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

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

Solve for x

$$x=\frac{2\sqrt{5}}{5}\approx 0.894427191$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Add 1 and 1 to get 2.

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

Combine x and -x to get 0.

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

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

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

Subtract 2 from both sides of the equation.

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

Subtract 2 from 2 to get 0.

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

Square both sides of the equation.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

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

Use the distributive property to multiply 4 by 1+x.

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

Apply the distributive property by multiplying each term of 4+4x by each term of 1-x.

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

Combine -4x and 4x to get 0.

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

Subtract x^{2} from both sides.

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

Combine -4x^{2} and -x^{2} to get -5x^{2}.

$$4-5x^{2}=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-5x^{2}=-4$$

Divide both sides by -5.

$$x^{2}=\frac{-4}{-5}$$

Fraction \frac{-4}{-5} can be simplified to \frac{4}{5} by removing the negative sign from both the numerator and the denominator.

$$x^{2}=\frac{4}{5}$$

Take the square root of both sides of the equation.

$$x=\frac{2\sqrt{5}}{5} x=-\frac{2\sqrt{5}}{5}$$

Substitute \frac{2\sqrt{5}}{5} for x in the equation \sqrt{1+x}+\sqrt{1-x}=\sqrt{x+2}.

$$\sqrt{1+\frac{2\sqrt{5}}{5}}+\sqrt{1-\frac{2\sqrt{5}}{5}}=\sqrt{\frac{2\sqrt{5}}{5}+2}$$

Simplify. The value x=\frac{2\sqrt{5}}{5} satisfies the equation.

$$\left(1+\frac{2}{5}\times 5^{\frac{1}{2}}\right)^{\frac{1}{2}}+\left(1-\frac{2}{5}\times 5^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(\frac{2}{5}\times 5^{\frac{1}{2}}+2\right)^{\frac{1}{2}}$$

Substitute -\frac{2\sqrt{5}}{5} for x in the equation \sqrt{1+x}+\sqrt{1-x}=\sqrt{x+2}.

$$\sqrt{1-\frac{2\sqrt{5}}{5}}+\sqrt{1-\left(-\frac{2\sqrt{5}}{5}\right)}=\sqrt{-\frac{2\sqrt{5}}{5}+2}$$

Simplify. The value x=-\frac{2\sqrt{5}}{5} does not satisfy the equation.

$$\left(1-\frac{2}{5}\times 5^{\frac{1}{2}}\right)^{\frac{1}{2}}+\left(1+\frac{2}{5}\times 5^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(-\frac{2}{5}\times 5^{\frac{1}{2}}+2\right)^{\frac{1}{2}}$$

Substitute \frac{2\sqrt{5}}{5} for x in the equation \sqrt{1+x}+\sqrt{1-x}=\sqrt{x+2}.

$$\sqrt{1+\frac{2\sqrt{5}}{5}}+\sqrt{1-\frac{2\sqrt{5}}{5}}=\sqrt{\frac{2\sqrt{5}}{5}+2}$$

Simplify. The value x=\frac{2\sqrt{5}}{5} satisfies the equation.

$$\left(1+\frac{2}{5}\times 5^{\frac{1}{2}}\right)^{\frac{1}{2}}+\left(1-\frac{2}{5}\times 5^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(\frac{2}{5}\times 5^{\frac{1}{2}}+2\right)^{\frac{1}{2}}$$

Equation \sqrt{x+1}+\sqrt{1-x}=\sqrt{x+2} has a unique solution.

$$x=\frac{2\sqrt{5}}{5}$$