How do you solve the equation and identify any extraneous solutions for #sqrt(3x-8) = 2#?

1 Answer
Jun 26, 2015

Answer:

#x=4#

Explanation:

Obviously, we are going raise both sides of this equation into the second degree. This is not an invariant (preserving the same solution) transformation. So, we have to be careful and make sure we will not come up with extraneous solutions.

The first line of defense against extraneous solutions is to identify a domain of possible solutions. Since we have a square root in this equation, the expression under it must be non-negative.
So, the set of values where a solution might be from is determined by an inequality:
#3x-8 >= 0#
or, equivalently,
#x >= 8/3#

That being taken care of, we can square both sides of this equation getting
#3x-8=4#
or, solving this for #x#,
#x=4#

This value satisfies the condition #x >= 8/3#.
Now let's perform the checking:
#sqrt(3*4-8)= sqrt(4)=2#
Check!

So, the only soluton we've found is the real solution of an original equation.