Question

Find the roots of the following equation, if they exist, by applying the quadratic formula:

3a²x² + 8abx + 4b² = 0, a ≠ 0

Answer 1

Given: 

3a²x² + 8abx + 4b² = 0 

On comparing it with Ax² + Bx + C = 0, we get: 

A = 3a², B = 8ab and C = 4b² 

Discriminant D is given by: 

D = (B² – 4AC) 

= (8ab)² – 4 × 3a² × 4b² 

= 16a²b² > 0 

Hence, the roots of the equation are real.

Roots α and β are given by:  

`α = (-b + sqrt(D))/(2a)`

= `(-8ab + sqrt(16a^2b^2))/(2 xx 3a^2)`

= `(-8ab + 4ab)/(6a^2)`

= `(-4ab)/(6a^2)`

= `(-2b)/(3a) ` 

`β = (-b - sqrt(D))/(2a)`

= `(-8ab - sqrt(16a^2b^2))/(2 xx 3a^2)`

= `(-8ab - 4ab)/(6a^2)`

= `(-12ab)/(6a^2)`

= `(-2b)/(a)` 

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