Question

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

Answer 1

This equation doesn't have real solutions.

Explanation:

I would first rearrange it as:

`sqrt(x^2+9)=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=+-sqrt(-8)`
`x=+-sqrt(-1*8)=sqrt(8)*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)color(red)(i)=+2sqrt(2)color(red)(i)`
`x_2=-sqrt(8)color(red)(i)=-2sqrt(2)color(red)(i)`

Now you can substitute back these two solutions into your original equation and they work (considering that `(sqrt(-1))^2=-1`)!