Free PDF Download of CBSE Class 10 Maths Chapter 4 Quadratic Equations Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Quadratic Equations MCQs with Answers to know their preparation level.

### Class 10 Maths MCQs Chapter 4 Quadratic Equations

1. Which of the following is not a quadratic equation
(a) xÂ² + 3x - 5 = 0
(b) xÂ² + x3 + 2 = 0
(c) 3 + x + xÂ² = 0
(d) xÂ² - 9 = 0

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)
â‡’ xÂ² + x + 8 = xÂ² - 4
â‡’ xÂ² + x + 8- xÂ² + 4 = 0
â‡’ x + 12 = 0, which is a linear equation.

6. The equation (x - 2)Â² + 1 = 2x - 3 is a
(a) linear equation
(c) cubic equation

Explaination:Reason: We have (x - 2)Â² + 1 = 2x - 3
â‡’ xÂ² + 4 - 2 Ã— x Ã— 2 + 1 = 2x - 3
â‡’ xÂ² - 4x + 5 - 2x + 3 = 0
âˆ´ xÂ² - 6x + 8 = 0, which is a quadratic equation.

9. The quadratic equation whose one rational root is 3 + âˆš2 is
(a) xÂ² - 7x + 5 = 0
(b) xÂ² + 7x + 6 = 0
(c) xÂ² - 7x + 6 = 0
(d) xÂ² - 6x + 7 = 0

Explaination:Reason: âˆµ one root is 3 + âˆš2
âˆ´ other root is 3 - âˆš2
âˆ´ Sum of roots = 3 + âˆš2 + 3 - âˆš2 = 6
Product of roots = (3 + âˆš2)(3 - âˆš2) = (3)Â² - (âˆš2)Â² = 9 - 2 = 7
âˆ´ Required quadratic equation is xÂ² - 6x + 7 = 0

10. The equation 2xÂ² + kx + 3 = 0 has two equal roots, then the value of k is
(a) Â±âˆš6
(b) Â± 4
(c) Â±3âˆš2
(d) Â±2âˆš6

Explaination:Reason: Here a = 2, b = k, c = 3
Since the equation has two equal roots
âˆ´ bÂ² - 4AC = 0
â‡’ (k)Â² - 4 Ã— 2 Ã— 3 = 0
â‡’ kÂ² = 24
â‡’ k = Â± âˆš24
âˆ´ k= Â± $\pm \sqrt{4 \times 6}$ = Â± 2âˆš6

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

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

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 - Î±) (x - Î²)
â‡’ (x - Î±) (x - Î²) - c = (x - a) (x - b)
This shows that roots of (x - Î±) (x - Î²) - 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
âˆ´ xÂ² - (10)x + 9 = 0
â‡’ xÂ² - 10x + 9 = 0
â‡’ xÂ² - 9x - x + 9
â‡’ x(x - 9) - 1(x - 9) = 0
â‡’ (x-9) (x-l) = 0 .
â‡’ Correct roots are 9 and 1.

19. If a and p are the roots of the equation 2xÂ² - 3x - 6 = 0. The equation whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is
(a) 6xÂ² - 3x + 2 = 0
(b) 6xÂ² + 3x - 2 = 0
(c) 6xÂ² - 3x - 2 = 0
(d) xÂ² + 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 = Â±2
(d) p = 2

Explaination:Reason: here Î± = $\frac{1}{Î²}$
âˆ´ Î±Î² = 1
â‡’ $\frac{2}{p}$ = 1
âˆ´ p = 2

21. If one root of the quadratic equation 2xÂ² + 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 2xÂ² + kx -6 = 0
âˆ´ 2(2)Â² +k(2) - 6 = 0
â‡’ 8 + 2k - 6 = 0
â‡’ 2k = -2
âˆ´ 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 7xÂ² + 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
âˆ´ D = bÂ² - 4ac = (1)Â² - 4 Ã— 2 Ã— (-1) = 1 + 8 = 9 > 0
âˆ´ Roots of the given equation are real and distinct.

24. The equation 12xÂ² + 4kx + 3 = 0 has real and equal roots, if
(a) k = Â±3
(b) k = Â±9
(c) k = 4
(d) k = Â±2

Explaination:Reason: Here a = 12, b = 4k, c = 3
Since the given equation has real and equal roots
âˆ´ bÂ² - 4ac = 0
â‡’ (4k)Â² - 4 Ã— 12 Ã— 3 = 0
â‡’ 16kÂ² - 144 = 0
â‡’ kÂ² = 9
â‡’ k = Â±3

25. If -5 is a root of the quadratic equation 2xÂ² + 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 2xÂ² + px -15 = 0
âˆ´ 2(-5)Â² + p (-5) - 15 = 0
â‡’ 50 - 5p -15 = 0
â‡’ 5p = 35
â‡’ p = 7

26. If the roots of the equations axÂ² + 2bx + c = 0 and bxÂ² - 2âˆšac 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 â‰¥ 0 and D2 â‰¥ 0
(2b)Â² - 4ac > 0 and (-2âˆšac)Â² - 4b.b â‰¥ 0
4bÂ² - 4ac â‰¥ 0 and 4ac - 4b2 > 0
bÂ² â‰¥ ac and ac â‰¥ bÂ²
â‡’ bÂ² = ac

