I've been trying to figure out this question but i dont know where im going wrong when i plug back in the solutions i get, i get a bunch of wacky numbers...

question : x^2 + (x + 2) ^2 = 290

my work:
= x^2 + (x +2) + (x + 2) = 290
= 2x^2 + 4x + 4 = 290
= 2x + 4x - 286 = 0
= 2 (x^2 + 2x - 143) = 0
= (x - 11) ( x + 13)
x = 11
x = -13 .. why is this wrong when i check?

2. Well, the first thing you have done wrong is (x+2)^2 does not equal to (x+2)+(x+2), it's (x+2)*(x+2)
so x^2+(x+2)*(x+2)=290
x^2+x^2+4x+4=290
2x^2+4x+4=290
2x^2+4x-286=0
now, you can either solve this by the quadradic formula, or factoring
QF: x=[-4±√(16+2288)]/4
x=(-4±48)/4
x=11 or -13
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if you did x^2 + (x +2) + (x + 2) = 290, it should be x^2+2x+4, how did you get 2x^2 + 4x + 4 = 290?
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Check: plug in 11
121+169=290
169+121=290

3. $\displaystyle 121 + (13)^2 = 290$

$\displaystyle 121+ 169 = 290$

$\displaystyle 290 = 290$

NOthing wrong with it...

$\displaystyle 169 + 121 = 290$

4. Originally Posted by jgv115
$\displaystyle 121 + (13)^2 = 290$

$\displaystyle 121+ 169 = 290$

$\displaystyle 290 = 290$

NOthing wrong with it...

$\displaystyle 169 + 121 = 290$