SOLUTION: find the real number solutions for the equation 3x^4+15^2-72=0

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Question 272153: find the real number solutions for the equation
3x^4+15^2-72=0
Found 2 solutions by Alan3354, oberobic:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
3x^4+15^2-72=0
x%5E4+%2B+5x%5E2+-+24+=+0
%28x%5E2+%2B+8%29%2A%28x%5E2+-+3%29+=+0
x^2 = - 8
x = ± 2*i*sqrt(2) (complex numbers)
----------------
x^2 = 3
x = ± sqrt(3) (real numbers)

Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
I assume the problem is to solve:
3x%5E4+%2B15x%5E2+-72+=+0
.
Factoring, we arrive at:
%283x%5E2+-+24%29%28x%5E2+%2B+3%29+=+0
.
Real solutions will be found where
%283x%5E2+-24%29+=+0
or
%28x%5E2+%2B3+%29+=+0
.
3x%5E2+-+24+=+0
3x%5E2+=+24
x%5E2+=+8
x+=+sqrt%288%29+=+2%2Asqrt%282%29
or
x+=+-sqrt%288%29+=+-2%2Asqrt%282%29
.
x%5E2+%2B+3+=+0
x%5E2+=+-3
There are no real solutions for this part.