SOLUTION: I am doing something wrong here if you could please assist... sqrt 4x + 8, + 8=0 I dropped the zero, and minused (-8)^2 and I came up with 4.

Algebra ->  Radicals -> SOLUTION: I am doing something wrong here if you could please assist... sqrt 4x + 8, + 8=0 I dropped the zero, and minused (-8)^2 and I came up with 4.      Log On


   

Question 49364This question is from textbook
: I am doing something wrong here if you could please assist... sqrt 4x + 8, + 8=0
I dropped the zero, and minused (-8)^2 and I came up with 4. This question is from textbook

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
I am only guessing that the comma is an 'x':
sqrt(4x) + 8x + 8 = 0
sqrt(4x) = -8x - 8
4x = (-8x - 8)^2
4x = 64x^2 + 128x + 64
0 = 64x^2 + 124x + 64
0 = 16x^2 + 31x + 16
x+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-31+%2B-+sqrt%28+961+-+1024+%29%29%2F%2832%29+ Since the discriminant is a negative number, there is no solution.
graph%28600%2C600%2C-2%2C15%2C-2%2C200%2Csqrt%284x%29%2B8x%2B8%2C0%29
or do you mean:
sqrt(4x + 8) + 8 = 0
sqrt(4x + 8) = -8
4x + 8 = 64
4x = 56
x = 14
If you plug fourteen in, you get 16 = 0. That is not true, so the answer is no solution.