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) (Show Source): You can put this solution on YOUR website!
Let u = x^2.
.
u^2 +8u +15 = 0
.
factor
.
(u+3)(u+5)=0
.
substitute u = x^2
.
(x^2 + 3)(x^2 + 5) = 0
.
So x^2 = -3 or -5. That means there are no real roots.
.
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