How do you solve #sqrt( 3x+1) - sqrt(x-1) = 2#?

1 Answer
Oct 13, 2015

Answer:

#x = 1#

Explanation:

You're dealing with radical terms, which means that you need to make sure that the expressions that are unde the square root are always positive.

This means that you need

#3x + 1 >= 0 implies x >= -1/3#

#x - 1 >= 0 implies x >= 1#

Combining these two conditions will get you #x >= 1#. This means that any value of #x# that is smaller than #1# will cannot be a solution to the equation.

With this in mind, start by squaring both sides of the equation

#(sqrt(3x+1) - sqrt(x-1))^2 = 2^2#

#(sqrt(3x+1))^2 - 2sqrt((3x+1)(x-1)) - (sqrt(x-1))^2 = 4#

#3x + 1 - 2sqrt((3x+1)(x-1)) - x +1 = 4#

#2x - 2sqrt((3x+1)(x-1)) + 2 = 4#

Isolate the remaining square root on one side of the equation

#sqrt((3x+1)(x-1)) = (4-2-2x)/2 = 1 - x#

Once again, square both sides of the equation

#(sqrt((3x+1)(x-1)))^2 = (1-x)^2#

#(3x+1)(x-1) = 1 - 2x - x^2#

#3x^2 - color(red)(cancel(color(black)(2x))) - 1 = 1 - color(red)(cancel(color(black)(2x))) - x^2#

Rearrange the quadratic equation by moving all the terms on one side of the equation

#4x^2 -2 = 0#

This is equivalent to

#2(x^2-1) = 0 implies x^2 - 1 = 0 <=> {(x=-1), (x=1) :}#

Since you need #x# to satisfy the condition #x>=1#, only #x=1# will be a valid solution; #x = -1# will be an extraneous solution.

Do a quick check to make sure that the calculations are correct

#sqrt(3 * (1) +1) - sqrt((1)-1) = 2#

#sqrt(4) - 0 = 2#

#2 = 2color(white)(x)color(green)(sqrt())#