# 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. - Mathematics

Find the value of p for which the quadratic equation

$\left( p + 1 \right) x^2 - 6(p + 1)x + 3(p + 9) = 0, p \neq - 1$ has equal roots. Hence, find the roots of the equation.

Disclaimer: There is a misprinting in the given question. In the question 'q' is printed instead of 9.

#### Solution

The given quadratic equation  $\left( p + 1 \right) x^2 - 6(p + 1)x + 3(p + 9) = 0$,

has equal roots.

Here,

$a = p + 1, b = - 6p - 6 \text { and } c = 3p + 27$.

As we know that $D = b^2 - 4ac$

Putting the values of $a = p + 1, b = - 6p - 6\text { and } c = 3p + 27$.

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

$= 36( p^2 + 2p + 1) - 12( p^2 + 10p + 9)$

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

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

The given equation will have real and equal roots, if D = 0

Thus,

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

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

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

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

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

$\Rightarrow p + 1 = 0 \text { or } p - 3 = 0$

$\Rightarrow p = - 1 \text { or } p = 3$

It is given that p ≠ −1, thus p = 3 only.
Now the equation becomes

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

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

$\Rightarrow x^2 - 3x - 3x + 9 = 0$

$\Rightarrow x(x - 3) - 3(x - 3) = 0$

$\Rightarrow (x - 3 )^2 = 0$

$\Rightarrow x = 3, 3$

​Hence, the root of the equation is 3.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths