If \(x^8+25=\displaystyle\frac{x^4}{y^2}\left(24y^2-y^4-49\right)\), then find the value of \(\displaystyle\frac{2x^4-y^2}{x^4y^2}\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{3}{35}\)
Dividing by \(x^4\) and moving every term to one side gives a sum of two squares equal to zero. Each square must then be zero, which fixes \(x^4\) and \(y^2\).
\(x\neq0\) (otherwise the left side is 25 and the right side is 0), so divide by \(x^4\): \(x^4+\displaystyle\frac{25}{x^4}=24-y^2-\frac{49}{y^2}\)
So \(\left(x^4+\displaystyle\frac{25}{x^4}-10\right)+\left(y^2+\frac{49}{y^2}-14\right)=0\)
That is \(\left(x^2-\displaystyle\frac{5}{x^2}\right)^2+\left(y-\frac{7}{y}\right)^2=0\)
Both squares are zero: \(x^4=5\) and \(y^2=7\)
\(\displaystyle\frac{2x^4-y^2}{x^4y^2}=\frac{2\times5-7}{5\times7}=\frac{3}{35}\)