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

To do:

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

Solution:
Let $f(x)= px^2+5x+r$

If $(x-2)$ is a factor of $f(x)$, then by factor theorem $f(2)=0$ ⇒$ p(2)^2+5(2)+r=0$  

⇒$4p+r+10=10$......(i)

$\Rightarrow \mathrm{p}(2{)}^2+5(2)+\mathrm{r}=0$

$$
$\Rightarrow \mathrm{p}(2)^{2}+5(2)+\mathrm{r}=0$

If $(x-\frac{1}{2})$ is a factor of $f(x)$, then by factor theorem 

$f(\frac{1}{2})=0$  |$x-\frac{1}{2}=0$ ⇒ $x=\frac{1}{2}$|

$\left(x-\frac{1}{2}\right)$

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

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

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

Subtracting ( 2 ) from ( 1 ), we get $3 p-3 r=0$

$\Rightarrow p=r$

Therefore , Proved

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Updated on: 10-Oct-2022

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