Question

How do you solve `-3x^2 - 9x = 0`?

Answer 1

`x = -3` and `x = 0`

Explanation:

There are two ways to do this:

First Way: Factoring

Now this one is quite nice as we can just remove a common factor from both of them... they are both divisible by `-x`:

`-x(3x + 9) = 0`

Now we need to make this equation `0`, therefore `-x` or `3x +9` must equal `0` as any number multiplied by `0 = 0`...

So we know one solution for `x` must therefore be `0` (as `-(0) * x = 0`, `x` being any number)

we also know that `3x + 9 = 0`, so we can solve this:

`3x + 9 = 0 ->` subtract 9

`3x = -9 ->` divide by 3

`x = -3`

So the two solutions for `x` are `0` and `-3`

Second Way: Quadratic Formula

This way is a bit harder and provides unneccesary difficulty for this question, though shows how the formula can be applied on any quadratic equation:

We know for any equation in the form:

`ax^2 + bx + c = 0`

`x` can be found by the equation:

`x=(-b+- sqrt(b^2 -4ac))/(2a)`

Now in the equation it is less obvious to see as there is no c value but:

`a = -3 ->` as `-3x^2`

`b = -9 ->` as `-9x`

`c = 0 ->` as not specified

Therefore substituting in:

`x=(-(-9)+- sqrt((-9)^2 -4(-3)(0)))/(2(-3))`

`x=(9+- sqrt(81))/(-6)`

`x =(9+-9)/(-6)`

`x=(18)/-6 = -3` and `x=(0)/-6=0`

So the two solutions for `x` are `0` and `-3`

Checking

We can test these solutions by adding them back in:

`-3(0)^2 - 9(0) = 0`

`0 - 0 = 0` ✓

and

`-3(-3)^2 - 9(-3) = 0`

`(-3)(9) + 27 = 0`

`-27 + 27 = 0` ✓

Therefore we know our answers are right.