Find the 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.

Given:

Given quadratic equation is $(p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0, p ≠ -1$.

To do:

We have to find the value of p for which the given quadratic equation has equal roots.

Solution:

 $(p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0, p ≠ -1$

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=p+1, b=-6(p+1)$ and $c=3(p+9)$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

$D=[-6(p+1)]^2-4(p+1)(3(p+9))$

$D=36(p+1)^2-12(p+1)(p+9)$

$D=36(p^2+2p+1)-12(p^2+p+9p+9)$

$D=36p^2+72p+36-12p^2-120p-108$

$D=24p^2-48p-72$

The given quadratic equation has equal roots if $D=0$.

Therefore,

$24p^2-48p-72=0$

$24(p^2-2p-3)=0$

$p^2-2p-3=0$

$p^2-3p+p-3=0$

$p(p-3)+1(p-3)=0$

$(p+1)(p-3)=0$

$p+1=0$ or $p-3=0$

$p=-1$ or $p=3$

The values of p are $-1$ and $3$.  

For $p = -1$,

$(p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0$

$(-1+1)x^2-6(-1+1)x + 3(-1+9) = 0$

$24=0$ which is not possible.

For $p = 3$,

$(p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0$

$(3 + 1)x^2 - 6(3 + 1)x + 3(3+9) = 0$

$4x^2 - 24x + 36 = 0$

$4(x^2-6x+9)=0$

$x^2-6x+9=0$

$(x - 3)^2 = 0$

$x-3=0$

$x=3$

Therefore, for $p=3$ the roots of the given quadratic equation are $3$ and $3$.

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Updated on: 10-Oct-2022

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