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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?
umm...so you mean the square root of X equals the square root of (x-8) plus two?
(Just making sure.)
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
:D
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?
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.
Ah! I get it now! Thanks!