SOLUTION: x(to the fourth) +8x(squared)+15=0

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Question 531041: x(to the fourth) +8x(squared)+15=0
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
Let u = x^2.
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u^2 +8u +15 = 0
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factor
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(u+3)(u+5)=0
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substitute u = x^2
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(x^2 + 3)(x^2 + 5) = 0
.
So x^2 = -3 or -5. That means there are no real roots.
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+graph%28500%2C500%2C-10%2C10%2C-10%2C100%2Cx%5E4%2B8%2Ax%5E2%2B15%29+