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

##### 1 Answer

#### Answer:

#### 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 **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

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#