Find the real solution(s) of #sqrt(3-x)-sqrt(x+1) > 1/3# ?

1 Answer
Sep 4, 2016

Answer:

#x in [-1, 1 - sqrt(71)/18)#

Explanation:

First find values of #x# where:

#sqrt(3-x)-sqrt(x+1) = 1/3#

Squaring both sides (which may introduce spurious solutions), we get:

#(3-x)-2sqrt((3-x)(x+1))+(x+1) = 1/9#

which simplifies to:

#4-2 sqrt(3+2x-x^2) = 1/9#

Multiply both sides by #9# to get:

#36-18 sqrt(3+2x-x^2) = 1#

Hence:

#35 = 18 sqrt(3+2x-x^2)#

Square both sides to get:

#1225 = 324(3+2x-x^2) = 972+648x-324x^2#

Rearrange to get:

#324x^2-648x+253 = 0#

Divide through by #324# to get:

#0 = x^2-2x+253/324 = (x-1)^2-71/324#

Hence #x = 1+-sqrt(71)/18#

Note that if #x = 1+sqrt(71)/18# then #3-x < 2#, #x+1 > 2# and:

#sqrt(3-x)-sqrt(x+1) < 0#

If #x = 1-sqrt(71)/18# then:

#sqrt(3-x)-sqrt(x+1) = sqrt(2+sqrt(71)/18)-sqrt(2-sqrt(71)/18) > 0#

and we find:

#(sqrt(3-x)-sqrt(x+1))^2 = (sqrt(2+sqrt(71)/18)-sqrt(2-sqrt(71)/18))^2#

#color(white)((sqrt(3-x)-sqrt(x+1))^2) = (2+sqrt(71)/18)-2sqrt((2+sqrt(71)/18)(2-sqrt(71)/18))+(2-sqrt(71)/18)#

#color(white)((sqrt(3-x)-sqrt(x+1))^2) = 4-2sqrt(4-71/324)#

#color(white)((sqrt(3-x)-sqrt(x+1))^2) = 4-2sqrt(1225/324)#

#color(white)((sqrt(3-x)-sqrt(x+1))^2) = 4-2*35/18#

#color(white)((sqrt(3-x)-sqrt(x+1))^2) = 36/9-35/9#

#color(white)((sqrt(3-x)-sqrt(x+1))^2) = 1/9#

So this is a valid solution.

Now if #-1 <= x < 1-sqrt(71)/18# then both square roots are defined and #sqrt(3-x)-sqrt(x+1) > 1/3#