Question

How do you solve `x ^ { 2} + \sqrt { 3} x - 6= 0`?

Answer 1

`x=-2sqrt3` or `sqrt3`

Explanation:

To solve `x^2+sqrt3x-6=0`, there could be two ways, one, by factorizing the quadratic polynomial and two, using completing square method.

For factorizing we should split te middle term `sqrt3` in two parts whose product is `-6`. As `sqrt3`, would be involved these are `2sqrt3` and `-sqrt3`, and hence above becomes

`x^2+2sqrt3x-sqrt3x-6=0`

or `x(x+2sqrt3)-sqrt3(x+2sqrt3)=0`

i.e. `(x-sqrt3)(x+2sqrt3)=0`

i.e. `x=sqrt3` or `-2sqrt3`

Other way could be

`x^2+2xxsqrt3/2xx x-6=0`

or `(x^2+2xxsqrt3/2xx x+(sqrt3/2)^2)-(sqrt3/2)^2-6=0`

or `(x+sqrt3/2)^2-27/4=0`

or `(x+sqrt3/2)^2-((3sqrt3)/2)^2=0`

or `(x+sqrt3/2+(3sqrt3)/2)(x+sqrt3/2-(3sqrt3)/2)=0`

or `(x+2sqrt3)(x-sqrt3)=0`

i.e. `x=-2sqrt3` or `sqrt3`