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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.

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    $\begingroup$ What have you tried? Show the steps you of your attempts and people can better help you. $\endgroup$ – Matthew Conroy Jul 26 '12 at 17:08
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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 $x=-2$

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    $\begingroup$ $\LaTeX$ tip: use \sqrt{2x+5} to get $\sqrt{2x+5}$. Please, edit your answer so you can avoid ambiguities in OP's mind. $\endgroup$ – Ian Mateus Jul 26 '12 at 18:30
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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 $\left[-\frac{5}{2}, \frac{19}{4}\right]$

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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.

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

Careful gazing gives $x=-2$.

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$\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 $\rm\:0 = (\color{#C00}{5\!-\!2b})^2\!-2b^2\!+\!7\, =\, 2\,(b\!-\!8)\,(b\!-\!2)\:$

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