How do you solve #sqrt( 3x-4) - sqrt(2x-7 )=3#?
1 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
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#