## Class 10 Maths MCQs Chapter 4 Quadratic Equations

1. Which of the following is not a quadratic equation
(a) x² + 3x – 5 = 0
(b) x² + x3 + 2 = 0
(c) 3 + x + x² = 0
(d) x² – 9 = 0

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Explaination:Reason: Since it has degree 3.

2. The quadratic equation has degree
(a) 0
(b) 1
(c) 2
(d) 3

Explaination:Reason: A quadratic equation has degree 2.

3. The cubic equation has degree
(a) 1
(b) 2
(c) 3
(d) 4

Explaination:Reason: A cubic equation has degree 3.

4. A bi-quadratic equation has degree
(a) 1
(b) 2
(c) 3
(d) 4

Explaination:Reason: A bi-quadratic equation has degree 4.

5. The polynomial equation x (x + 1) + 8 = (x + 2) {x – 2) is
(a) linear equation
(c) cubic equation

Explaination:Reason: We have x(x + 1) + 8 = (x + 2) (x – 2)
⇒ x² + x + 8 = x² – 4
⇒ x² + x + 8- x² + 4 = 0
⇒ x + 12 = 0, which is a linear equation.

6. The equation (x – 2)² + 1 = 2x – 3 is a
(a) linear equation
(c) cubic equation

Explaination:Reason: We have (x – 2)² + 1 = 2x – 3
⇒ x² + 4 – 2 × x × 2 + 1 = 2x – 3
⇒ x² – 4x + 5 – 2x + 3 = 0
∴ x² – 6x + 8 = 0, which is a quadratic equation.

7. The roots of the quadratic equation 6x² – x – 2 = 0 are

Explaination:Reason: We have 6×2 – x – 2 = 0
⇒ 6x² + 3x-4x-2 = 0
⇒ 3x(2x + 1) -2(2x + 1) = 0
⇒ (2x + 1) (3x – 2) = 0
⇒ 2x + 1 = 0 or 3x – 2 = 0
∴ x =$$-\frac{1}{2}$$, x = $$\frac{2}{3}$$

8. The quadratic equation whose roots are 1 and
(a) 2x² + x – 1 = 0
(b) 2x² – x – 1 = 0
(c) 2x² + x + 1 = 0
(d) 2x² – x + 1 = 0

9. The quadratic equation whose one rational root is 3 + √2 is
(a) x² – 7x + 5 = 0
(b) x² + 7x + 6 = 0
(c) x² – 7x + 6 = 0
(d) x² – 6x + 7 = 0

Explaination:Reason: ∵ one root is 3 + √2
∴ other root is 3 – √2
∴ Sum of roots = 3 + √2 + 3 – √2 = 6
Product of roots = (3 + √2)(3 – √2) = (3)² – (√2)² = 9 – 2 = 7
∴ Required quadratic equation is x² – 6x + 7 = 0

10. The equation 2x² + kx + 3 = 0 has two equal roots, then the value of k is
(a) ±√6
(b) ± 4
(c) ±3√2
(d) ±2√6

Explaination:Reason: Here a = 2, b = k, c = 3
Since the equation has two equal roots
∴ b² – 4AC = 0
⇒ (k)² – 4 × 2 × 3 = 0
⇒ k² = 24
⇒ k = ± √24
∴ k= ± $$\pm \sqrt{4 \times 6}$$ = ± 2√6

11. The roots of the quadratic equation $$x+\frac{1}{x}=3$$, x ≠ 0 are.

Explaination:Reason: We have $$x+\frac{1}{x}=3$$
⇒ $$\frac{x^{2}+1}{x}=3$$
⇒ x² + 1 = 3x
On comparing with ax² + bx + c = 0
∴ a = 1, b = – 3, c = 1
⇒ D = b² – 4ac = (-3)² – 4 × (1) × (1) = 9 – 4 = 5

12. The roots of the quadratic equation 2x² – 2√2x + 1 = 0 are

Explaination:Reason: Here a = 2, b = -2√2 , c = 1
∴ D = b² – 4ac = (-2√2 )² – 4 × 2 × 1 = 8 – 8 = 0

13. The sum of the roots of the quadratic equation 3×2 – 9x + 5 = 0 is
(a) 3
(b) 6
(c) -3
(d) 2

Explaination:Reason: Here a = 3, b = -9, c = 5
∴ Sum of the roots $$=\frac{-b}{a}=-\frac{(-9)}{3}=3$$

14. If the roots of ax2 + bx + c = 0 are in the ratio m : n, then
(a) mna² = (m + n) c²
(b) mnb² = (m + n) ac
(c) mn b² = (m + n)² ac
(d) mnb² = (m – n)² ac

Explaination:

15. If one root of the equation x² + px + 12 = 0 is 4, while the equation x² + px + q = 0 has equal roots, the value of q is

Explaination:Reason: Since 4 is a root of x² + px + 12 = 0
∴ (4)² + p(4) + 12 = 0
⇒ p = -7
Also the roots of x² + px + q = 0 are equal, we have p² – 4 x 1 x q = 0
⇒ (-7)² -4q = 0
$$\therefore q=\frac{49}{4}$$

16. a and p are the roots of 4x² + 3x + 7 = 0, then the value of $$\frac{1}{\alpha}+\frac{1}{\beta}$$ is

Explaination:

17. If a, p are the roots of the equation (x – a) (x – b) + c = 0, then the roots of the equation (x – a) (x – P) = c are
(a) a, b
(b) a, c
(c) b, c
(d) none of these

