Begin by finding the restrictions on the domain
5+x>=0 and 3-x>=05+x≥0and3−x≥0
Simplify
x>=-5 and -x>=-3x≥−5and−x≥−3
x>=-5 and x<=3x≥−5andx≤3
-5<=x<=3−5≤x≤3
Append the restrictions to the original equation:
sqrt(5+x)-sqrt(3-x)=2;-5<=x<=3√5+x−√3−x=2;−5≤x≤3
Square both sides:
(sqrt(5+x)-(sqrt(3-x))^2=2^2;-5<=x<=3(√5+x−(√3−x)2=22;−5≤x≤3
Expanding the square of the left side may appear to be a bit daunting but the pattern, (a-b)^2 = a^2-2ab+b^2(a−b)2=a2−2ab+b2, makes it quite simple:
Let a = sqrt(5+x) and b = sqrt(3-x)a=√5+xandb=√3−x, then a^2=5+x,b^2=3-x, and -2ab=-2sqrt((3-x)(5+x))a2=5+x,b2=3−x,and−2ab=−2√(3−x)(5+x)
5+x -2sqrt((3-x)(5+x)) + 3-x=4;-5<=x<=35+x−2√(3−x)(5+x)+3−x=4;−5≤x≤3
Collect everything but the radical term to one side:
-2sqrt((3-x)(5+x)) = -4;-5<=x<=3−2√(3−x)(5+x)=−4;−5≤x≤3
Divide both sides by -2:
sqrt((3-x)(5+x)) = 2;-5<=x<=3√(3−x)(5+x)=2;−5≤x≤3
Square both sides:
(3-x)(5+x)=4;-5<=x<=3(3−x)(5+x)=4;−5≤x≤3
Expand the left side, using the F.O.I.L. method:
15+3x-5x-x^2=4;-5<=x<=315+3x−5x−x2=4;−5≤x≤3
Combine like terms
11-2x-x^2=0;-5<=x<=311−2x−x2=0;−5≤x≤3
Multiply both side by -1:
x^2+2x-11=0;-5<=x<=3x2+2x−11=0;−5≤x≤3
Check the discriminant:
d=b^2-4(a)(c) = 2^2-4(1)(-11)=4+44=48d=b2−4(a)(c)=22−4(1)(−11)=4+44=48
Using the quadratic formula:
x = (-b+-sqrt(d))/(2a)x=−b±√d2a
x = (-2+-sqrt(48))/2x=−2±√482
x = -1+-sqrt(12)x=−1±√12
x = -1+-2sqrt(3)x=−1±2√3
If you substitute the negative root into the original equation you obtain, -2=2−2=2, which indicates an extraneous root caused by squaring; therefore, it must be discarded:
x = 2sqrt(3)-1x=2√3−1 is the only solution.
