We have given,
\(\left(\frac{2x}{x - 5}\right)^2 + 5 \left(\frac{2x}{x - 5}\right) - 24 = 0\)
Let \(\frac{2x}{(x - 5)}\) be y
\(\therefore y^2 + 5y - 24 = 0\)
Now factorise,
\(y^2 + 8y - 3y - 24 =0\)
\(y(y + 8) - 3(y + 8) = 0\)
\((y + 8) (y - 3) =0\)
\(y = 3, -8\)
Putting y = 3
\(\frac {2x}{x - 5} = 3\)
\(2x = 3x - 15\)
\(x = 15\)
Putting y = -8
\(\frac{2x}{x - 5} = -8\)
\(2x = -8x + 40\)
\(10x = 40\)
\(x =4\)
Hence, x is 15, 4.
