91Ó°ÊÓ

Consider the equation $\sqrt{x}+2=\sqrt{3x+4}$.

The radical on the right is isolated. Now we square both sides and simplify.

role="math" localid="1644902345068" $\begin{array}{rcl}{(\sqrt{x}+2)}^2& =& {\left(\sqrt{3x+4}\right)}^2\\ {\left(\sqrt{x}\right)}^2+2·\sqrt{x}·2+2^2& =& 3x+4\\ x+4\sqrt{x}+4& =& 3x+4\end{array}$

Isolate the radical on the left hand side by subtracting $x+4$ from both sides.

$\begin{array}{rcl}x+4\sqrt{x}+4-(x+4)& =& 3x+4-(x+4)\\ x+4\sqrt{x}+4-x-4& =& 3x+4-x-4\\ 4\sqrt{x}& =& 2x\end{array}$

Now, remove the radical by squaring both sides.

$\begin{array}{rcl}{\left(4\sqrt{x}\right)}^2& =& {\left(2x\right)}^{2}16x=4x^2\end{array}$

Set the left hand side equal to zero by subtracting $16x$from both sides.

$\begin{array}{rcl}16x& =& 4x^2\\ 16x-16x& =& 4x^2-16x\\ 0& =& 4x^2-16x\end{array}$

Factor the right hand side and solve for $x$

$$\begin{gathered} 0=4x·x-4x·4 \\ 0=4x(x-4) \end{gathered}$$

So the solutions are $x=0$and

$x=4$.

Substitute $0$for $x$in the original equation.

$\begin{array}{rcl}\sqrt{x}+2& =& \sqrt{3x+4}\\ \sqrt{0}+2& =& \sqrt{3·0+4}\\ 2& =& \sqrt{4}\\ 2& =& 2\end{array}$

As the statement is true, so the found solution is correct.

Substitute $4$for $x$in the original equation.

$\begin{array}{rcl}\sqrt{x}+2& =& \sqrt{3x+4}\\ \sqrt{4}+2& =& \sqrt{3·4+4}\\ 2+2& =& \sqrt{12+4}\\ 4& =& \sqrt{16}\\ 4& =& 4\end{array}$

As the statement is true, so the found solution is correct.