# Important Questions for Chapter 4 - Quadratic Equations

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**Important questions based on NCERT syllabus for Chapter 4 - Quadratic Equations:**

*Question 1*: If one root of the quadratic equation 2x^2 + kx − 6 = 0 is 2, find the value of k. Also, find the other root.

*Solution*:

Since x = 2 is a root of the equation

2x^2 + kx − 6 = 0

∴2×2^2 + 2k − 6 =0

⇒ 8 + 2k − 6 = 0

⇒ 2k + 2 = 0

⇒ k = −1

Substituting k=−1 in the equation 2x^2 + kx − 6 = 0, we get

2x^2 − x − 6 = 0

⇒ 2x^2 − 4x + 3x − 6 = 0

⇒ 2x(x−2) + 3(x−2) = 0

⇒ (x−2)(2x+3) = 0

⇒ x−2 = 0,2x + 3 = 0

⇒ x = 2, x = −3/2

Hence, the other root is −3/2

*Question 2*: Using quadratic formula solve the following quadratic equation:

p^2 x^2 + (p^2 − q^2)x − q^2 = 0, p ≠ 0

*Solution*:

We have, p^2 x^2 + (p^2 − q^2)x − q^2 = 0,p = 0

Comparing this equation with ax2+bx+c=0, we have

a = p^2, b = p^2 − q^2 and c = −q^2

∴ D = b^2 − 4ac = (p^2 − q^2)^2 − 4 × p^2 × −q^2

⇒ D = (p^2 − q^2)^2 + 4p^2q^2

⇒ D = (p^2 + q^2)^2

⇒ D > 0

So, the given equation has real roots given by

α = −b+√D/2a = −(p^2 − q^2) + (p^2 + q^2)2p^2 = q^2p^2

and, β = −b−√D/2a = −(p^2 − q^2) − (p^2 + q^2)/2p^2 = −1

*Question 3*: Seven years ago Varun's age was five times the square of Swati's age. Three years hence Swati's age will be two fifth of Varun's age. Find their present ages.

*Solution*:

Seven years ago, let Swati's age be x years, Varun's present age =(5x2+7) years

Three years hence, we have

Swati's age =(x+7+3) years=(x+10) years

Varun's age =(5x2+7+3) years=(5x2+10) years

It is given that three years hence Swati's age will be 25 of Varun's age.

∴ x + 10 = 25(5x^2 + 10)

⇒ x + 10 = 2x^2 + 4

⇒ 2x^2 − x − 6 = 0

⇒ 2x^2 − 4x + 3x − 6 = 0

⇒ 2x(x−2) + 3(x−2) = 0

⇒ (2x+3)(x−2) = 0

⇒ x − 2 = 0 [∵2x+3≠0 as x>0]

⇒ x = 2

Hence, Swati's pressent age = (2+7) years = 9 years

Varun's present age =(5×22+7) years = 27 years.