Asked by Ariana
3. Solve for x in 8x^2 + 2x - 4 = 0
I have tried factoring, taking out a GCF then factoring, completing the square, and got a bunch of wrong answers. I thought we were supposed to complete the square, so maybe I'm just doing that wrong. I'm not looking for the answer to the problem - just a way to solve it.
Here's what I tried to do for completing the square:
i. (b/2)^2 = (2/2)^2 = 1
ii. 8x^2 + 2x + 1 = 5
iii. what? It's not a perfect square.
I have tried factoring, taking out a GCF then factoring, completing the square, and got a bunch of wrong answers. I thought we were supposed to complete the square, so maybe I'm just doing that wrong. I'm not looking for the answer to the problem - just a way to solve it.
Here's what I tried to do for completing the square:
i. (b/2)^2 = (2/2)^2 = 1
ii. 8x^2 + 2x + 1 = 5
iii. what? It's not a perfect square.
Answers
Answered by
Steve
You'll need to use the quadratic formula on that one. Or, completing the square,
8x^2+2x-4 = 0
8(x^2 + x/4) = 4
8(x^2 + x/4 + 1/64) = 4 + 8/64
8(x + 1/8)^2 = 4 + 1/8
(x + 1/8)^2 = 33/64
x + 1/8 = ±√33 / 8
x = -1/8 ±√33/8
x = (-1±√33)/8
8x^2+2x-4 = 0
8(x^2 + x/4) = 4
8(x^2 + x/4 + 1/64) = 4 + 8/64
8(x + 1/8)^2 = 4 + 1/8
(x + 1/8)^2 = 33/64
x + 1/8 = ±√33 / 8
x = -1/8 ±√33/8
x = (-1±√33)/8
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