Question

Find whether the value x = `(1)/(a^2)` and x = `(1)/(b^2)` are the solution of the equation:
a²b²x² - (a² + b²) x + 1 = 0, a ≠ 0, b ≠ 0.

Answer 1

a²b²x² - (a² + b²) x + 1 = 0; x = `(1)/(a^2), x = (1)/(b^2)`

By putting x = `(1)/(a^2)` in L.H.S. of equation

L.H.S. = `a^2b^2 xx (1/a^2)^2 - (a^2 + b^2) xx (1)/(a^2) + 1` 

= `b^2/a^2 - 1 - b^2/a^2 + 1` = 0 = R.H.S.

By Putting x = `(1)/b^2`, in L.H.S. of equation

L.H.S. = `a^2b^2 xx (1/b^2)^2 - (a^2 + b^2) xx (1)/(b^2) + 1` 

= `a^2/b^2 - a^2/b^2 - 1 + 1` = 0 = R..H.S.
Hence, x = `(1)/a^2,(1)/b^2` are the solution of the equation.