Solve: (x-3)^2 + (x+2)^2 = 17
A ashmarissa New member Joined Sep 12, 2012 Messages 6 Sep 12, 2012 #2 Yea but I keep ending up with what I start with. I guess I don't know how to make that 17 fit into the problem.
Yea but I keep ending up with what I start with. I guess I don't know how to make that 17 fit into the problem.
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,582 Sep 12, 2012 #3 ashmarissa said: Yea but I keep ending up with what I start with. I guess I don't know how to make that 17 fit into the problem. Click to expand... How are you multiplying out, combining like terms, moving the 17 over, simplifying, and then ending up with what you'd started with? :shock: Please reply showing your work, so the volunteers can try to find where things are going awry. Thank you! :wink:
ashmarissa said: Yea but I keep ending up with what I start with. I guess I don't know how to make that 17 fit into the problem. Click to expand... How are you multiplying out, combining like terms, moving the 17 over, simplifying, and then ending up with what you'd started with? :shock: Please reply showing your work, so the volunteers can try to find where things are going awry. Thank you! :wink:
A ashmarissa New member Joined Sep 12, 2012 Messages 6 Sep 12, 2012 #4 (X-3)^2 + (x+2)^2=17 (X-3)(x-3) + (x+2)(x+2) -17=0 (X^2-3x-3x+9) + (x^2+2x+2x+4) -17 (X^2-6x+9) + (x^2+4x+4) -17 That is as far as I have gotten.
(X-3)^2 + (x+2)^2=17 (X-3)(x-3) + (x+2)(x+2) -17=0 (X^2-3x-3x+9) + (x^2+2x+2x+4) -17 (X^2-6x+9) + (x^2+4x+4) -17 That is as far as I have gotten.