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