Question

Find the sum of all values of $p$ for which the roots are real and equal of the following equation:
$px(x-3)+9=0$

Answer 1

$px(x-3)+9=0$

$\Rightarrow$ $p{x}^2-3px+9=0$

On comparing above obtained equation with $a{x}^2+bx+c=0$ we get,

$a=p,b=-3p,c=9$

It is given that roots are real and equal which implies that the discriminant is equal to $0.$ So,

$b^2-4ac=0$

$\Rightarrow$ $(-3p{)}^2-4(p\left)\right(9)=0$

$\Rightarrow$ $9p^2-36p=0$

$\Rightarrow$ $9p(p-4)=0$

$\Rightarrow$ $9p=0$ and $p-4=0$

$\Rightarrow$ $p=0$ and $p=4$

Hence, the sum of both values of $p$ is equal to $4.$