If \(a^2+b^2=4b+6a-13\), then find the value of \((a+b)^2-10\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 15
Moving every term to one side and completing the squares gives a sum of two squares equal to zero. A square is never negative, so each square must be zero.
\(a^2-6a+b^2-4b+13=0\)
\(13=9+4\), so \(\left(a^2-6a+9\right)+\left(b^2-4b+4\right)=0\)
\((a-3)^2+(b-2)^2=0\), so \(a=3\) and \(b=2\)
\((a+b)^2-10=5^2-10=25-10=15\)