Question

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

3a²x² + 8abx + 4b² = 0, a ≠ 0

Answer 1

We have been given, 3a²x² + 8abx + 4b² = 0

Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:

D = b² - 4ac

Now, according to the equation given to us, we have,a = 3a², b = 8ab and c = 4b².

Therefore, the discriminant is given as,

D = (8ab)² - 4(3a²)(4b²)

= 64a²b² - 48a²b²

= 16a²b²

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<sup>2</sup>b<sup>2</sup>)/(2(3a<sup>2</sup>))`

`=(-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`.