find x intercepts of 4x^2-8x+4

Question

campbell_st · Answer

let your quadratic equal zero and solve. 

$$4x^2 - 8x + 4 = 0$$

you can take out a factor of 4

$$4(x^2 - 2x + 1) = 0$$

just factor the quadratic to then solve for x to find the x-intercept

anonymous · Answer

I'm getting that there are two x intercepts correct?

campbell_st · Answer

well yes there are but you'll find its a repeated value. 

you are factoring a perfect square

anonymous · Answer

Ok, I am a little confused.  Can you show me by working it out in steps, maybe then I'll see it and understand it.

campbell_st · Answer

ok, what are the factors of 1 that add to -2, they are both negative.

anonymous · Answer

1+-3

campbell_st · Answer

if you multiply them you get - 3 

its -1 * -1 = 1             -1 + -1 = -2

does that make sense?

anonymous · Answer

Yes that makes more sense, thank you.

anonymous · Answer

I was confused when the answers did not offer a repeated value.

campbell_st · Answer

well I think you are looking at 

$$4(x -1)^2 = 0$$

what value of x makes the equation true, that is the x- intercept.

anonymous · Answer

ah...ok