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.