SOLUTION: I'm trying to find the x-intercepts for the equation Y=16-4x^2. What I've tried to do is change the y to 0, and factor the rest of the equation, but I keep getting left with a 4 t

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> SOLUTION: I'm trying to find the x-intercepts for the equation Y=16-4x^2. What I've tried to do is change the y to 0, and factor the rest of the equation, but I keep getting left with a 4 t      Log On

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Question 211249This question is from textbook
: I'm trying to find the x-intercepts for the equation Y=16-4x^2. What I've tried to do is change the y to 0, and factor the rest of the equation, but I keep getting left with a 4 that I can't seem to get rid of. My Algebra skills are a little rusty...please help.
0=4(4-x^2)
0=4(2+x)(2-x)
0=-2
0=2
(-2,0) and (2,0) are the x intercepts and the answers at the solutions part of the book, but why is that 4 still there?
This question is from textbook

Answer by Earlsdon(6287) About Me  (Show Source):
You can put this solution on YOUR website!
Remeber the "zero product" rule?
If a%2Ab+=+0 the either a+=+0 or b+=+0 or both.
Your work so far is immpecable. When you get to the line:
0+=+4%282%2Bx%29%282-x%29 Apply the zero product rule:
4+=+0 Clearly not true.
2%2Bx+=+0 thus +x+=+-2
2-x+=+0 thus x+=+2
You can ignore the 4 as it is not pertinent to the x-intercepts.
Another way to look at doing this problem (if the 4 still bothers you) is:
-4x%5E2%2B16+=+0 Divide both sides by -4.
x%5E2-4+=+0 Factor.
%28x-2%29%28x%2B2%29+=+0 Apply the zero product rule.
x-2+=+0 or x%2B2%29=+0 so...
x+=+2 or x+=+-2 and these are the x-intercepts.
graph%28400%2C400%2C-5%2C5%2C-5%2C20%2C-4x%5E2%2B16%29