Question

What are the possible values of x in 6x2 + 432 = 0?

Answer 1

`6x^2 + 432 = 0\\ x^2+72=0\\ x^2=-72`

No real solutions.

Complex solutions

`x=-√(-72) \vee x=√(-72)\\ x=-6\sqrt 2 i \vee x=6\sqrt 2 i`

Answer 2

The possible values of x are:

`x=6√(2)i\ and\ x=-6√(2)i`

Explanation:

We are given a quadratic equation in terms of variable ''x'' as:

`6x^2+432=0`

This equation could also be written as:

`6(x^2+72)`

Since we take out 6 common both the terms as both the terms are a multiple of ''6''.

Hence, we get:

`x^2+72=0\\\\i.e.\\\\x^2=-72\\\\\\i.e.\\\\\\x=\pm √(-72)\\\\\\i.e.\\\\\\x=\pm 6√(2)i`

Hence, the possible values of x are:

`x=6√(2)i\ and\ x=-6√(2)i`