How do you solve #sqrt( 3x-4) - sqrt(2x-7 )=3#?

1 Answer
Oct 18, 2015

Answer:

#x = 3.58125" "# or #" "x = 80.41875#

Explanation:

Since you're dealing with square roots, it's alway a good idea to start by determining the intervals that the possible solutions must fall in.

You know tha for real numbers, you can only take the square root of positive numbers. This means that you need

#3x - 4 >= 0 implies x >= 4/3#

#2x - 7 >= 0 implies x >= 7/2#

Merge these two conditions to get #x >= 7/2#. This means that any value of #x# that does not satisfy this condition will not be a valid solution to the original equation.

Next, sqaure both sides of the equation to reduce the number of radical terms

#(sqrt(3x-4) - sqrt(2x-7))^2 = 3^2#

#(sqrt(3x-4))^2 - 2sqrt((3x-4)(2x-7)) + (sqrt(2x-7))^2 = 9#

#3x - 4 - 2sqrt((3x-4)(2x-7)) + 2x - 7 = 9#

Isolate the remaining radical term on one side of the equation

#-2sqrt((3x-4)(2x-7)) = 9 - 5x + 11#

#-2sqrt((3x-4)(2x-7)) = 5 * (4-x)#

Now square both sides of the equation again

#(-2sqrt((3x-4)(2x-7)))^2 = [5(4-x)]^2#

#4 * (3x-4)(2x-7) = 25 * (16 - 8x + x^2)#

#24x^2 - 116x + 112 = 400 - 200x + 25x^2#

Move all the terms on one side of the equation to get

#x^2 - 84x + 288 = 0#

Use the quadratic formula to get

#x_(1,2) = (-(84) +- sqrt((-84)^2 - 4 * 1 * 288))/(2 * 1)#

#x_(1,2) = (84 +- sqrt(5904))/2 = (84 +- 76.8375)/2#

The two solutions will be

#x_1 = (84 + 76.8375)/2 = 80.41875#

and

#x_2 = (84 - 76.8375)/2 = 3.58125#

SInce both solutions satisfy the initial condition #x >= 7/2#, they will are both potential solutions to the original equation.

Plug these values into the original equation

#sqrt(3.58125 * 3 - 4) - sqrt(3.58125 * 2 - 7) = 2.999986 ~~ 3#

and

#sqrt(80.41875 * 3 - 4) - sqrt(80.41875 * 2 - 7) = 3.0000001 ~~ 3#