How do you solve sqrt( 3x-4) - sqrt(2x-7 )=3√3x−4−√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 >= 4/33x−4≥0⇒x≥43
2x - 7 >= 0 implies x >= 7/22x−7≥0⇒x≥72
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(√3x−4−√2x−7)2=32
(sqrt(3x-4))^2 - 2sqrt((3x-4)(2x-7)) + (sqrt(2x-7))^2 = 9(√3x−4)2−2√(3x−4)(2x−7)+(√2x−7)2=9
3x - 4 - 2sqrt((3x-4)(2x-7)) + 2x - 7 = 93x−4−2√(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 + 11−2√(3x−4)(2x−7)=9−5x+11
-2sqrt((3x-4)(2x-7)) = 5 * (4-x)−2√(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(−2√(3x−4)(2x−7))2=[5(4−x)]2
4 * (3x-4)(2x-7) = 25 * (16 - 8x + x^2)4⋅(3x−4)(2x−7)=25⋅(16−8x+x2)
24x^2 - 116x + 112 = 400 - 200x + 25x^224x2−116x+112=400−200x+25x2
Move all the terms on one side of the equation to get
x^2 - 84x + 288 = 0x2−84x+288=0
Use the quadratic formula to get
x_(1,2) = (-(84) +- sqrt((-84)^2 - 4 * 1 * 288))/(2 * 1)x1,2=−(84)±√(−84)2−4⋅1⋅2882⋅1
x_(1,2) = (84 +- sqrt(5904))/2 = (84 +- 76.8375)/2x1,2=84±√59042=84±76.83752
The two solutions will be
x_1 = (84 + 76.8375)/2 = 80.41875x1=84+76.83752=80.41875
and
x_2 = (84 - 76.8375)/2 = 3.58125x2=84−76.83752=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√3.58125⋅3−4−√3.58125⋅2−7=2.999986≈3
and
sqrt(80.41875 * 3 - 4) - sqrt(80.41875 * 2 - 7) = 3.0000001 ~~ 3√80.41875⋅3−4−√80.41875⋅2−7=3.0000001≈3
