Square root equation

Question

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 ?

Answer 1

You probably mean

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

The steps:

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

Answer 2

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.