Question

`3a^2x^2+8abx+4b^2=0`

Answer 1

Given: 

`3a^2x^2+8abx+4b^2=0` 

On comparing it with` Ax^2+Bx+C=0` we get: 

`Aa=3a^2, B=8ab and C=4b^2` 

Discriminant D is given by: 

`D=(B^2-4AC) ` 

=`(8ab)^2-4xx3a^2xx4b^2` 

=` 16a^2b^2>0` 

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

`α =(-b+sqrt(D))/(2a)=(-8ab+sqrt16a^2b^2)/(2xx3a^2)=(-8ab+4ab)/(6a^2)=(-4ab)/(6a^2)=(-2b)/(3a) ` 

`β=(-b-sqrt(D))/(2a)=(-8ab- sqrt16a^2b^2)/(2xx3a^2)=(-8ab-4ab)/(6a^2)=(-12ab)/(6a^2)=(-2b)/(a)` 

Thus, the roots of the equation are `(-2b)/(3a) and  (-2b)/a`