Eyes on Final Fantasy Forums > Reference > Study Hall > Solving Equalities with Radicals?

[Goldenboko]

Stilling trying to learn all this stuff I'll need to get into Pre-Calculus.

One of the problem's has me beat.

Its.

Radical(X) = Radical(X-8) +2

I have no idea how to do this. I keep getting stuff like 4=2.
Does that mean the correct answer would be no solution?

[Jessweeee♪]

umm...so you mean the square root of X equals the square root of (x-8) plus two?

(Just making sure.)

[The Summoner of Leviathan]

sqrt(x) = sqrt(x-8)+2

Square both sides.

x= x-8+4sqrt(x-8)+4

Simplify

x= x-4+4sqrt(x-8)

Simplify more

4=4sqrt(x-8)

Divide everything by 4.

1=sqrt(x-8)

Square it again.

1=x-8

Add 8 to both sides.

9=x

[Goldenboko]

Yes. I don't know a real method to solving this, because we're suppose to learn independently (how stupid).


EDIT: Woah, lemme read.

I don't see how this is working.

I know if you square, the square root of x, you'll get x.
I don't see how the square of the square root of (x-8) +2 would equal what you said. It would equal x-8+4
Which would say x=x-8+4 which isn't right... Dx What am I misunderstanding here?

[The Summoner of Leviathan]

I squared sqrt(x-8)+2. For the purpose of this explanation let sqrt(x-8) equal y.

So here we go.

sqrt(x-8)+2; sqrt(x-8)=y

Substitute.

y+2

Square it.

(y+2)(y+2)=y^2+4y+4

Substitute again.

[sqrt(x-8)]^2+4[sqrt(x-8)]+4

Simplify.

(x-8)+4sqrt(x-8)+4

Simplify more.

x-4+4sqrt(x-8).

Basically, when you square sqrt(x-8)+2, you have to treat it like you would squaring y+2, in other words, you square the y, add the product of 2 times y and 2 times y and square the 2.

[Goldenboko]

Ah! I get it now! Thanks!