what are all ordered pairs $(a,b)$ for which $17a^2-10ab+2b^2-4a+2=0$

Question

what are all ordered pairs $(a,b)$ for which $17a^2-10ab+2b^2-4a+2=0$

I am having a hard time solving this problem.

I started by plugging $a=0$ and $a=b=1$ but that doesn't help. So the answer is not so obvious meaning there is a strategy required however I just dont see it.

I can't factor anything since there is no common factors and that $17$ is being annoying as well.

Any ideas on how to proceed with this problem?

Answer 1

If we write $$17a^2-a(5b+2)+2(b^2+1)=0$$ the quadratic formula still applies.

Therefore

$$b=\frac{10\pm \sqrt{100a^2-8(17a^2-4a+2)}}{4}$$

$$=\frac{1}{2}\left( 5a\pm\sqrt{-9a^2+8a-4} \right)$$

However, $-9a^2+8a-4<0$ so there are no real solutions, but infinitely many complex ones.