If \(a^2+b^2+10a-8b+41=0\), then find the value of \(a^2-b^2\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9
Completing the squares turns the equation into a sum of two squares equal to zero. A square is never negative, so each square must be zero.
\(41=25+16\), so \(\left(a^2+10a+25\right)+\left(b^2-8b+16\right)=0\)
\((a+5)^2+(b-4)^2=0\), so \(a=-5\) and \(b=4\)
\(a^2-b^2=25-16=9\)