Question

How do you solve this square root problem?

Answer 1

Could I ^4 it?

Answer 2

I don't think that works, bakon . . .

Answer 3

well nvm then

Answer 4

You have to take it one operation at a time. Solving an equation is a matter of doing order of operations in reverse. It's like untying a knot: undo the operations that were done.

Answer 5

You have a square-root isolated on the left side, so you can square both sides to start.
Then to isolate the inner square-root, you have to subtract x.
Then square both sides again. At that point you'll have a quadratic equation that you can simplify and put into standard form.

Answer 6

So \[\sqrt{\sqrt{x+4}+x}^{2} =4^2?\]

Answer 7

yep

Answer 8

So that would be \[\sqrt{x+4} +x =16?\]

Answer 9

Or would the x be ^2 also?

Answer 10

No, it's correct as is. Now subtract x and square both sides again.

Answer 11

\[(x+4)^2 =(16-x)\]

Answer 12

??

Answer 13

The right needs to be ^2 too

Answer 14

and the x+4 does not beed to be

Answer 15

Yes, then you can expand the right to (16-x)(16-x) and FOIL

Answer 16

So x^2 -32x+256 = X+4

Answer 17

So far, so good. After putting it into standard form, you should see that it is factorable.

Answer 18

To put it in standard form, do I move it all to one side?

Answer 19

just subtract the X+4?

Answer 20

Yep. ax^2 + bx + c = 0

Answer 21

Ok. \[x^2 -33x-252=0\]

Answer 22

+252

Answer 23

Oh yea. Thanks!

Answer 24

Since it's quadratic, you're guaranteed two solutions, but the original equation is a radical equation, which often have extraneous solutions, so make sure you check both solutions in the original equation to see if they both work.

Answer 25

Thanks!

Answer 26

Note that this is a general method for this kind of equation, but also not that due to the simple nature of the original equation, you might have also been able to reason through it with some guessing and checking to find the solution. Make sure to train your number sense and intuition as well as mechanical algebraic skills.