Question

Find the values of p for which the quadratic equation $(p+1)x^2-6(p+1)x+3(p+9)=0,p\ne -1$ has equal roots. Hence, find the roots of the equation.

Answer 1

Given quadratic equation $(p+1)x^2-6(p+1)x+3(p+9)$=0.....(i)
Roots has equal then discriminant equal to 0

$\Rightarrow d=$$b^2-4ac$ $=0$

Compare equation (i) with $a{x}^2+bx+c=0$

We get,

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

So discriminant,

$(6p+6{)}^2-4(p+1\left)\right(3p+27)$ =0...(ii)
$36p^2+36+72p-4(3p^2+30p+27)$ =0
$36p^2+36+72p-12p^2-120p-108$ =0
$24p^2-48p-72$ =0
divide by 24
$p^2-2p-3$=0
$p^2+p-3p-3$ =0
$p(p+1)-3(p+1)$ =0
$(p-3)(p+1)$ =0
p = 3 or -1
Since $p\ne -1$
Therefore, $p=3$.
Case (i): When $p=3$
$\Rightarrow (3+1)x^2-6(3+1)x+3(3+9)$=0 [From (i)]
$\Rightarrow 4x^2-24x+36=0$
$\Rightarrow x^2-6x+9=0$
$\Rightarrow x^2-2\left(3\right)x+(3{)}^2=0$
$\Rightarrow (x-3{)}^2=0$ [Since, $(a-b{)}^2=a^2-2ab+b^2$]
$\therefore x=3$