#### Question

In the following, determine whether the given quadratic equation have real roots and if so, find the roots:

3a^{2}x^{2} + 8abx + 4b^{2} = 0, a ≠ 0

#### Solution

We have been given, 3a^{2}x^{2} + 8abx + 4b^{2} = 0

Now we also know that for an equation ax^{2} + bx + c = 0, the discriminant is given by the following equation:

D = b^{2} - 4ac

Now, according to the equation given to us, we have,a = 3a^{2}, b = 8ab and c = 4b^{2}.

Therefore, the discriminant is given as,

D = (8ab)^{2} - 4(3a^{2})(4b^{2})

= 64a^{2}b^{2} - 48a^{2}b^{2}

= 16a^{2}b^{2}

Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

`x=(-b+-sqrtD)/(2a)`

Therefore, the roots of the equation are given as follows,

`x=(-(8ab)+-sqrt16a^{2}b^{2})/(2(3a^{2}))`

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

`=(-4b+-2b)/(3a)`

Now we solve both cases for the two values of x. So, we have,

`x=(-4b+2b)/(3a)`

`=-(2b)/(3a)`

Also,

`x=(-4b-2b)/(3a)`

`=(-2b)/a`

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