Question

If $p=\frac{\sqrt{q}+z}{r^2+z}$, find the value of $z$ in terms of $p$, $q$, and $r$.

• A. $\frac{p-r^2\sqrt{q}}{p-1}$
• B. $\frac{p{r}^2-\sqrt{q}}{r^2}$
• C. $\sqrt{\frac{p+qr}{r^2}}$
• D. $\frac{r^2\sqrt{q}-p}{p+1}$
• E. $\frac{\sqrt{q}-p{r}^2}{p-1}$

Answer 1

The correct option is E $\frac{\sqrt{q}-p{r}^2}{p-1}$

Given, $p=\frac{\sqrt{q}+z}{r^2+z}$

$\Rightarrow p(r^2+z)=\sqrt{q}+z$

$\Rightarrow p{r}^2+pz=\sqrt{q}+z$

$\Rightarrow pz-z=\sqrt{q}-p{r}^2$

$\Rightarrow z(p-1)=\sqrt{q}-p{r}^2$

$\Rightarrow z=\frac{\sqrt{q}-p{r}^2}{p-1}$