# How do you solve this radical equation $\sqrt{2x+5} + 2\sqrt{x+6} = 5$?

I have a radical equation $$\sqrt{2x+5} + 2\sqrt{x+6} = 5$$ and I am having trouble calculating an answer. I keep on getting weird numbers that are not correct as my answer. How do you solve this? A step-by-step procedure would be highly appreciated.

• What have you tried? Show the steps you of your attempts and people can better help you. – Matthew Conroy Jul 26 '12 at 17:08

In fact my answer is the practical explanation of Old John's answer: $$[\sqrt{2x+5}+2 \sqrt{x+6}]^{2} = (2x+5)+4\sqrt{(2x+5)(x+6)} +4(x+6) = 25$$ $$-4-6x = 4\sqrt{(2x+5)(x+6)}$$ $$16+48x+36{x}^{2} = 32{x}^{2}+272x+480$$ $$x= -2$$ $$x= 58$$

Then you check if they are solutions to the initial equation. You will get $x=-2$

• $\LaTeX$ tip: use \sqrt{2x+5} to get $\sqrt{2x+5}$. Please, edit your answer so you can avoid ambiguities in OP's mind. – Ian Mateus Jul 26 '12 at 18:30

First we require $x \geq -\frac{5}{2}$ (for both radicals to be well defined). Now we have that $\sqrt{2x+5}=5-2\sqrt{x+6}$.

If $5-2\sqrt{x+6}\ge 0\Rightarrow -\frac{5}{2}\le x\le \frac{19}{4}$ then \begin{equation}\sqrt{2x+5}^2=(5-2\sqrt{x+6})^2\Leftrightarrow 2x+5=25+4(x+6)-20\sqrt{x+6}\end{equation} Can you continue from here? Do you understand why we make the hypothesis?

Here is some of the rest:

\begin{equation} 2x+5=25+4(x+6)-20\sqrt{x+6}\Leftrightarrow 20\sqrt{x+6}=44+2x\Leftrightarrow 10\sqrt{x+6}=x+22\end{equation} Since $x+22\ge 0$ (remember that $-\frac{5}{2}\le x\le \frac{19}{4}$) we have that

\begin{equation} (10\sqrt{x+6})^2=(x+22)^2\Leftrightarrow 100(x+6)=x^2+44x+484\Leftrightarrow x^2-56x-116=0\end{equation}

Solve the last equation and choose only the $x$ that lie in $\left[-\frac{5}{2}, \frac{19}{4}\right]$

Step 1: square both sides, possibly after moving one of the square roots to the other side.

You should now have an equation with only one square root in it.

Step 2: isolate the remaining square root on its own on one side.

Step 3: square both sides to get rid of the remaining square root.

Step 4: solve the polynomial equation you now have.

Step 5: check which of the solutions you obtained in step 4 really are solutions of the original equation. This step is necessary , since squaring both sides at various steps will have introduced extra solutions which are not solutions of the equation you started with.

Left hand side of the equation is increasing, so there is at most one root.

Careful gazing gives $x=-2$.

$\rm\begin{eqnarray}\rm a^2 &=&\rm 2x\!+\!5 \\ \rm b^2 &=&\rm\ \: x\!+\!6 \end{eqnarray}\:\Rightarrow\: a^2\!-\!2b^2 = -7.\$ But $\rm\:\color{#C00}a\!+\!2b = 5\:$ so $\rm\:0 = (\color{#C00}{5\!-\!2b})^2\!-2b^2\!+\!7\, =\, 2\,(b\!-\!8)\,(b\!-\!2)\:$