# How do you solve #sqrt (x+3)= x-3# and find any extraneous solutions?

##### 1 Answer

#### Answer:

#### Explanation:

You're dealing with a *square root* here, so right from the start you know two things

#x+3 >=0 implies x >= -3# When working with

real numbers, i.e.#x in RR# , the square root can only be taken frompositive numbers. This of course implies that#x+3# must be positive or at best equal to zero.

#x - 3 >=0 implies x >= 3# Similarly, when working in

#RR# , the square root of a number isalwayspositive. This implies that#x-3# must be positive or at best equal to zero.

Combine these two restrictions to get

#color(purple)(|bar(ul(color(white)(a/a)color(black)(x >= 3)color(white)(a/a)|))) -># any value of#x# that acts as solution to the original equationmustsatisfy this condition

Now, your goal here is to get rid of the square root. You can do that by *squaring* both sides of the equation

#(sqrt(x+3))^2 = (x-3)^2#

This will get you

#x + 3 = x^2 - 6x + 9#

Rearrange to quadratic equation form

#x^2 - 7x + 6 = 0#

Now, two numbers that add up to give

#(x-1)(x-6) = 0 implies {(x_1 = 1),(x_2 = 6) :}#

Now, notice that the first possible solution **does not** satisfy the required condition

#x = 1 color(red)(cancel(color(black)(>))) =3 -># this means that#x=1# is anextraneous solution

As a result, the only valid solution to the *original equation* is

#x=6: " "sqrt(6+3) = 6 -3 <=> 3 = 3 " "color(green)(sqrt())#

As you can see,

#x = 1: " " sqrt(1 + 3) = 1 - 3 <=> 2 != -2" "color(red)(xx)#