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

Jun 25, 2016

$x = - 3$ 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 $- x$:

$- x \left(3 x + 9\right) = 0$

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

So we know one solution for $x$ must therefore be $0$ (as $- \left(0\right) \cdot x = 0$, $x$ being any number)

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

$3 x + 9 = 0 \to$ subtract 9

$3 x = - 9 \to$ divide by 3

$x = - 3$

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

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:

$a {x}^{2} + b x + c = 0$

$x$ can be found by the equation:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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

$a = - 3 \to$ as $- 3 {x}^{2}$

$b = - 9 \to$ as $- 9 x$

$c = 0 \to$ as not specified

Therefore substituting in:

$x = \frac{- \left(- 9\right) \pm \sqrt{{\left(- 9\right)}^{2} - 4 \left(- 3\right) \left(0\right)}}{2 \left(- 3\right)}$

$x = \frac{9 \pm \sqrt{81}}{- 6}$

$x = \frac{9 \pm 9}{- 6}$

$x = \frac{18}{-} 6 = - 3$ and $x = \frac{0}{-} 6 = 0$

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

### Checking

We can test these solutions by adding them back in:

$- 3 {\left(0\right)}^{2} - 9 \left(0\right) = 0$

$0 - 0 = 0$

and

$- 3 {\left(- 3\right)}^{2} - 9 \left(- 3\right) = 0$

$\left(- 3\right) \left(9\right) + 27 = 0$

$- 27 + 27 = 0$

Therefore we know our answers are right.