91Ó°ÊÓ

Consider the equation $\sqrt{x-1}+2=\sqrt{2x+6}$.

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

$\begin{array}{rcl}{(\sqrt{x-1}+2)}^2& =& {\left(\sqrt{2x+6}\right)}^2{\left(\sqrt{x-1}\right)}^2+2·\sqrt{x-1}·2+2^2=2x+6\\ x-1+4\sqrt{x-1}+4& =& 2x+6\\ x+3+4\sqrt{x-1}& =& 2x+6\end{array}$

Isolate the radical on the left hand side by subtracting $x+3$from both sides.

role="math" localid="1644901794343" $\begin{array}{rcl}x+3+4\sqrt{x-1}-(x+3)& =& 2x+6-(x+3)\\ x+3+4\sqrt{x-1}-x-3& =& 2x+6-x-3\\ 4\sqrt{x-1}& =& x+3\end{array}$

Now, remove the radical by squaring both sides.

role="math" localid="1644901918217" $\begin{array}{rcl}{\left(4\sqrt{x-1}\right)}^2& =& {(x+3)}^2\\ 16(x-1)& =& x^2+2·x·3+3^2\\ 16x-16& =& x^2+6x+9\end{array}$

Set the left-hand side equal to zero by adding $16-16x$both sides

$\begin{array}{rcl}16x-16+16-16x& =& x^2+6x+9+16-16x\\ 0& =& x^2-10x+25\end{array}$

Now the trinomial in the right hand side is a perfect square trinomial, so factor it and find the solution.

$\begin{array}{rcl}0& =& x^2-10x+25\\ 0& =& x^2-2·x·5+5^2\\ 0& =& {(x-5)}^{2}0=x-5\\ 0+5& =& x-5+5\\ 5& =& x\end{array}$

Thus the solution is $x=5$.

Substitute $5$for $x$in the original equation.

$\begin{array}{rcl}\sqrt{x-1}+2& =& \sqrt{2x+6}\\ \sqrt{5-1}+2& =& \sqrt{2·5+6}\\ \sqrt{4}+2& =& \sqrt{10+6}\\ 2+2& =& \sqrt{16}\\ 4& =& 4\end{array}$

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