplification, and option D is not a polynomial.
MCQ 2. If 2 is a root of the equation x² - kx + 6 = 0, then the value of k is:
- A. A) 5
- B. B) -5
- C. C) 2
- D. D) -2
Answer: A. A) 5 — Substitute x = 2 into the equation: (2)² - k(2) + 6 = 0. This simplifies to 4 - 2k + 6 = 0, so 10 - 2k = 0, which gives 2k = 10, and thus k = 5.
MCQ 3. The nature of the roots of the quadratic equation 2x² - 4x + 3 = 0 is:
- A. A) Real and distinct
- B. B) Real and equal
- C. C) No real roots
- D. D) Cannot be determined
Answer: C. C) No real roots — The discriminant D = b² - 4ac. For the given equation, a=2, b=-4, c=3. So, D = (-4)² - 4(2)(3) = 16 - 24 = -8. Since D < 0, the equation has no real roots.
MCQ 4. The roots of the quadratic equation x² - 3x - 10 = 0 are:
- A. A) 5, -2
- B. B) -5, 2
- C. C) 5, 2
- D. D) -5, -2
Answer: A. A) 5, -2 — By factorization, x² - 5x + 2x - 10 = 0, which means x(x-5) + 2(x-5) = 0. So, (x-5)(x+2) = 0. The roots are x = 5 and x = -2.
MCQ 5. For the quadratic equation 3x² - 7x + 4 = 0, the sum of its roots is:
- A. A) 7/3
- B. B) -7/3
- C. C) 4/3
- D. D) -4/3
Answer: A. A) 7/3 — For a quadratic equation ax² + bx + c = 0, the sum of roots is -b/a. Here, a=3, b=-7, so the sum of roots is -(-7)/3 = 7/3.
MCQ 6. A quadratic equation whose roots are 2 and -3 is:
- A. A) x² - x - 6 = 0
- B. B) x² + x - 6 = 0
- C. C) x² - x + 6 = 0
- D. D) x² + x + 6 = 0
Answer: B. B) x² + x - 6 = 0 — If α and β are roots, the equation is x² - (α+β)x + αβ = 0. Here, α=2, β=-3. So, sum (α+β) = 2 + (-3) = -1. Product (αβ) = 2 * (-3) = -6. The equation is x² - (-1)x + (-6) = 0, which simplifies to x² + x - 6 = 0.
MCQ 7. If the quadratic equation kx² + 4x + 1 = 0 has real and equal roots, then the value of k is:
- A. A) 4
- B. B) -4
- C. C) 0
- D. D) 2
Answer: A. A) 4 — For real and equal roots, the discriminant D = b² - 4ac must be 0. Here, a=k, b=4, c=1. So, (4)² - 4(k)(1) = 0, which means 16 - 4k = 0. Therefore, 4k = 16, and k = 4.
MCQ 8. The product of two consecutive positive integers is 306. The quadratic equation representing this situation is:
- A. A) x² + x + 306 = 0
- B. B) x² - x + 306 = 0
- C. C) x² + x - 306 = 0
- D. D) x² - x - 306 = 0
Answer: C. C) x² + x - 306 = 0 — Let the two consecutive positive integers be x and x+1. Their product is x(x+1) = 306. Expanding this gives x² + x = 306. Rearranging to standard form gives x² + x - 306 = 0.
MCQ 9. If α and β are the roots of the equation 2x² - 3x + 5 = 0, then the value of α² + β² is:
- A. A) -11/4
- B. B) 1/4
- C. C) 19/4
- D. D) 29/4
Answer: A. A) -11/4 — For 2x² - 3x + 5 = 0, α+β = -(-3)/2 = 3/2 and αβ = 5/2. We know α² + β² = (α+β)² - 2αβ. So, (3/2)² - 2(5/2) = 9/4 - 5 = 9/4 - 20/4 = -11/4.
MCQ 10. For what value of k does the quadratic equation x² - 2x + k = 0 have no real roots?
- A. A) k < 1
- B. B) k > 1
- C. C) k = 1
- D. D) k = 0
Answer: B. B) k > 1 — For no real roots, the discriminant D = b² - 4ac must be less than 0. Here, a=1, b=-2, c=k. So, (-2)² - 4(1)(k) < 0, which means 4 - 4k < 0. This implies 4 < 4k, or k > 1.
MCQ 11. The roots of the equation x² - 4x - 1 = 0 are:
- A. A) 2 ± √5
- B. B) -2 ± √5
- C. C) 2 ± √3
- D. D) -2 ± √3
Answer: A. A) 2 ± √5 — Using the quadratic formula x = [-b ± √(b² - 4ac)] / 2a: x = [4 ± √((-4)² - 4(1)(-1))] / 2(1) = [4 ± √(16 + 4)] / 2 = [4 ± √20] / 2 = [4 ± 2√5] / 2 = 2 ± √5.
