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

1 Answer
Aug 10, 2015

Answer:

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#)!