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

$$\text{If }(x-2)\text{ is a factor: }p(2)^2+5(2)+r=0$$

$$\Rightarrow 4p + 10 + r = 0$$

$$\Rightarrow 4p + r = -10 \quad \text{...(i)}$$

If $(x - \frac{1}{2})$ is a factor:

$$\begin{aligned} p\left(\frac{1}{4}\right) + \frac{5}{2} + r &= 0 \\ \Rightarrow \quad \frac{p}{4} + r &= \frac{-5}{2} \end{aligned}$$

Multiply by 4:

$$p + 4r = -10 \quad \text{...(ii)}$$

From (i) and (ii),

$$4p + r = p + 4r$$

$$\Rightarrow 3p = 3r$$

$$\Rightarrow p = r$$