Expand $(x-7)^2 + x^2 = (x+2)^2$ algebraically

Question

$(x-7)^2 + x^2 = (x+2)^2$

$(x-7)(x-7) + x^2 = (x+2)(x+2)$

$x^2 -7x -7x + 49 + x^2 = x^2 + 4x + 4$

$x^2 + 18x - 45 = x^2 + x^2$

From that point on, everything I do is incorrect. I don't know what to do with the three $x^2$.

Answer 1

You don't need to expand the squares. Use $a^2-b^2 = (a+b)(a-b)$ to get $$ x^2=(x+2)^2-(x-7)^2=(x+2+x-7)(x+2-x+7)=9(2x-5)=18x-45 $$

Answer 2

Expand the lhs and you get $lhs=2x^2-14x+49$. Expand the rhs and get $rhs=x^2+4x+4$.So $$lha-rhs=(2x^2-14x+49)-(x^2+4x+4)=x^2-18x+45$$ What you wrote is correct but you did not finish. You ended with $$x^2 + 18x - 45 = x^2 + x^2$$ so substract the last lhs from the last rhs of your post and you are done.