If \(16x^2+9y^2+4z^2=24(x-y+z)-61\), then find the value of \(2(xy+z)\).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4
Moving every term to one side lets the expression be written as a sum of three perfect squares equal to zero. A square is never negative, so each square must be zero, and that fixes \(x\), \(y\) and \(z\).
\(16x^2-24x+9y^2+24y+4z^2-24z+61=0\)
Splitting 61 as \(9+16+36\): \((16x^2-24x+9)+(9y^2+24y+16)+(4z^2-24z+36)=0\)
\((4x-3)^2+(3y+4)^2+(2z-6)^2=0\)
\(4x-3=0\Rightarrow x=\displaystyle\frac{3}{4}\); \(3y+4=0\Rightarrow y=-\displaystyle\frac{4}{3}\); \(2z-6=0\Rightarrow z=3\)
\(xy=\displaystyle\frac{3}{4}\times\left(-\frac{4}{3}\right)=-1\)
\(2(xy+z)=2(-1+3)=4\)