Question 163137
Let {{{A=sqrt(2x+2)}}} and {{{B=sqrt(x-3)}}}


So the equation {{{sqrt(2x+2)- sqrt(x-3)=2}}} then becomes {{{A-B=2}}}


{{{A-B=2}}} Start with the given equation



{{{A=2+B}}} Add "B" to both sides



{{{A^2=(2+B)^2}}} Square both sides. Remember, squaring undoes the square root (which is in both A and B)



{{{A^2=4+4B+B^2}}} FOIL the right side



{{{(sqrt(2x+2))^2=4+4*sqrt(x-3)+(sqrt(x-3))^2}}} Plug in {{{A=sqrt(2x+2)}}} and {{{B=sqrt(x-3)}}}



{{{2x+2=4+4*sqrt(x-3)+x-3}}} Square {{{sqrt(2x+2)}}} to get {{{2x+2}}}. Square {{{sqrt(x-3)}}} to get {{{x-3}}}. 



{{{2x+2=1+x+4*sqrt(x-3)}}} Combine like terms.



{{{2x+2-1-x=4*sqrt(x-3)}}} Subtract 1 from both sides. Subtract x from both sides. 



{{{x+1=4*sqrt(x-3)}}} Combine like terms.



{{{(x+1)^2=16(x-3)}}} Square both sides. This will eliminate the square root.



{{{x^2+2x+1=16(x-3)}}} FOIL the left side



{{{x^2+2x+1=16x-48}}} Distribute



{{{x^2+2x+1- 16x+48=0}}} Get all terms to the left side



{{{x^2-14x+49=0}}} Combine like terms



{{{(x-7)^2=0}}} Factor the left side



{{{x-7=0}}} Take the square root of both sides (to get rid of the square)



{{{x=7}}} Add 7 to both sides.



So the solution is {{{x=7}}}