SOLUTION: I cannot solve x^4-5x^2-8=0 My professor told me to substitute y for x but I still cannot fathom what is going on. I have tried for the past several hours please help.

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Question 58382: I cannot solve x^4-5x^2-8=0
My professor told me to substitute y for x but I still cannot fathom what is going on. I have tried for the past several hours please help.

Found 2 solutions by checkley71, Nate:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
SET X^2=Y WE GET
Y^2-5Y-8
USING THE QUADRATIC EQUATION WE GET
Y=(5+-SQRT(25-4*-8))/2
Y=(5+-SQRT(25+32))/2
Y=(5+-(SQRT57))/2
Y=(5+-7.55)/2
Y=(5+7.55)/2
Y=12.55/2
Y=6.275
&
Y=(5-7.55)/2
Y=-2.55/2
Y=-1.275
X^2=Y OR
X^2=6.275
X=SQRT6.275
X=2.505
&
X^2=-1.275
X=SQRT-1.275
X=1.129i

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
Depending on your math status, I could lead you onto what your professor was indicating.
y = x^2
leads to ~>
(x^2)^2 - 5(x^2) - 8 = 0
(y)^2 - 5(y) - 8 = 0
y^2 - 5y - 8 = 0
y+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
y+=+%285+%2B-+sqrt%28+25+%2B+32+%29%29%2F%282%29+
y+=+%285+%2B-+sqrt%28+57+%29%29%2F%282%29+
Now, we have the partial answer. Since the original equation supported x, we must have the variable x equal something to develop the answer.
sqrt%28y%29+=+x
x+=+sqrt%28%285+%2B-+sqrt%28+57+%29%29%2F%282%29%29
About:
x = 2.5050
x = -2.5050
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E4+-+5x%5E2+-+8%29