Find the real solution(s) of #sqrt(3-x)-sqrt(x+1) > 1/3# ?
1 Answer
Explanation:
First find values of
#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
#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
#0 = x^2-2x+253/324 = (x-1)^2-71/324#
Hence
Note that if
#sqrt(3-x)-sqrt(x+1) < 0#
If
#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