# How do you solve 9r^2-3=-152?

Mar 29, 2017

No real root

But… $r = \pm \frac{i \sqrt{149}}{3}$

#### Explanation:

Add $3$ to both sides: $9 {r}^{2} = - 149$. Now divide both sides by $9$: ${r}^{2} = - \frac{149}{9}$.

Now, take the square root: $r = \pm \sqrt{- \frac{149}{9}} = \pm \frac{\sqrt{- 149}}{\sqrt{9}} = \pm \frac{\sqrt{- 149}}{3} = \setminus \cdots$

Wait… What is the square root of a negative number? It can't be negative since the square of a negative is positive. Thus, there are no real roots to this equation!

Well… We can define that the square root of negative one to be some number. Mathematicians call it imaginary and symbolize it with $i$. Thus, we write $\pm \frac{\sqrt{- 149}}{3} = \pm \frac{\sqrt{- 1 \cdot 149}}{3} = \pm \frac{\sqrt{- 1} \sqrt{149}}{3}$. Remember that we defined the square root of $- 1$ to be $i$. Thus, the answer is $\pm \frac{i \sqrt{149}}{3}$.