Question

`2x^2+6sqrt3x-60=0` 

 

Answer 1

The given equation is  `2x^2+6sqrt3x-60=0`  

Comparing it with `ax^2+bx+c=0` 

`a=2, b=6sqrt3 and c=-60` 

∴ Discriminant, `D=b^2-4ac=(6sqrt3)^2-4xx2xx(-60)=180+480=588>0`  

So, the given equation has real roots.
Now, `sqrt(D)=sqrt588=14sqrt3` 

∴`α =(-b+sqrt(D))/(2a)=(-6sqrt(3)+14sqrt(3))/(2xx2)=(8sqrt(3))/4=2sqrt(3)` 

β=`(-b+sqrt(D))/(2a)=(-6sqrt(3)+14sqrt(3))/(2xx2)=(-20sqrt(3))/4=-5sqrt(3)`

 Hence, `2sqrt(3)` and `-5sqrt(3)` are the root of the given equation.