SOLUTION: I am trying to solve (x-3)^2+(x+2)^2=17. So I get (x-3)(x-3)+ (x+2)(x+2)=17 I tried the FOIL method here to get x^2-3x-3x+9=17 and x^2+2x+2x+4=17. For the first equation I did thi

Question 208164This question is from textbook Elementary and intermediate algebra
: I am trying to solve (x-3)^2+(x+2)^2=17. So I get (x-3)(x-3)+ (x+2)(x+2)=17
I tried the FOIL method here to get x^2-3x-3x+9=17 and x^2+2x+2x+4=17. For the first equation I did this: x^2-6x+9-17=17-17, which is x^2-6x-8 which I cannot factor unless I make -6x and -8 into positive integers. For the second equation I did this: x^2+4x+4-17=17-17, which is x^2+4x-13 which I cannot factor. Please tell me what I am doing wrong. This question is from textbook Elementary and intermediate algebra

Found 2 solutions by Alan3354, rapaljer:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
(x-3)^2+(x+2)^2=17
---------
x^2-6x+9 + x^2+4x+4 = 17 You have to add them up, not take the 1 at a time.
2x^2-2x+13 = 17
2x^2-2x+4 = 0
x^2-x+2 = 0
(x-2)*(x+1) = 0
x = 2
x = -1

Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
Don't make two equations out of this one.

(x-3)^2+(x+2)^2=17
x^2 -6x+9 + x^2 +4x +4 = 17
2x^2 -2x + 13 = 17

Set the equation equal to zero:
2x^2 -2x -4=0

Factor the common factor of 2:
2(x^2 -x-2)=0
2(x-2)(x+1)=0

Therefore there are two solutions:
(x-2)=0
x=2

(x+1)=0
x=-1

If you need additional help with either factoring or solving quadratic equations, please see my website by clicking on my tutor name "Rapaljer" anywhere in algebra.com. Look for the link on my homepage "Basic, Intermediate and College Algebra: One Step at a Time." Look for "Basic Algebra", then "Chapter 2" and select the topic that you need from the list of sections that appears. Also, see my page "MATH IN LIVING COLOR" and also my new page of videos on Factoring. From my homepage look for "Rapalje Videos".

R^2

Dr. Robert J. Rapalje
Seminole Community College
Altamonte Springs, Florida
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