Question

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

`sqrt3x^2+10x-8sqrt3=0`

Answer 1

We have been given, `sqrt3x^2+10x-8sqrt3=0`

Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:

D = b² - 4ac

Now, according to the equation given to us, we have,`a=sqrt3`, b = 10 and `c=-8sqrt3`.

Therefore, the discriminant is given as,

`D=(10)^2-4(sqrt3)(-8sqrt3)`

= 100 + 96

= 196

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 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=(-(10)+-sqrt196)/(2(sqrt3))`

`=(-10+-14)/(2sqrt3)`

`=(-5+-7)/sqrt3`

Now we solve both cases for the two values of x. So, we have,

`x=(-5+7)/sqrt3`

`=2/sqrt3`

Also,

`x=(-5-7)/sqrt3`

`=-4sqrt3`

Therefore, the roots of the equation are `2/sqrt3` and `-4sqrt3`.