SOLUTION: Could you please help me with this problem: Solve for x: {{{2x^4-26x^2+72=0}}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Could you please help me with this problem: Solve for x: {{{2x^4-26x^2+72=0}}}      Log On


   

Question 21481: Could you please help me with this problem:
Solve for x: 2x%5E4-26x%5E2%2B72=0
Found 2 solutions by venugopalramana, Earlsdon:
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x: 2x%5E4-26x%5E2%2B72=0
X^4-13X^2+36=0
X^4-4X^2-9X^2+36=0
X^2(X^2-4)-9(X^2-4)=0
(X^2-4)(X^2-9)=0
X^2=4......................OR..............X^2=9
X=2..OR..-2............OR.......X=3...OR......-3

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
2x%5E4+-+26x%5E2+%2B+72+=+0 Factor out a 2.
2%28x%5E4+-+13x%5E2+%2B+36%29+=+0 Factor the parentheses.
2%28x%5E2+-+4%29%28x%5E2+-+9%29+=+0 Factor the parentheses again.
2%28x+%2B+2%29%28x+-+2%29%28x+%2B+3%29%28x+-+3%29+=+0 Apply the zero product principle.
%28x+%2B+2%29+=+0 and/or %28x+-+2%29+=+0 and/or %28x+%2B+3%29+=+0 and/or %28x+-+3%29+=+0
The roots are:
x = -2
x = 2
x = -3
x = 3