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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. - CBSE Class 10 - Mathematics

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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.

Solution

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=b24ac=0

[6(p+1)]24×(p+1)×3(p+9)=0

36(p+1)212(p+1)(p+9)=0

12(p+1)[3(p+1)(p+9)]=0

12(p+1)(2p6)=0

p+1=0 or 2p6=0

p+1=0

p=1

This is not possible as p1

2p6=0

p=3

So, the value of p is 3.

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

(3+1)x26(3+1)x+3(3+9)=0

4x224x+36=0

4(x26x+9)=0

4(x3)2=0

x=3

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

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Solution 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. Concept: Nature of Roots.
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