Solve `sqrt(2+x)+sqrt(2-x)=x` :

Begin by squaring both sides; note that this may introduce extraneous solutions, so we must check all solutions in the original equation.

`(sqrt(2+x)+sqrt(2-x))^2=x^2`

`2+x+2sqrt((2+x)(2-x))+2-x=x^2`

`2sqrt(4-x^2)=x^2-4` Square both sides again:

`4(4-x^2)=(x^2-4)^2`

`(x^2-4)^2+4(x^2-4)=0` **`4(4-x^2)=-4(x^2-4)` **

`(x^2-4)(x^2-4+4)=0`

`x^2=0 ==> x=0`

`x^2-4=0 ==> x=+-2`

Check the solutions:

If x=0 we have `sqrt(2+0)+sqrt(2-0)=2sqrt(2)!=0` , so x=0 is not a solution.

If x=-2 we have `sqrt(2+(-2))+sqrt(2-(-2))=0+sqrt(4)=2!=-2` so x=-2 is not a solution.

If x=2 we have `sqrt(2+2)+sqrt(2-2)=2+0=2` so x=2 is a solution.

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The solution is x=2

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The graph of the left side in black, the right side in red. The solution is the point of intersection:

(The grapher does not show that the black curve extends to and ends at x=2.)