#### Question

In the following, determine whether the given quadratic equation have real roots and if so, find the roots:

`2x^2-2sqrt6x+3=0`

#### Solution

We have been given, `2x^2-2sqrt6x+3=0`

Now we also know that for an equation ax^{2} + bx + c = 0, the discriminant is given by the following equation:

D = b^{2} - 4ac

Now, according to the equation given to us, we have,a = 2, `b=-2sqrt6` and c = 3.

Therefore, the discriminant is given as,

`D=(-2sqrt6)^2-4(2)(3)`

= 24 - 24

= 0

Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real and equal roots.

Now, the roots of an equation is given by the following equation,

`x=(-b+-sqrtD)/(2a)`

Therefore, the roots of the equation are given as follows,

`x=(-(2sqrt6)+-sqrt0)/(2(2))`

`=(-sqrt6+-0)/2`

`=-sqrt(3/2)`

Therefore, the roots of the equation are real and equal and its value is `-sqrt(3/2)`.