SOLUTION: Solve by making appropriate substitution. x^4 - 14x^2 - 32 = 0

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Question 84579: Solve by making appropriate substitution.
x^4 - 14x^2 - 32 = 0
Found 2 solutions by checkley75, rapaljer:
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
X^4-14X^2-32=0
(X^2-16)(X^2+2)=0
(X-4)(X+4)(X^2+2)=0

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
The method you are asked to use for this problem is the method of substitution. You should begin by noticing that in this equation, if you let u=x%5E2, then u%5E2+=+%28x%5E2%29%5E2=+x%5E4

Now, make these substitutions and you get:
x%5E4+-+14x%5E2+-+32+=+0
u%5E2+-+14u+-+32+=+0+

This factors into
%28u-16%29%28u%2B2%29+=+0
u=16 or u=+-2

Don't stop here! You must still solve for the original variable which is x. Substitute these values of u back into the formula u=x%5E2
u=16 or u=+-2
x%5E2=16 or x%5E2=+-2

Now, are you supposed to give REAL solutions, or are you supposed to also include imaginary (complex) solutions? IF COMPLEX solutions are allowed, then
x%5E2=16 or x%5E2=+-2
x=+0%2B-sqrt%2816%29 or x=+0%2B-sqrt%28-2%29+
x=+4, x=-4 or x=+0%2B-i%2Asqrt%282%29+

If you are finding ONLY REAL solutions, then the answer is only x=4 or x=-4.

Note: You don't really need the 0 in the answers above. It's just that in the algebra.com format, you can't write this + or - notation without using the 0! Sorry about that.

R^2 at SCC