# Square root equation

I have the equation $\sqrt{(7-x)} - \sqrt {(x+13)} = 2$ The square root should be expanded so it is square root of $7-x$ - square root of $x+13 = 2$. When i square both sides i get: $7-x - x-13 = 4$

then i clean on the LHS so i get $-2x-6 = 4$ which leads to $(-2x-6)^2 = 4^2$ and after working that out i get:

$4x^2 + 24 x + 36 = 16$.

Next step will then be $4x^2 +24x +20 = 0$

Using the pq formula i get $x^2 = -6/2$ +- $\sqrt {6 \over 2}^2 -5$

And after that i get $x^2 = -6/2 +- 2$ which leads to $x^2 = -1$ or $x^2= -5$.

The answer according to the textbook is $-9$. Do you know how to solve this ?

• when you square both sides, check again what do you get – Dimitri C Jul 27 '15 at 6:39
• The first square root is defined for $x\leq 7$ and the second for $x\geq 13$, so the equation has no (real) solution. Are you sure you typed the question correctly. – mickep Jul 27 '15 at 6:42
• Sorry it should be $\sqrt {(x+13)}$ not$\sqrt {(x-13)}$ – addde Jul 27 '15 at 6:49

You probably mean

$$\sqrt{7-x} - \sqrt{x+13} = 2.$$

The steps:

$$\big( \sqrt{7-x} - \sqrt{x+13} \big)^2 = 2^2\\ \Downarrow\\ \sqrt{7-x}^2 + \sqrt{x+13}^2 - 2 \sqrt{7-x} \sqrt{x+13} = 4\\ \Downarrow\\ \big(7-x\big) + \big(x+13\big) - 2 \sqrt{7-x} \sqrt{x+13} = 4\\ \Downarrow\\ 20 - 2 \sqrt{7-x} \sqrt{x+13} = 4\\ \Downarrow\\ \sqrt{7-x} \sqrt{x+13} = 8\\ \Downarrow\\ \big(7-x\big)\big(x+13\big) = 64\\ \Downarrow\\ \big(10-3-x\big)\big(10+3+x\big) = 64\\ \Downarrow\\ 100-\big(3+x\big)^2 = 64\\ \Downarrow\\ \big(3+x\big)^2 = 36\\ \Downarrow\\ 3+x = \pm 6\\ \Downarrow\\ \bbox[16px,border:2px solid #800000] {x = -9 \vee x = 3}$$

• Thanks you very much i will try that! – addde Jul 27 '15 at 6:53
• One quibble: In your penultimate step, you meant $3 + x = \pm \color{red}{6}$. I like how you rewrote the product $(7 - x)(x + 13)$ as a difference of squares. – N. F. Taussig Jul 27 '15 at 16:33
• Indeed, a type. You know that $(x+y)(x-y) = x^2-y^2$? – johannesvalks Jul 27 '15 at 16:46
• $x = 3$ is not a solution of $\sqrt{7 - x} - \sqrt{x + 13} = 2$. – Paolo Jul 28 '15 at 13:52
• @johannesvalks $x=3$ is not a solution, which can be easily verified; when squaring the whole equation we've generated a non-existent solution. I downvoted only to point this out, could you please include this in your answer and I will happily undo the downvote. – Theta Jul 7 '17 at 19:12

Rembember that $(a-b)^2 = a^2 + b^2 -2ab$ which isn't $a^2 - b^2$. So $\sqrt{7-x} - \sqrt{x-13}$ squared is $$(7-x) + (x-13) - 2\sqrt{(7-x)(x-13)}$$

Now you can isolate the square root term and square again.