Important Questions Class 10 Maths Chapter 4 Quadratic Equations

In this article explore, Important Questions Class 10 Maths Chapter 4 Quadratic Equations. Quadratic equations have a very long history. Often ancient Indian mathematicians like Shri Brahmagupta(C.E.598–665) and Shri Sridharacharya (C.E. 1025) are given credit for developing it. It is the method of completing the square for solving a polynomial equation. These equations are useful in solving our day-to-day problems as well as complex activities such as launching a satellite. A basic quadratic equation can be written as Ax2 By C = 0 here ‘A’, ‘B’, and ‘C’ is the constant term, ‘Ax2′ is the quadratic term, ‘By’ is the linear term and ‘A’ should not be zero.

Students preparing for important questions class 10 maths Chapter 4 Quadratic Equations should pay attention to the following topics:

• Finding roots of quadratic equations with the factorisation method
• Finding roots of quadratic equations by completing the square method
• Quadratic formula
• Distinct real roots in quadratic equations

Important Questions Chapter 4 – Quadratic Equations

Q1. Find the value of p so that the quadratic equation px(x – 3) + 9 = 0 has two equal roots. 

Solution:

We have, px (x – 3) + 9 = 0

px2 – 3px + 9 = 0 Here a = p, b = -3p,

D = 0

b2 – 4ac = 0 ⇒ (-3p)2 – 4(p)(9) = 0

⇒ 9p2 – 36p = 0

⇒ 9p (p – 4) = 0

⇒ 9p = 0 or p – 4= 0

p = 0 (rejected) or p = 4

∴ p = 4 ……..(∵ Coeff. of x2 cannot be zero)

Q2. Find the value of m so that the quadratic equation mx (x – 7) + 49 = 0 has two equal roots. 

Solution:

We have, mx (x – 7) + 49 = 0

mx^2 – 7mx + 49 = 0

Here, a = m, b = – 7m, c = 49

D = b^2 – 4ac = 0 …[For equal roots

⇒ (-7m)^2 – 4(m) (49) = 0

⇒ 49m^2 – 4m (49) = 0

⇒ 49m (m – 4) = 0

⇒ 49m = 0 or m – 4 = 0

m = 0 (rejected) or m = 4

∴ m = 4

Q3. Solve the following quadratic equation for x: x^2 – 2ax – (4b^2 – a^2) = 0

Solution:

Given quadratic equation can be written as

x^2 – 2ax – 4b^2 + a^2 = 0.

(x^2 – 2ax + a^2) – 4b^2 = 0 or (x – a)^2 – (2b)^2 = 0

As we know,

[a^2 – b^2 = (a + b)(a – b)]

∴ (x – a + 2b) (x – a – 2b) = 0

⇒ x – a + 2b = 0 or x – a – 2b = 0

⇒ x = a – 2b or x = a + 2b

⇒ x = a – 2b and x = a + 2b

Q4. Find the value(s) of k so that the quadratic equation 3x^2 – 2kx + 12 = 0 has equal roots. 

Solution:

Given: 3x^2 – 2kx + 12 = 0

Here a = 3, b = -2k, c = 12

D = 0 … [Since roots are equal As

b2 – 4ac = 0

∴ (-2k)^2 – 4(3) (12) = 0

⇒ 4k^2 – 144 = 0 ⇒ k^2 = 144/4 = 36

∴ k = ±√36=±6

Q5. For what value of k, are the roots of the quadratic equation y^2 + k^2 = 2 (k + 1)y equal? 

Solution:

y^2 + k^2 = 2(k + 1)y

y^2 – 2(k + 1)y + k^2 = 0

Here a = 1, b = -2(k + 1), c = k^2

D = 0 … [Roots are equal

b^2 – 4ac = 0

∴ [-2(k + 1)]2 – 4 × (1) × (k2) = 0

⇒ 4(k2 + 2k + 1) – 4k2 = 0

⇒ 4k2 + 8k + 4 – 4k2 = 0

⇒ 8k + 4 = 0

⇒ 8k = -4 ∴ k = −48=−12

Q6. Find that value of p for which the quadratic equation (p + 1)x^2 – 6(p + 1)x + 3 (p + 9) = 0, p ≠ -1 had equal roots. 

