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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`.
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Solution
`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.
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