equation is: $(x + y)^2 = (x + 3)(y − 3)$
I'm not asking for a solution, but an approach. How do I prove this kind of question? I have tried to arrange it so that it
is $x + y$ = ....
But I still get nothing, nothing intuitive at least. What is a way to tackle this problem?
Hint:
By expanding we have \begin{align*} (x+y)^2&=(x+3)(y-3)\\ \iff x^2+2xy+y^2&=xy-3x+3y-9\\ \iff x^2+xy+y^2+3x-3y+9&=0\\ \iff \tfrac12(x+y)^2+\tfrac12(x+3)^2+\tfrac12(y-3)^2&=0 \end{align*} So we must have $$x+y=x+3=y-3=0$$
Use the inequality: $ab \le \dfrac{(a+b)^2}{4}$ with $a = x+3, b = y-3$ you have: $(x+y)^2 \le \dfrac{(x+y)^2}{4}\implies (x+y)^2 \le 0 \implies x+y = ...$
Hint by dxiv:
Set $u: = x +3; v: = y- 3$; then
$\star)$ $u^2 + uv + v^2 = 0$.
Consider the square:
$(u +v)^2 = u^2 + 2uv +v^2 \ge 0$.
Subtract $\star$ :
$\Rightarrow$ $uv \ge 0$.
Hence $\star$ implies :
$v= u = 0$, I.e. $x +3 = y - 3 = 0$.
