Solve sqrt{x+2}-sqrt{2-x}=2

$$\sqrt { x + 2 } - \sqrt { 2 - x } = 2$$

Solve for x

$$x=2$$

Solution Steps

Subtract -\sqrt{2-x} from both sides of the equation.

$$\sqrt{x+2}=2+\sqrt{2-x}$$

Square both sides of the equation.

$$\left(\sqrt{x+2}\right)^{2}=\left(2+\sqrt{2-x}\right)^{2}$$

Calculate \sqrt{x+2} to the power of 2 and get x+2.

$$x+2=\left(2+\sqrt{2-x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{2-x}\right)^{2}.

$$x+2=4+4\sqrt{2-x}+\left(\sqrt{2-x}\right)^{2}$$

Calculate \sqrt{2-x} to the power of 2 and get 2-x.

$$x+2=4+4\sqrt{2-x}+2-x$$

Add 4 and 2 to get 6.

$$x+2=6+4\sqrt{2-x}-x$$

Subtract 6-x from both sides of the equation.

$$x+2-\left(6-x\right)=4\sqrt{2-x}$$

To find the opposite of 6-x, find the opposite of each term.

$$x+2-6+x=4\sqrt{2-x}$$

Subtract 6 from 2 to get -4.

$$x-4+x=4\sqrt{2-x}$$

Combine x and x to get 2x.

$$2x-4=4\sqrt{2-x}$$

Square both sides of the equation.

$$\left(2x-4\right)^{2}=\left(4\sqrt{2-x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-4\right)^{2}.

$$4x^{2}-16x+16=\left(4\sqrt{2-x}\right)^{2}$$

Expand \left(4\sqrt{2-x}\right)^{2}.

$$4x^{2}-16x+16=4^{2}\left(\sqrt{2-x}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$4x^{2}-16x+16=16\left(\sqrt{2-x}\right)^{2}$$

Calculate \sqrt{2-x} to the power of 2 and get 2-x.

$$4x^{2}-16x+16=16\left(2-x\right)$$

Use the distributive property to multiply 16 by 2-x.

$$4x^{2}-16x+16=32-16x$$

Subtract 32 from both sides.

$$4x^{2}-16x+16-32=-16x$$

Subtract 32 from 16 to get -16.

$$4x^{2}-16x-16=-16x$$

Add 16x to both sides.

$$4x^{2}-16x-16+16x=0$$

Combine -16x and 16x to get 0.

$$4x^{2}-16=0$$

Divide both sides by 4.

$$x^{2}-4=0$$

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

To find equation solutions, solve x-2=0 and x+2=0.

$$x=2 x=-2$$

Substitute 2 for x in the equation \sqrt{x+2}-\sqrt{2-x}=2.

$$\sqrt{2+2}-\sqrt{2-2}=2$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Substitute -2 for x in the equation \sqrt{x+2}-\sqrt{2-x}=2.

$$\sqrt{-2+2}-\sqrt{2-\left(-2\right)}=2$$

Simplify. The value x=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

Substitute 2 for x in the equation \sqrt{x+2}-\sqrt{2-x}=2.

$$\sqrt{2+2}-\sqrt{2-2}=2$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Equation \sqrt{x+2}=\sqrt{2-x}+2 has a unique solution.

$$x=2$$