Algerbra Help!!!!?

Question

FInd the x intercepts

Y= X^2 - 4x + 4

Solve for x using the square root method.

X^2+2=8

I think that it goes like this
X^2=8-2
X^2=6

Answer 1

0=(x-2)^2 or x=2 is the only root
For the second one, you are right, x^2=6 is right: so x=+sqrt(6) or -sqrt(6).

Answer 2

To solve this you have to factor, or find out what you multiply to get x^2-4x+4 so x times x = x^2 and 2 times 2=4 so...

(x-2)(x-2) are the binonomial factors of x. (you subtract 2 because you have a negative 4x) Multiply it out with FOIL and you'll see it works.

To find the intercepts, you have to set each binomial equal to 0

x-2=0

add two to both sides and x=2

2 is your x intercept
____________________________

x^2 does equal 6 in the second one so x= the square root of 6.

Answer 3

X^2 - 4x + 4 is in the form of the identity, (a-b)^2=a^2-2ab+b^2
So,X^2 - 4x + 4=(0 ==>(x-2)^2=0
==>x=2
==> 2 is the intercept

For second qn,
you are correct till there, now you take the sqr on bothsides,
then x=+ or - sqr 6

Answer 4

0 = x^2 - 4x + 4
0 = (x - 2 )^2
0 = x = 2

on the last one, all you need to do is take a square root for both sides
x = +/- sqr(6)

Answer 5

Y=X^2-4X+4
(X-2)(X-2)

There is only one intercept at x=2

X^2+2=8
X^2=6
X=sqrt(6)

Answer 6

no effing idea