Find that value of *p* for which the quadratic equation (*p* + 1)*x*^{2} − 6(*p* + 1)*x* + 3(*p* + 9) = 0, *p* ≠ − 1 has equal roots. Hence find the roots of the equation.

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#### Solution

It is given that the quadratic equation (*p* + 1)*x*^{2} − 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=b^{2}−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)x^{2}−6(3+1)x+3(3+9)=0

⇒4x^{2}−24x+36=0

⇒4(x^{2}−6x+9)=0

⇒4(x−3)^{2}=0

⇒x=3

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

Concept: Nature of Roots

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