Question

`sqrt3x^2+10x-8sqrt3=0` 

Answer 1

Given: 

`sqrt3x^2+10x-8sqrt3=0` 

On comparing it with `ax^2+bx+x=0` we get; 

`a=sqrt3,b=10  and c=-8sqrt3` 

Discriminant D is given by: 

`D=(b^2-4ac)` 

=`(10)^2-4xxsqrt3xx(-8sqrt3)` 

=`100+96` 

=`196>0` 

Hence, the roots of the equation are real.
Roots α  and β are given by:  

`α=(-b+sqrt(D))/(2a)=(-10+sqrt(196))/(2sqrt(3))=(-10+14)/(2sqrt(3))=4/(2sqrt3)=2/sqrt(3)=2/sqrt(3)xxsqrt3/sqrt3=(2sqrt(3))/3` 

β=`(-b+sqrt(D))/(2a)=(-10+sqrt(196))/(2sqrt(3))=(-10+14)/(2sqrt(3))=(-24)/(2sqrt(3))=(-12)/(2sqrt(3))(-12)/sqrt(3)xxsqrt3/sqrt3=(-12sqrt(3))/3=4sqrt(3)` 

Thus, the roots of the equation are `(2sqrt3)/3` and `-4sqrt3`