Explaination:Reason: By given condition, (x – a) (x – b) + c = (x – α) (x – β)
⇒ (x – α) (x – β) – c = (x – a) (x – b)
This shows that roots of (x – α) (x – β) – c are a and b

18. Mohan and Sohan solve an equation. In solving Mohan commits a mistake in constant term and finds the roots 8 and 2. Sohan commits a mistake in the coefficient of x. The correct roots are
(a) 9,1
(b) -9,1
(c) 9, -1
(d) -9, -1

Explaination:Reason: Correct sum = 8 + 2 = 10 from Mohan
Correct product = -9 x -1 = 9 from Sohan
∴ x² – (10)x + 9 = 0
⇒ x² – 10x + 9 = 0
⇒ x² – 9x – x + 9
⇒ x(x – 9) – 1(x – 9) = 0
⇒ (x-9) (x-l) = 0 .
⇒ Correct roots are 9 and 1.

19. If a and p are the roots of the equation 2x² – 3x – 6 = 0. The equation whose roots are $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$ is
(a) 6x² – 3x + 2 = 0
(b) 6x² + 3x – 2 = 0
(c) 6x² – 3x – 2 = 0
(d) x² + 3x-2 = 0

Explaination:

20. If the roots of px2 + qx + 2 = 0 are reciprocal of each other, then
(a) P = 0
(b) p = -2
(c) p = ±2
(d) p = 2

Explaination:Reason: here α = $$\frac{1}{β}$$
∴ αβ = 1
⇒ $$\frac{2}{p}$$ = 1
∴ p = 2

21. If one root of the quadratic equation 2x² + kx – 6 = 0 is 2, the value of k is
(a) 1
(b) -1
(c) 2
(d) -2

Explaination:Reason: Scice x = 2 is a root of the equation 2x² + kx -6 = 0
∴ 2(2)² +k(2) – 6 = 0
⇒ 8 + 2k – 6 = 0
⇒ 2k = -2
∴ k = -1

22. The roots of the quadratic equation

(a) a, b
(b) -a, b
(c) a, -b
(d) -a, -b

Explaination:

23. The roots of the equation 7x² + x – 1 = 0 are
(a) real and distinct
(b) real and equal
(c) not real
(d) none of these

Explaination:Reason: Here a = 2, b = 1, c = -1
∴ D = b² – 4ac = (1)² – 4 × 2 × (-1) = 1 + 8 = 9 > 0
∴ Roots of the given equation are real and distinct.

24. The equation 12x² + 4kx + 3 = 0 has real and equal roots, if
(a) k = ±3
(b) k = ±9
(c) k = 4
(d) k = ±2

Explaination:Reason: Here a = 12, b = 4k, c = 3
Since the given equation has real and equal roots
∴ b² – 4ac = 0
⇒ (4k)² – 4 × 12 × 3 = 0
⇒ 16k² – 144 = 0
⇒ k² = 9
⇒ k = ±3

25. If -5 is a root of the quadratic equation 2x² + px – 15 = 0, then
(a) p = 3
(b) p = 5
(c) p = 7
(d) p = 1

Explaination:Reason: Since – 5 is a root of the equation 2x² + px -15 = 0
∴ 2(-5)² + p (-5) – 15 = 0
⇒ 50 – 5p -15 = 0
⇒ 5p = 35
⇒ p = 7

26. If the roots of the equations ax² + 2bx + c = 0 and bx² – 2√ac x + b = 0 are simultaneously real, then
(a) b = ac
(b) b2 = ac
(c) a2 = be
(d) c2 = ab

Explaination:Reason: Given equations have real roots, then
D1 ≥ 0 and D2 ≥ 0
(2b)² – 4ac > 0 and (-2√ac)² – 4b.b ≥ 0
4b² – 4ac ≥ 0 and 4ac – 4b2 > 0
b² ≥ ac and ac ≥ b²
⇒ b² = ac

27. The roots of the equation (b – c) x² + (c – a) x + (a – b) = 0 are equal, then
(a) 2a = b + c
(b) 2c = a + b
(c) b = a + c
(d) 2b = a + c

Explaination:Reason: Since roots are equal
∴ D = 0 => b² – 4ac = 0
⇒ (c – a)² -4(b – c) (a – b) = 0
⇒ c² – b² – 2ac -4(ab -b² + bc) = 0 =>c + a-2b = 0 => c + a = 2b
⇒ c² + a² – 2ca – 4ab + 4b² + 4ac – 4bc = 0
⇒ c² + a² + 4b² + 2ca – 4ab – 4bc = 0
⇒ (c + a – 2b)² = 0
⇒ c + a – 2b = 0
⇒ c + a = 2b

28. A chess board contains 64 equal squares and the area of each square is 6.25 cm². A border round the board is 2 cm wide. The length of the side of the chess board is
(a) 8 cm
(b) 12 cm
(c) 24 cm
(d) 36 cm

29. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Their present ages are
(a) 7 years, 49 years
(b) 5 years, 25 years
(c) 1 years, 50 years
(d) 6 years, 49 years

30. The sum of the squares of two consecutive natural numbers is 313. The numbers are
(a) 12, 13
(b) 13,14
(c) 11,12
(d) 14,15

We hope the given MCQ Questions for Class 10 Maths Quadratic Equations with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 4 Quadratic Equations Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.

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