Question

Find the value of $p$ so that the quadratic equation $px(x-3)+9=0$ has equal roots.

Answer 1

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

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

It has equal roots and let the roots be $\alpha$.

We know that for a general quadratic equation $a{x}^2+bx+c$,

Product of roots $=\frac{c}{a}$

Sum of roots $=\frac{-b}{a}$

$\therefore \alpha +\alpha =2\alpha =\frac{3p}{p}$ and ${\alpha }^2=\frac{9}{p}$ {For the given equation}

It gives, $\alpha =\frac{3p}{2p}=\frac{3}{2}$

Also, ${\alpha }^2=\frac{9}{p}=\frac{3^2}{2^2}$

$\therefore p=\frac{4\times 9}{9}=4$