27. The roots of the equation (b - c) xÂ² + (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
âˆ´ D = 0 => bÂ² - 4ac = 0
â‡’ (c - a)Â² -4(b - c) (a - b) = 0
â‡’ cÂ² - bÂ² - 2ac -4(ab -bÂ² + bc) = 0 =>c + a-2b = 0 => c + a = 2b
â‡’ cÂ² + aÂ² - 2ca - 4ab + 4bÂ² + 4ac - 4bc = 0
â‡’ cÂ² + aÂ² + 4bÂ² + 2ca - 4ab - 4bc = 0
â‡’ (c + a - 2b)Â² = 0
â‡’ c + a - 2b = 0
â‡’ c + a = 2b

28. A chess board contains 64 equal squares and the area of each square is 6.25 cmÂ². 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

31. Which of the following is not a quadratic equation? [NCERT Exemplar Problems]
(a) 2(x - 1)Â² = 4xÂ² - 2x + 1
(b) 2x - xÂ² = xÂ² + 5
(c) (âˆš2x + âˆš3 )Â² + xÂ² = 3xÂ² - 5x
(d) (xÂ² + 2x)Â² = x4 + 3 + 4x3

Explaination:
2xÂ² + 3 + 2âˆš6 x + xÂ² = 3xÂ² - 5x
2âˆš6x + 5x + 3 = 0

32. If (x - a) is one of the factors of the polynomial axÂ² + bx + c, then one of the roots of axÂ² + bx + c = 0 is
(a) 1
(b) c
(c) a
(d) none of these

Explaination:
âˆµ x - a is one of the factors one root = a.

33. Which of the following are the roots of the quadratic equation, xÂ² - 9x + 20 = 0 by factorisation?
(a) 3, 4
(b) 4, 5
(c) 5, 6
(d) 6, 1

Explaination:
Given equation is xÂ² - 9x + 20 = 0
â‡’ xÂ² - 5x - 4x + 20 = 0
â‡’ x(x - 5) - 4(x - 5) = 0
â‡’ (x - 5) (x - 4) = 0
â‡’ either x - 5 = 0 and x - 4 = 0
â‡’ x = 5 and x = 4
âˆ´ x = 4 and 5 are the roots/solution of the given quadratic equation.

34. If (1 - p) is a root of the equation xÂ² + px + 1 -p = 0, then roots are
(a) 0, 1
(b) -1, 1
(c) 0, -1
(d) - 1, 2

Explaination:
(1 -p) is a root
âˆ´ (1 - p)Â² + p(1 - p)+ 1 - p = 0
â‡’ (1 - p)[1 - p + p + 1] = 0
â‡’ (1 - p)(2) = 0
â‡’ p=1
xÂ²+x = 0
One root = 0 and another root = - 1
âˆ´ roots are 0 and - 1.

35. If a, P are roots of the equation xÂ² + 5x + 5 = 0, then equation whose roots are a + 1 and p + 1 is
(a) xÂ² + 5x - 5 = 0
(b) xÂ² + 3x + 5 = 0
(c) xÂ² + 3x + 1 = 0
(d) none of these

Explaination:
Î± + Î² = -5, Î±Î² = 5.
Required equation is xÂ² - (Î± + 1 + Î² + 1)x + (Î± + 1) (Î² + 1) = 0
â‡’ xÂ² - (Î± + Î² + 2)x + (Î±Î² + Î± + Î² +1) = 0
â‡’ xÂ² - (-5 + 2)x + (5 - 5 + 1) = 0
â‡’ xÂ² + 3x + 1 = 0

36. Which of the following equations has two distinct real roots? [NCERT Exemplar Problems]
(a) 2xÂ² - 3âˆš2x + $\frac{9}{4}$ =0
(b) xÂ² + x - 5 = 0
(c) xÂ² + 3x + 2âˆš2 = 0
(d) 5xÂ² - 3x + 1 = 0

Explaination: (b) D > 0

37. Which of the following equations has no real roots ? [NCERT Exemplar Problems]
(a) xÂ² - 4x + 3âˆš2 =0
(b) xÂ² + 4x - 3âˆš2 = 0
(c) xÂ² - 4x - 3âˆš2 = 0
(d) 3xÂ² + 4âˆš3x + 4 = 0

Explaination: (a) D < 0

38. (xÂ² + 1)Â² - xÂ² = 0 has [NCERT Exemplar Problems]
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real root

Explaination: (c) no real roots

39. If the difference of the roots of the equation xÂ² - bx + c = 0 be 1, then
(a) bÂ² - 4c + 1 = 0
(b) bÂ² + 4c = 0
(c) bÂ² - 4c - 1 - 0
(d) bÂ² - 4c = 0

Explaination:
Let roots are Î± and Î²
â‡’ Î± - Î² = 1
âˆµ (Î± - Î²)Â² = (Î± + Î²)Â² - 4Î±Î²
â‡’ 1 = bÂ² - 4c
â‡’ bÂ² - 4c - 1 = 0

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