If both $x-2$ and $x-\frac{1}{2}$ are factors of $p x^{2}+5 x+r,$ show that $p=r$.

(A) Let $f(x) = p x^{2} + 5 x + r$.
Since $(x-2)$ is a factor of $f(x)$,by the Factor Theorem,$f(2) = 0$.
$p(2)^{2} + 5(2) + r = 0$
$4p + 10 + r = 0$
$4p + r = -10$ --- $(1)$
Since $(x - \frac{1}{2})$ is a factor of $f(x)$,by the Factor Theorem,$f(\frac{1}{2}) = 0$.
$p(\frac{1}{2})^{2} + 5(\frac{1}{2}) + r = 0$
$\frac{1}{4}p + \frac{5}{2} + r = 0$
Multiply the entire equation by $4$:
$p + 10 + 4r = 0$
$p + 4r = -10$ --- $(2)$
Comparing equations $(1)$ and $(2)$,since both are equal to $-10$:
$4p + r = p + 4r$
$4p - p = 4r - r$
$3p = 3r$
$p = r$
Hence,it is proved.