Question

Solve for $x:$
$\frac{x-1}{\sqrt{x}+1}=4+\frac{\sqrt{x}-1}{2}$

Answer 1

We rationalise the denominator of $\frac{x-1}{\sqrt{x}+1}$ as shown below:

$\frac{x-1}{\sqrt{x}+1}\times \frac{\sqrt{x}-1}{\sqrt{x}-1}=\frac{(x-1)(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}=\frac{(x-1)(\sqrt{x}-1)}{(\sqrt{x}{)}^2-(1{)}^2}=\frac{(x-1)(\sqrt{x}-1)}{x-1}=\sqrt{x}-1$

Now, consider the given problem $\frac{x-1}{\sqrt{x}+1}=4+\frac{\sqrt{x}-1}{2}$ that can be simplified as follows:

$\frac{x-1}{\sqrt{x}+1}=4+\frac{\sqrt{x}-1}{2}\Rightarrow \sqrt{x}-1=4+\frac{\sqrt{x}-1}{2}\Rightarrow \sqrt{x}-1-\frac{\sqrt{x}-1}{2}=4\Rightarrow \frac{2(\sqrt{x}-1)-\sqrt{x}-1}{2}=4\Rightarrow \frac{\sqrt{x}-1}{2}=4\Rightarrow \sqrt{x}-1=4\times 2\Rightarrow \sqrt{x}-1=8\Rightarrow \sqrt{x}=8+1\Rightarrow \sqrt{x}=9\Rightarrow (\sqrt{x}{)}^2=9^2\Rightarrow x=81$

Hence, $x=81$