SOLUTION: how to find the real or imaginary solutions of each equation by factoring {{{x^4-3x^2=2x^2}}} show work

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Question 710767: how to find the real or imaginary solutions of each equation by factoring
x%5E4-3x%5E2=2x%5E2
show work
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E4-3x%5E2=2x%5E2
Factoring will generally not help solve an equation unless one side of the equation is zero. So We start by subtracting both sides by 2x%5E2:
x%5E4-5x%5E2=0

Now we factor. Always start factoring by factoring out the greatest common factor (GCF). (Note: We rarely (but not never) factor out the GCF if it is 1.) The GCF here is not 1:
x%5E2%28x%5E2-5%29=0
Neither factor will factor further so we are finished factoring.

Now we use the Zero Product Property which tells that that a product can be zero only if one (or more) of the factors is zero. So:
x%5E2+=+0 or x%5E2-5=0
Solving the first equation is fairly simple. The only number you can square and get 0 is 0. We have a couple of ways to solve the second equation:
  • Use the Quadratic Formula on the second equation as it is; or...
  • Add 5 and then find the square root of each side.
I'm going to use the formula because it is easy to forget the two square roots, positive and negative, when you find the square root of each side of an equation. (The formula, with its "plus or minus" (+), handles the two square roots automatically.

Using the Quadratic Formula on the second equation we get:
x+=+%28-%280%29+%2B-+sqrt%28%280%29%5E2-4%281%29%28-5%29%29%29%2F2%281%29
Simplifying...
x+=+%28-%280%29+%2B-+sqrt%280-4%281%29%28-5%29%29%29%2F2%281%29
x+=+%28-%280%29+%2B-+sqrt%280%2B20%29%29%2F2%281%29
x+=+%28-%280%29+%2B-+sqrt%2820%29%29%2F2%281%29
x+=+%280+%2B-+sqrt%2820%29%29%2F2
[Note: I am keeping the zero because Algebra.com's software will not allow the "plus or minus" symbol without something in front of it.]
x+=+%280+%2B-+sqrt%284%2A5%29%29%2F2
x+=+%280+%2B-+sqrt%284%29%2Asqrt%285%29%29%2F2
x+=+%280+%2B-+2%2Asqrt%285%29%29%2F2
which is short for:
x+=+%280+%2B+2%2Asqrt%285%29%29%2F2 or x+=+%280+-+2%2Asqrt%285%29%29%2F2
[Now that the "plus or minus" is gone I don't need the zero anymore.]
x+=+%282%2Asqrt%285%29%29%2F2 or x+=+%28-2%2Asqrt%285%29%29%2F2
which reduces to:
x+=+sqrt%285%29 or x+=+-sqrt%285%29
These, plus the solution of 0 we found earlier, are the solutions to our equation. (Note: Since x was a factor twice in x%5E2%28x%5E2-5%29=0, it counts twice as a root. So the "4" solutions are: 0, 0, sqrt%285%29, or -sqrt%285%29 All four of these are real numbers.