# Thread: how to solve this equation

1. ## how to solve this equation

Solve:

5 - √(x^2 - 7) = x - 2

If I square both sides, I end up with x^2 and I don't know how to get rid of the ^2.

(If answer key cannot be reached, I think x^2 could be x^7 as the printing of the exponent is a superscript which is too small to read)

2. Hello, shenton!

I have the chilling feeling that you're doing it all wrong . . .

Solve: . $5 - \sqrt{x^2-7} \:=\:x - 2$

If I square both sides, I end up with $x^2$ and I don't know how to get rid of the $^2.$
Two issue arise here:
. . (1) You've never solved a quadratic before?

If you square both sides: . $\left(5 - \sqrt{x^2 - 7}\right)^2\;=\;(x-2)^2$

. . you get: . $25 - 10\sqrt{x^2-7} + x^2 - 7 \;=\;x^2 - 4x + 4$ . . . no help!

You must isolate the radical before you square:

. . . . . . . . . . . . . . . . $7 - x \;=\;\sqrt{x^2 - 7}$

Square: . . . . . . . . $(7 - x)^2\;=\;\left(\sqrt{x^2-7}\right)^2$

and you get: . $49 - 14x + x^2\;=\;x^2 - 7$

. . . . . . . . . . . . . . . $-14x \;=\;-56$

. . . . . . . . . . . . . . . . . . $x \;=\;4$

3. Thanks, Soroban for the help!

I have the same chilling too, that's why I came to MHF.

The teacher has not taught quadratic and the topic is on radicals.

I understood your workings (thanks for the detailed workings).

Cheers.