SOLUTION: I have the solutions to this problem from the back of my book. I just don't know how to get the answer. This problem is also listed on my testguide, for a test I'm taking tonight

Question 101557This question is from textbook Blitzer Algebra and Trigonometry
: I have the solutions to this problem from the back of my book. I just don't know how to get the answer. This problem is also listed on my testguide, for a test I'm taking tonight. Please help!!
(x^2-x)^2 - 14x(x^2-x) + 24 = 0
I've tried everything I can think of. I know it's not a quadratic equation. I've tried factoring...but in the end I end up with 5 terms, but like I said I'm not even sure if what I did up to that point is right.
I squared what is in the 1st set of parentheses first then performed the multiplication of the next set of parentheses, then combined like terms. Beyond that I just don't know what to do. This question is from textbook Blitzer Algebra and Trigonometry

Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
(x^2-x)^2 - 14x(x^2-x) + 24 = 0
x^4-2x^3+x^2-14x^3+14x^2+24=0
x^4-16x^3+15x^2+24=0
I was able to graph it at about x=1.66, x=14.99
Ed

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