Question

In each of the following, determine whether the given values are solution of the given equation or not:
`a^2x^2 - 3abx + 2b^2 = 0; x = a/b, x = b/a`.

Answer 1

`a^2x^2 - 3abx + 2b^2 = 0; x = a/b, x = b/a`.
Now on substituting x = `a/b` in L.H.S.
L.H.S. = a²x² − 3abx + 2b²
= `a^2 xx (a/b)^2 - 3ab xx a/b + 2b^2`
= `a^4/b^2 - 3a^2 + 2b^2`
= `(a^4 - 3a^2b^2 + 2b^2)/b^2`
= a⁴ − 3a²b² + 2b⁴ ≠ 0 ≠ R.H.S.
∴ x = `a/b` is not a solution of the equation
Put x = `b/a` in L.H.S. of given equation
L.H.S. = `a^2 xx (b/a)^2 - 3ab xx b/a + 2b^2`
= b² − 3b² + 2b²
= 3b² − 3b² 
= 0
= R.H.S.
∴ x = `b/a` is a solution of the given equation.