$\displaystyle \sqrt{2x+6}-\sqrt{x+4}=1$
solve for x.

Attempts:

$\displaystyle (\sqrt{2x+6}-\sqrt{x+4})\sqrt{2x+6}-\sqrt{x+4}=1(\sqrt{2x+6}-\sqrt{x+4})$
$\displaystyle 2x+6+x+4=\sqrt{2x+6}-\sqrt{x+4}$

$\displaystyle 2x+6+x+4+\sqrt{x+4}=\sqrt{2x+6}$
$\displaystyle 2x+6=(3x+10)^2+x+4$
$\displaystyle 2x+6=9x^2+61x+104$
$\displaystyle 0=9x^2+59x+98$

$\displaystyle x=\frac{-59\pm\sqrt{47}i}{18}$

But, the answer sheet says that x=5. Any help?

2. $\displaystyle \begin{gathered} \sqrt {2x + 6} = 1 + \sqrt {x + 4} \hfill \\ 2x + 6 = 1 + 2\sqrt {x + 4} + x + 4 \hfill \\ x + 1 = 2\sqrt {x + 4} \hfill \\ x^2 + 2x + 1 = 4\left( {x + 4} \right) \hfill \\ \end{gathered}$

3. note that if the solution exists, then it must lie on $\displaystyle x\ge-3.$

write the inequality as $\displaystyle \sqrt{2x+6}=1+\sqrt{x+4},$ since both sides are positive, we can square them and get $\displaystyle 2x+6=1+2\sqrt{x+4}+x+4$ then $\displaystyle x+1=2\sqrt{x+4},$ in order to square we require also $\displaystyle x\ge-1,$ and of course the previous condition, so we need $\displaystyle x\ge-1$ in order to make these things work, thus $\displaystyle x^2+2x+1=4x+16$ then $\displaystyle x^2-2x-15=0\implies(x+3)(x-5)=0,$ we require that $\displaystyle x\ge-1,$ so $\displaystyle x=5$ is the solution to our problem.