Question

Find that value of p for which the quadratic equation (p + 1)x² − 6(p + 1)x + 3(p + 9) = 0, p ≠ − 1 has equal roots. Hence find the roots of the equation.

Answer 1

It is given that the quadratic equation (p + 1)x² − 6(p + 1)x + 3(p + 9) = 0, p ≠ − 1 has equal roots.

Therefore, the discriminant of the quadratic equation is 0.

Here, 

a=(p+1)

b=−6(p+1)

c=3(p+9)

∴D=b²−4ac=0

⇒[−6(p+1)]²−4×(p+1)×3(p+9)=0

⇒36(p+1)²−12(p+1)(p+9)=0

⇒12(p+1)[3(p+1)−(p+9)]=0

⇒12(p+1)(2p−6)=0

⇒p+1=0 or 2p−6=0

p+1=0

⇒p=−1

This is not possible as p≠−1

2p−6=0

⇒p=3

So, the value of p is 3.

Putting p = 3 in the given quadratic equation, we get

(3+1)x²−6(3+1)x+3(3+9)=0

⇒4x²−24x+36=0

⇒4(x²−6x+9)=0

⇒4(x−3)²=0

⇒x=3

Thus, the root of the given quadratic equation is 3.