SOLUTION: This is the problem I am trying to solve: (x-2)^2 + x^2=10. I have tried every rule and method I can think of. I know that (x-2)^2 equals x^2 -4x +4, so then would it be x^4 -4x +4

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: This is the problem I am trying to solve: (x-2)^2 + x^2=10. I have tried every rule and method I can think of. I know that (x-2)^2 equals x^2 -4x +4, so then would it be x^4 -4x +4      Log On


   

Question 142538This question is from textbook University of Phoenix
: This is the problem I am trying to solve: (x-2)^2 + x^2=10. I have tried every rule and method I can think of. I know that (x-2)^2 equals x^2 -4x +4, so then would it be x^4 -4x +4?? I'm not even sure if this pertains to this problem. I am at a loss. I really appreciate any help you can offer.. Thanks in advance :0) This question is from textbook University of Phoenix

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-2%29%5E2+%2B+x%5E2=10 Start with the given equation


x%5E2-4x%2B4+%2B+x%5E2=10 Foil


x%5E2-4x%2B4+%2B+x%5E2-10=0 Subtract 10 from both sides


2x%5E2-4x-6=0 Combine like terms. This is where you made an error: x%5E2 plus x%5E2 is not x%5E4



2%28x-3%29%28x%2B1%29=0 Factor the left side (note: if you need help with factoring, check out this solver)



Now set each factor equal to zero:
x-3=0 or x%2B1=0

x=3 or x=-1 Now solve for x in each case


So our answer is
x=3 or x=-1


Notice if we graph y=2x%5E2-4x-6 we can see that the roots are x=3 and x=-1 . So this visually verifies our answer.


+graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C0%2C+2x%5E2-4x-6%29+