Question

Let $x$ and $y$ be real numbers satisfying the inequality $5x^2+y^2-4xy+24\le 10x-1.$ Find the value of $x^2+y^2-29.$
(correct answer + 2, wrong answer 0)

Answer 1

$5x^2+y^2-4xy+24\le 10x-1\Rightarrow (x^2-10x+25)+(4x^2+y^2-4xy)\le 0\Rightarrow (x-5{)}^2+(2x-y{)}^2\le 0$

Hence, we must have
$x-5=0$ and $2x-y=0$
$\Rightarrow x=5,y=10$
$\therefore x^2+y^2-29=96$