SOLUTION: Solve the equation by making an appropriate substitution. (Don't forget imaginary solutions) x(exponent 4) + 2x(exponent 2) -8 =0

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Question 887314: Solve the equation by making an appropriate substitution. (Don't forget imaginary solutions)
x(exponent 4) + 2x(exponent 2) -8 =0
Found 3 solutions by josgarithmetic, lwsshak3, algebrapro18:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E4%2B2x%5E2-8=0
Use x%5E2 as if it were the variable. (or use a substitution for x%5E2).
%28x%5E2-2%29%28x%5E2%2B4%29=0

Either x%5E2-2=0, which gives highlight%28x=0%2B-+sqrt%282%29%29
OR
x%5E2%2B4=0, which will give x%5E2=-4, x=0%2B-+sqrt%28-4%29, meaning highlight%28x=0%2B-+2i%29

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
x(exponent 4) + 2x(exponent 2) -8 =0
x^4+2x^2-8=0
(x^2+4)(x^2-2)=0
..
x^2+4=0
x^2=-4
x=±√-4=±2i
or
x^2-2=0
x^2=2
x=±√2

Answer by algebrapro18(249) About Me  (Show Source):
You can put this solution on YOUR website!
So first we can make a substitution. Lets let s = x^2. Then the equation becomes:

s^2+2x-8 = 0

Now we can factor the quadratic.

s^2+2x-8 = 0
(s-2)(s+4)=0

Now we can plug in our value for s and set each factor equal to 0.

(s-2)(s+4)=0
(x^2-2)(x^2+4)=0
x^2-2 = 0 and x^2+4 = 0
x^2 = 2 and x^2 = -4

NOTE:Remember that the square root of a negative number is the same as the square root of the positive number times i.

x = sqrt%282%29, -sqrt%282%29 and x = 2i,-2i