This is the same as solving #sqrt(x+3)+sqrt(2x-1) = 5#
The requirement #x >= 1/2# is there only to assure us that we are not working with imaginary numbers. That is, it is there to make sure that #2x-1# is non-negative. At the same time #x+3# will be non-negative, So we can take square roots of both without involving imaginary numbers.
#sqrt(x+3)+sqrt(2x-1) = 5#
I prefer to start by isolating one of the square roots on one side:
#sqrt(x+3) = 5 - sqrt(2x-1)#
Square both sides (this may introduce extraneous solutions.)
#x+3 = 25 - 10sqrt(2x-1) + 2x-1#
Isolate the #10sqrt(2x-1)#
#10sqrt(2x-1) = x + 21#
Square both sides again. (this may introduce extraneous solutions.)
#100(2x-1) = x^2+42x+441#
Write the quadratic equation in standard form:
#x^2 - 158x+541 =0#
Try to factor it (unsuccessfully), then use the quadratic formula or complete the square:
#x^2 - 158x" "" " =-541#
#x^2 - 158x+79^2 =-541 + 6241#
#(x-79)^2 = 5700#
#x-79 = +- sqrt5700 = +- 10 sqrt57#
#x = 79+-10sqrt57#
The value: #79+10sqrt57# is too large to solve
#sqrt(x+3)+sqrt(2x-1) = 5#
(when we take these two square roots and add the results we will not get #5# -- use a calculator if you have one available.)
So that is an extraneous solution.
The other value, #79-10sqrt57# does solve the equation
#sqrt57 ~~ 7.5#, so this solution is about #4# And it is reasonable to think that
#sqrt(4+3)+sqrt(2(4)-1)# which is #2sqrt7# should be near #5#
(use a calculator to check if you have one available.)
