Question

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

Answer 1

Step 1. Compare the quadratic equation with standard form.

We know that if $\alpha ,\beta$ are the roots of a quadratic equation $a{x}^2+bx+c=0$ then

1. $\alpha +\beta =-\frac{b}{a}$
2. $\alpha \beta =\frac{c}{a}$

Now comparing this with the given quadratic equation $p{x}^2+5x+r$ we get

1. $a=p$
2. $c=r$

Now $x-2$ and $x-\frac{1}{2}$ are the factors of the polynomial .

$$\begin{gathered} \left(x-2\right)\left(x-\frac{1}{2}\right)=0 \\ \therefore \left(x-2\right)=0 \\ \Rightarrow x=2 \\ \left(2x+1\right)=0 \\ \Rightarrow x=\frac{1}{2} \end{gathered}$$

$\therefore$ The roots of the polynomial is $2$and $\frac{1}{2}$

Step 2. Prove $p=r$

Let $\alpha =2$ and $\beta =\frac{1}{2}$

$$\begin{gathered} \therefore \frac{c}{a}=\alpha \beta \\ \Rightarrow \frac{c}{a}=2·\frac{1}{2} \\ \Rightarrow \frac{r}{p}=1\left[\because c=r\mathrm{and}\mathrm{a}=\mathrm{p}\right] \\ \Rightarrow r=p \end{gathered}$$

Hence proved, $p=r$.