If both $\left( x-2 \right)$ and $\left( x-\dfrac{1}{2} \right)$ are factors of $p{{x}^{2}}+5x+r$, prove that $p=r$.

Hint: We are given the factors of a quadratic equation. So, we will first multiply these factors to obtain the quadratic equation. Then we will compare the obtained quadratic equation to the given quadratic equation. After comparing the corresponding coefficients, we will be able to conclude that $p=r$.

Complete step-by-step solution
The given factors of the quadratic equation are $\left( x-2 \right)$ and $\left( x-\dfrac{1}{2} \right)$. Let us denote the quadratic equation formed by these two factors by $f\left( x \right)$. So, we have
$f\left( x \right)=\left( x-2 \right)\left( x-\dfrac{1}{2} \right)$
Now, we will multiply these two factors in the following manner,
\[\begin{align}
  & f\left( x \right)=\left( x-2 \right)\left( x-\dfrac{1}{2} \right) \\
 & ={{x}^{2}}-2x-\dfrac{1}{2}x+2\cdot \dfrac{1}{2}
\end{align}\]
Simplifying the above expression, we will get the quadratic equation as follows,
$\begin{align}
  & f\left( x \right)={{x}^{2}}-2x-\dfrac{1}{2}x+2\cdot \dfrac{1}{2} \\
 & ={{x}^{2}}-\left( \dfrac{4x}{2}+\dfrac{x}{2} \right)+1 \\
 & ={{x}^{2}}-\dfrac{5x}{2}+1
\end{align}$
So we have obtained the quadratic equation $f\left( x \right)={{x}^{2}}-\dfrac{5x}{2}+1$. Now, we will multiply this obtained quadratic equation by $-2$. Therefore, we get $f\left( x \right)=2{{x}^{2}}-5x+2$.
Now, the given quadratic equation is $p{{x}^{2}}+5x+r$, and its factors are $\left( x-2 \right)$ and $\left( x-\dfrac{1}{2} \right)$. Comparing the corresponding coefficients of the given quadratic equation and $f\left( x \right)$, we get the following,
The coefficient of $x$ is 5 in both the quadratic equations;
The coefficient of ${{x}^{2}}$ in the given quadratic equation is $p$ and that in $f\left( x \right)$ is $-2$. So, $p= 2$;
The constant term in the given quadratic equation is $r$ and that in $f\left( x \right)$ is 2. So, $r=2$. Therefore, $p=r$. Hence, proved.

Note: We can also solve this question by an alternate method. We know that $\left( x-2 \right)$ and $\left( x-\dfrac{1}{2} \right)$ are factors of $p{{x}^{2}}+5x+r$. This means that the roots of the given quadratic equation are $\alpha =2$ and $\beta =\dfrac{1}{2}$. For the general quadratic equation $a{{x}^{2}}+bx+c=0$ with roots $\alpha $ and $\beta $, we know that $\alpha +\beta =-\dfrac{b}{a}$ and $\alpha \beta =\dfrac{c}{a}$. For the given quadratic equation, substituting $a=p$, $b=5$, $c=r$, $\alpha =2$ and $\beta =\dfrac{1}{2}$ we get $2+\dfrac{1}{2}=-\dfrac{5}{p}$ and $2\times \dfrac{1}{2}=\dfrac{r}{p}$. From the first equation, we have $p=-2$and from the second equation, we have $p=r$. Hence, proved.