Solution:

For the given quadratic equation to have equal roots, D = 0

Here a = (p + 1), b = -6(p + 1), c = 3(p + 9)

D = b^2 – 4ac

⇒ [-6(p + 1)]^2 – 4(p + 1).3 (p + 9) = 0

⇒ 36(p + 1)^2 – 12(p + 1) (p + 9) = 0

⇒ 12(p + 1) (3p + 3 – p – 9) = 0

⇒ 12(p + 1)(2p – 6) = 0

⇒ 24(p + 1)(p – 3) = 0

⇒ p + 1 = 0 or p – 3 = 0

⇒ p = -1 (rejected) or p = 3

∴ p = 3

Q7. If the roots of the quadratic equation (a – b)x^2 + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c

Solution:

Here’a’ = a – b, ‘b’ = b – c, ‘c’ = c – a

D = 0 ….[Roots are equal

b^2 – 4ac = 0

⇒ (b – c)^2 – 4(a – b)(c – a) = 0

⇒ b^2 + c^2 – 2bc – 4(ac – a^2 – bc + ab) = 0

⇒ b^2 + c^2 – 2bc – 4ac + 4a^2 + 4bc – 4ab = 0

⇒ 4a^2 + b2^2 + c^2 – 4ab + 2bc – 4ac = 0

⇒ (-2a)^2 + (b)^2 + (c)2^2 + 2(-2a)(b) + 2(b)(c) + 2(c)(-2a) = 0

⇒ [(-2a) + (b) + (c)]^2 = 0

….[∵ x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x + y + z)^2

Taking square-root on both sides

-2a + b + c = 0

⇒ b + c = 2a ∴ 2a = b + c

Q8. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers.

Solution:

Let three consecutive natural numbers are x, x + 1, x + 2.

According to the question,

(x + 1)^2 – [(x + 2)^2 – x^2] = 60

⇒ x^2 + 2x + 1 – (x^2 + 4x + 4 – x^2) = 60

⇒ x^2 + 2x + 1 – 4x – 4 – 60 = 0

⇒ x^2 – 2x – 63 = 0

⇒ x^2 – 9x + 7x – 63 = 0

⇒ x(x – 9) + 7(x – 9) = 0

⇒ (x – 9) (x + 7) = 0

⇒ x – 9 = 0 or x + 7 = 0

⇒ x = 9 or x = -7

Natural nos. can not be -ve, ∴ x = 9

∴ Numbers are 9, 10, 11.

Q9. If the sum of two natural numbers is 8 and their product is 15, find the numbers.

Solution:

Let the numbers be x and (8 – x).

According to the Question,

x(8 – x) = 15

⇒ 8x – x^2 = 15

⇒ 0 = x^2 – 8x + 15

⇒ x^2 – 5x – 3x + 15 = 0

⇒ x(x – 5) – 3(x – 5) = 0

⇒ (x – 3)(x – 5) = 0

x – 3 = 0 or x – 5 = 0

x = 3 or x = 5

When x = 3, numbers are 3 and 5.

When x = 5, numbers are 5 and 3.

Q10. Sum of the areas of two squares is 400 cm^2. If the difference of their perimeters is 16 cm, find the sides of the two squares.

Solution:

Let the side of Large square = x cm

Let the side of small square = y cm

According to the Question,

x^2 + y^2 = 400… (i) …[∵ area of square = (side)^2

4x – 4y = 16 …[∵ Perimeter of square = 4 sides

⇒ x – y = 4 … [Dividing both sides by 4

⇒ x = 4 + y …(ii)

Putting the value of x in equation (i),

(4 + y)^2 + y2^2 = 400

⇒ y2 + 8y + 16 + y^2 – 400 = 0

⇒ 2y^2 + 8y – 384 = 0

⇒ y^2 + 4y – 192 = 0 … [Dividing both sides by 2

⇒ y^2 + 16y – 12y – 192 = 0

⇒ y(y + 16) – 12(y + 16) = 0

⇒ (y – 12)(y + 16) = 0

⇒ y – 12 = 0 or y + 16 = 0

⇒ y = 12 or y = -16 … [Neglecting negative value

∴ Side of small square = y = 12 cm

and Side of large square = x = 4 + 12 = 16 cm