# How do you solve and check for extraneous solutions in 3sqrt(x^2+9)=3?

Aug 10, 2015

This equation doesn't have real solutions.

#### Explanation:

I would first rearrange it as:

$\sqrt{{x}^{2} + 9} = \frac{3}{3}$
$\sqrt{{x}^{2} + 9} = 1$
square both sides:
${x}^{2} + 9 = 1$
${x}^{2} = - 9 + 1$
${x}^{2} = - 8$

At this point you have a problem: there is no real number that can satisfy this equation: no real number squared can give you a negative value!
On the other hand you could use a different type of number although I am not sure you know it.

You can square root both sides and write your equation as:
$x = \pm \sqrt{- 8}$
$x = \pm \sqrt{- 1 \cdot 8} = \sqrt{8} \cdot \sqrt{- 1} =$
You can set $\sqrt{- 1} = i$ which is your "new" number called Imaginary Unit!!!
So the solutions for your original equation will be:
${x}_{1} = + \sqrt{8} \textcolor{red}{i} = + 2 \sqrt{2} \textcolor{red}{i}$
${x}_{2} = - \sqrt{8} \textcolor{red}{i} = - 2 \sqrt{2} \textcolor{red}{i}$

Now you can substitute back these two solutions into your original equation and they work (considering that ${\left(\sqrt{- 1}\right)}^{2} = - 1$)!