Solve sqrt{1-x}+sqrt{1+x}=sqrt{2}

$$\sqrt{ 1-x } + \sqrt{ 1+x } = \sqrt{ 2 }$$

Solve for x

$$x=1$$

$$x=-1$$

Solution Steps

Subtract \sqrt{1+x} from both sides of the equation.

$$\sqrt{1-x}=\sqrt{2}-\sqrt{1+x}$$

Square both sides of the equation.

$$\left(\sqrt{1-x}\right)^{2}=\left(\sqrt{2}-\sqrt{1+x}\right)^{2}$$

Calculate \sqrt{1-x} to the power of 2 and get 1-x.

$$1-x=\left(\sqrt{2}-\sqrt{1+x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}-\sqrt{1+x}\right)^{2}.

$$1-x=\left(\sqrt{2}\right)^{2}-2\sqrt{2}\sqrt{1+x}+\left(\sqrt{1+x}\right)^{2}$$

The square of \sqrt{2} is 2.

$$1-x=2-2\sqrt{2}\sqrt{1+x}+\left(\sqrt{1+x}\right)^{2}$$

Calculate \sqrt{1+x} to the power of 2 and get 1+x.

$$1-x=2-2\sqrt{2}\sqrt{1+x}+1+x$$

Add 2 and 1 to get 3.

$$1-x=3-2\sqrt{2}\sqrt{1+x}+x$$

Subtract 3+x from both sides of the equation.

$$1-x-\left(3+x\right)=-2\sqrt{2}\sqrt{1+x}$$

To find the opposite of 3+x, find the opposite of each term.

$$1-x-3-x=-2\sqrt{2}\sqrt{1+x}$$

Subtract 3 from 1 to get -2.

$$-2-x-x=-2\sqrt{2}\sqrt{1+x}$$

Combine -x and -x to get -2x.

$$-2-2x=-2\sqrt{2}\sqrt{1+x}$$

Square both sides of the equation.

$$\left(-2-2x\right)^{2}=\left(-2\sqrt{2}\sqrt{1+x}\right)^{2}$$

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

$$4+8x+4x^{2}=\left(-2\sqrt{2}\sqrt{1+x}\right)^{2}$$

Expand \left(-2\sqrt{2}\sqrt{1+x}\right)^{2}.

$$4+8x+4x^{2}=\left(-2\right)^{2}\left(\sqrt{2}\right)^{2}\left(\sqrt{1+x}\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$4+8x+4x^{2}=4\left(\sqrt{2}\right)^{2}\left(\sqrt{1+x}\right)^{2}$$

The square of \sqrt{2} is 2.

$$4+8x+4x^{2}=4\times 2\left(\sqrt{1+x}\right)^{2}$$

Multiply 4 and 2 to get 8.

$$4+8x+4x^{2}=8\left(\sqrt{1+x}\right)^{2}$$

Calculate \sqrt{1+x} to the power of 2 and get 1+x.

$$4+8x+4x^{2}=8\left(1+x\right)$$

Use the distributive property to multiply 8 by 1+x.

$$4+8x+4x^{2}=8+8x$$

Subtract 8 from both sides.

$$4+8x+4x^{2}-8=8x$$

Subtract 8 from 4 to get -4.

$$-4+8x+4x^{2}=8x$$

Subtract 8x from both sides.

$$-4+8x+4x^{2}-8x=0$$

Combine 8x and -8x to get 0.

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

Divide both sides by 4.

$$-1+x^{2}=0$$

Consider -1+x^{2}. Rewrite -1+x^{2} as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

To find equation solutions, solve x-1=0 and x+1=0.

$$x=1 x=-1$$

Substitute 1 for x in the equation \sqrt{1-x}+\sqrt{1+x}=\sqrt{2}.

$$\sqrt{1-1}+\sqrt{1+1}=\sqrt{2}$$

Simplify. The value x=1 satisfies the equation.

$$2^{\frac{1}{2}}=2^{\frac{1}{2}}$$

Substitute -1 for x in the equation \sqrt{1-x}+\sqrt{1+x}=\sqrt{2}.

$$\sqrt{1-\left(-1\right)}+\sqrt{1-1}=\sqrt{2}$$

Simplify. The value x=-1 satisfies the equation.

$$2^{\frac{1}{2}}=2^{\frac{1}{2}}$$

List all solutions of \sqrt{1-x}=-\sqrt{x+1}+\sqrt{2}.

$$x=1 x=-1$$