SOLUTION: What are real or imaginary solutions of the polynomials. x^4-52x^2+576=0.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: What are real or imaginary solutions of the polynomials. x^4-52x^2+576=0.      Log On


   

Question 1038065: What are real or imaginary solutions of the polynomials. x^4-52x^2+576=0.
Found 2 solutions by Boreal, josmiceli:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x^4-52x^2+576=0
(x^2-36)(x^2-16)=0
x=+/-6; +/-4graph%28300%2C200%2C-10%2C10%2C-100%2C1000%2Cx%5E4-52x%5E2%2B576%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +z+=+x%5E2+
then +x%5E4+=+z%5E2+
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Rewrite the equation:
+z%5E2+-+52z+%2B+576+=+0+
Complete the square:
+z%5E2+-+52z+%2B+%28+52%2F2+%29%5E2+=+%28+52%2F2+%29%5E2+-+576+
+z%5E2+-+52z+%2B+676+=+676+-+576+
+z%5E2+-+52z+%2B+676+=+100+
+%28+z+-+26+%29%5E2+=+10%5E2+
Take the square root of both sides
+z+-+26+=+10+
+z+=+36+
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also, using the negative square root of +100+,.
+z+-+26+=+-10+
+z+=+16+
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Now go back to
+x%5E2+=+z+
+x%5E2+=+36+
+x+=+6+
+x+=+-6+
and
+x%5E2+=+16+
+x+=+4+
+x+=+-4+
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I get 4 real solutions,
+x+=+4+
+x+=+-4+
+x+=+6+
+x+=+-6+
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You can check these by plugging back into equation
Here's the plot:
+graph%28+400%2C+400%2C+-10%2C+10%2C+-150%2C+700%2C+x%5E4+-+52x%5E2+%2B+576+%29+