Question

Find the values of $p$ for which the quadratic equation $px\times (x-3)+9=0$ has equal roots.

Answer 1

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

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

Given: The quadratic equation has equal roots

For a quadratic equation to have real roots: $D=0$

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

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

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

$\Rightarrow p^2-4p=0$

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

$\Rightarrow p=0$, $p=4$