Question

In each of the following, determine whether the given numbers are solutions of the given equation or not: `x^2 - 3sqrt(3)x + 6 = 0; sqrt(3), -2sqrt(3)`

Answer 1

`x^2 - 3sqrt(3)x + 6 = 0; sqrt(3), -2sqrt(3)`
(a) Substituting the value of x = `sqrt(3)`
L.H.S. = `x^2 - 3sqrt(3)x + 6`
= `(sqrt(3))^2 - 3sqrt(3) xx sqrt(3) + 6`
= 3 - 9 + 6
= 0
= R.H.S.
∴ x = `sqrt(3)` is its solution.
(b) x = `2sqrt(3)`
Substituting x = `2sqrt(3)`
L.H.S. = `x^2 -3sqrt(3)x + 6`
= `(-2sqrt(3))^2 - 3sqrt(3) (-2sqrt(3)) + 6`
= 12 + 18 + 6
= 36 ≠ 0
∵ x = `-2sqrt(3)` is not its solution.