Question 288676
# 1

{{{sqrt(x+2) = x}}} Start with the given equation.



{{{x+2 = x^2}}} Square both sides.



{{{0 = x^2-x-2}}} Get everything to one side.



{{{x^2-x-2=0}}} Rearrange the equation.



{{{(x-2)(x+1)=0}}} <a href="http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/change-this-name4450.solver">Factor</a>.



{{{x+2=0}}} or {{{x+1=0}}} Use the zero product property.



{{{x=2}}} or {{{x=-1}}} Solve for 'x' in each case.



So the possible solutions for {{{sqrt(x+2) = x}}} are {{{x=2}}} or {{{x=-1}}}. However, if you plug {{{x=-1}}} into {{{sqrt(x+2) = x}}}, you'll get {{{1=-1}}} which is NOT true. So {{{x=-1}}} is NOT a solution.



So the only solution is {{{x=2}}}


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# 2


{{{sqrt(2-x)=2-x}}} Start with the given equation.



{{{2-x=(2-x)^2}}} Square both sides.



{{{2-x=4-4x+x^2}}} FOIL



{{{0=4-4x+x^2+x-2}}} Get everything to one side.



{{{0=x^2-3x+2}}} Combine like terms.



{{{0=(x-2)(x-1)}}} <a href="http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/change-this-name4450.solver">Factor</a>.



{{{x-2=0}}} or {{{x-1=0}}} Use the zero product property.



{{{x=2}}} or {{{x=1}}} Solve for 'x' in each case.



So the possible solutions are {{{x=2}}} or {{{x=1}}}. If you plug in each solution to check, you'll find that they both work. 



So the solutions are {{{x=2}}} or {{{x=1}}}