MCQ 12. The degree of a quadratic equation is always:
- A. A) 0
- B. B) 1
- C. C) 2
- D. D) 3
Answer: C. C) 2 — A quadratic equation is defined as a polynomial equation of degree 2, meaning the highest power of the variable is 2.
MCQ 13. The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. If the breadth is 'x' metres, the quadratic equation representing this situation is:
- A. A) 2x² + x - 528 = 0
- B. B) 2x² - x - 528 = 0
- C. C) x² + 2x - 528 = 0
- D. D) x² - 2x - 528 = 0
Answer: A. A) 2x² + x - 528 = 0 — Let the breadth be x metres. The length is 2x + 1 metres. Area = length × breadth, so x(2x + 1) = 528. Expanding this gives 2x² + x = 528. Rearranging to standard form results in 2x² + x - 528 = 0.
MCQ 14. If one root of the quadratic equation ax² + bx + c = 0 is the reciprocal of the other, then:
- A. A) a = b
- B. B) b = c
- C. C) a = c
- D. D) a + c = 0
Answer: C. C) a = c — Let the roots be α and 1/α. The product of roots is c/a. So, α * (1/α) = c/a. This simplifies to 1 = c/a, which implies a = c.
MCQ 15. If one root of the equation x² + px - 4 = 0 is 4, then the value of p is:
- A. A) 3
- B. B) -3
- C. C) 0
- D. D) 1
Answer: B. B) -3 — If 4 is a root, substituting x = 4 into the equation should satisfy it: (4)² + p(4) - 4 = 0. This gives 16 + 4p - 4 = 0, so 12 + 4p = 0. Thus, 4p = -12, and p = -3.
MCQ 16. The solutions of x - 1/x = 3, (x ≠ 0) are:
- A. A) (3 ± √13)/2
- B. B) (-3 ± √13)/2
- C. C) (3 ± √5)/2
- D. D) (-3 ± √5)/2
Answer: A. A) (3 ± √13)/2 — Multiply the equation by x to remove the fraction: x² - 1 = 3x. Rearrange to standard form: x² - 3x - 1 = 0. Using the quadratic formula, x = [3 ± √((-3)² - 4(1)(-1))] / 2(1) = [3 ± √(9 + 4)] / 2 = (3 ± √13)/2.
MCQ 17. Find the value(s) of k for which the quadratic equation (k-12)x² + 2(k-12)x + 2 = 0 has real and equal roots.
- A. A) k = 12
- B. B) k = 14
- C. C) k = 12 or k = 14
- D. D) No such k exists.
Answer: B. B) k = 14 — For real and equal roots, the discriminant D must be 0, and the coefficient of x² must not be zero (k-12 ≠ 0). D = [2(k-12)]² - 4(k-12)(2) = 0. This simplifies to 4(k-12)² - 8(k-12) = 0. Factoring out 4(k-12), we get 4(k-12)[(k-12) - 2] = 0, so 4(k-12)(k-14) = 0. This gives k = 12 or k = 14. However, k cannot be 12 because then the equation would not be quadratic (coefficient of x² would be 0). Therefore, k = 14 is the only valid solution.
MCQ 18. If α and β are the roots of the equation x² - 5x + 6 = 0, then the quadratic equation whose roots are 1/α and 1/β is:
- A. A) 6x² - 5x + 1 = 0
- B. B) 6x² + 5x + 1 = 0
- C. C) x² - 5x + 6 = 0
- D. D) x² + 5x + 6 = 0
Answer: A. A) 6x² - 5x + 1 = 0 — For x² - 5x + 6 = 0, α+β = 5 and αβ = 6. For the new equation, sum of roots (1/α + 1/β) = (α+β)/αβ = 5/6. Product of roots (1/α * 1/β) = 1/αβ = 1/6. The new equation is x² - (5/6)x + (1/6) = 0. Multiplying by 6 gives 6x² - 5x + 1 = 0.
MCQ 19. A real number 'α' is a root of the quadratic equation ax² + bx + c = 0, a ≠ 0 if:
- A. A) aα² - bα + c = 0
- B. B) aα² + bα - c = 0
- C. C) aα² + bα + c = 0
- D. D) aα² + bα + c ≠ 0
Answer: C. C) aα² + bα + c = 0 — By definition, a number 'α' is a root of a polynomial equation if substituting 'α' for the variable makes the equation true. So, aα² + bα + c = 0.
MCQ 20. A quadratic equation can have at most:
- A. A) One root
- B. B) Two roots
- C. C) Three roots
- D. D) Infinitely many roots
Answer: B. B) Two roots — According to the Fundamental Theorem of Algebra, a polynomial equation of degree 'n' has exactly 'n' roots (counting multiplicity and complex roots). A quadratic equation has degree 2, so it has at most two distinct real roots.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.