Prove the following

Let $f(x)=p x^{2}+5 x+r$

Since, $x-2$ is a factor of $f(x)$ then $f(2)=0$

$\therefore \quad p(2)^{2}+5(2)+r=0$

$\Rightarrow \quad 4 p+10+r=0$     $\ldots$ (i)

Since, $\quad x-\frac{1}{2}$ is a factor of $f(x)$, then $f\left(\frac{1}{2}\right)=0$

$\therefore$ $p\left(\frac{1}{2}\right)^{2}+5\left(\frac{1}{2}\right)+r=0$

$\Rightarrow$ $p \times \frac{1}{4}+\frac{5}{2}+r=0$

$\Rightarrow$ $p+10+4 r=0$ .......(ii)

Since, $x-2$ and $x-\frac{1}{2}$ are factors of $f(x)=p x^{2}+5 x+r$

From Eqs. (i) and (ii), $\quad 4 p+10+r=p+10+4 r \Rightarrow 3 p=3 r$

$\therefore$ $p=r$