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Solve the equation

$$\sqrt{s+13}-\sqrt{7-s} = 2$$

I moved the $-\sqrt{7-s}$ to the right side

Thus, I had $$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$

I then squared both sides $$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$

Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$

I got $$s + 13 = 4 + 4\sqrt{7-s}+ 7 – s$$

I then combined like terms $$2s + 2= 4 \sqrt{7-s}$$

I’m stuck at this point. Does anyone have an idea how to solve this equation?

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  • $\begingroup$ (Divide by $2$ and then) square once more. (And take good note of @julien's observation!) $\endgroup$ – Andreas Caranti May 23 '13 at 12:15
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    $\begingroup$ You need to square once more. This will give you a quadratic equation. But don't forget to check if the solutions do satisfy the original equation. Squarng can make false solutions appear. $\endgroup$ – Julien May 23 '13 at 12:16
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    $\begingroup$ Please consider learning how to typeset mathematics properly on this site. This page can serve as an introduction to basic mathematical typesetting here. (You can also click on the "edit" button to see how this particular post was edited to display the mathematics.) $\endgroup$ – user642796 May 23 '13 at 12:18
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$$2s+2=4\sqrt{7-s}$$ Divide by $2$ both sides $$s+1=2\sqrt{7-s}$$ Square both sides $$s^2+1+2s=28-4s$$ Put everything on the left $$s^2+6s-27=0$$ Now solve by radicals $$s_{1,2}=\frac{-6\pm\sqrt{6^2-4(-27)}}{2}$$ giving you the two solutions $$s_1=3\qquad;\qquad s_2=-9$$

Finally discard $s_2$ since it is not a solution of your title equation

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From where you left off, squaring both sides again yields

$(2s + 2)^2 = (4 √(7-s))^2$ Which becomes

$4s^2 +8s +4 = 16(7-s)$

And you have a quadratic to solve, take all terms to one side and factorise or use the quadratic formula to find solutions to $s$, there will be two solutions of course. And then plug values of $s$ found back into your original equation to make sure you dont have a negative value under the square root.

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You're almost at the end result:

After achieving $ 2s + 2 = 4 \sqrt{7-s} $, square both sides again to generate a quadratic equation.

$$ 4 ( s + 1 )^2 = 16 ( 7 - s ) = 4 (s^2 + 2s+ 1)$$ Solving which is quite easy.

$$\begin{align} s^2 + 2s + 1 &= 28 - 4s \\ s^2 + 6s - 27 &= 0 \\ (s + 9) (s - 3) &= 0 \end{align}$$


NOTE

You have to neglect $ s = -9 $ from the solutions as it'll give you $ \sqrt{-9 + 13} - \sqrt{ 7 - (-9)} = \sqrt{16} = +2 - (+4) \neq 2 $

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Wolframalpha helps you solving equations and can even show you a step-by-step solution.

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Clearly $7\ge s\ge-13\iff7-(-3)\ge s+3\ge-13-(-3)$

WLOG $s+3=10\cos 2y$ where $0\le2y\le\pi$

$\implies\sqrt{s+13}=\sqrt{10(1+\cos2y)}=2\sqrt5\cos y,\sqrt{7-s}=2\sqrt5\sin y$

$\implies2\sqrt5\cos y-2\sqrt5\sin y=2$

Squaring we get $$20(1-\sin2y)=4\iff\sin2y=\dfrac45\implies\cos2y=\pm\sqrt{1-\sin^22y}=\cdots$$

Check which values of $s=10\cos2y-3$ satisfies the given equation.

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