Question

Solve for $a,b$:

$a^2+b^2-4a+16b+68=0$.

Answer 1

Now,

$a^2+b^2-4a+16b+68=0$

$(a^2-4a+4)+(b^2+16b+64)=0$

$(a-2{)}^2+(b+8{)}^2=0$

Now, sum of two positive numbers is zero unless they are identically zero individually.

so, (a-2) = 0 & (b+8) = 0

a = 2 & b = -8.