Question

Find the values of p for which the roots of the quadratic question $px\left(x-3\right)+9=0$ have two equal roots.

Answer 1

We have,

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

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

Since, roots are equal.

Let the roots $\alpha$ and $\alpha$.

$\alpha +\alpha =-\frac{-3p}{p}$

$2\alpha =3$

$\alpha =\frac{3}{2}$

$\alpha \times \alpha =\frac{9}{p}$

${\alpha }^2=\frac{9}{p}$

Therefore,

$\left(\frac{9}{4}\right)=\frac{9}{p}$

$p=4$

Hence, this is the answer.