In the following, determine whether the given values are solutions of the given equation or not:
`x^2 - 3sqrt3x+6=0`, `x=sqrt3`, `x=-2sqrt3`
Solution
We have been given that,
`x^2 - 3sqrt3x+6=0`, `x=sqrt3`, `x=-2sqrt3`
Now if `x=sqrt3` is a solution of the equation then it should satisfy the equation.
So, substituting `x=sqrt3` in the equation, we get
`x^2 - 3sqrt3x+6`
`=(sqrt3)^2-3sqrt3(sqrt3)+6`
= 3 - 9 + 6
= 0
Hence `x=sqrt3` is a solution of the quadratic equation.
Also, if `x=-2sqrt3` is a solution of the equation then it should satisfy the equation
So, substituting `x=-2sqrt3` in the equation, we get
`x^2 - 3sqrt3x+6`
`=(-2sqrt3)^2-3sqrt3(-2sqrt3)+6`
= 12 - 18 + 6
= 0
Hence `x=-2sqrt3` is a solution of the quadratic equation.
Therefore, from the above results we find out that `x=sqrt3` and `x=-2sqrt3` are the solutions of the given quadratic equation.
