Question

In each of the following, determine whether the given values are solution of the given equation or not:
x² - 3`sqrt(3)` + 6 = 0; x = `sqrt(3)`, x = -2`sqrt(3)`

Answer 1

x² - 3`sqrt(3)` + 6; x = `sqrt(3)`, x = -2`sqrt(3)`.
Now substitute x = `sqrt(3)` in L.H.S. of given equation.
L.H.S. = `(sqrt(3))^2 -3sqrt(3) xx sqrt(3) + 6 = 0`
= 3 - 9 + 6
= 0
= R.H.S.
x = `sqrt(3)` is a solution of the given equation.
Substitute x = `-2sqrt(3)` in L.H.S. of given equation
⇒ `(-2sqrt(3))^2 -3sqrt(3) xx -2sqrt(3) + 6 = 0`
⇒ L.H.S. = 12 + 18 + 6 ≠ 0 ≠ R.H.S.
x = `-2sqrt(3)` is not a solution of the given equation.