If \(\left(x-\displaystyle\frac{1}{3}\right)^2+(y-4)^2=0\), then find the value of \(\displaystyle\frac{y+x}{y-x}\).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{13}{11}\)
A square of a real number is never negative, so two squares add up to zero only when each square is zero. This fixes both \(x\) and \(y\).
\(x-\displaystyle\frac{1}{3}=0\), so \(x=\displaystyle\frac{1}{3}\)
\(y-4=0\), so \(y=4\)
\(y+x=4+\displaystyle\frac{1}{3}=\frac{13}{3}\) and \(y-x=4-\displaystyle\frac{1}{3}=\frac{11}{3}\)
\(\displaystyle\frac{y+x}{y-x}=\frac{13}{3}\div\frac{11}{3}=\frac{13}{11}\)