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

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

Solve for x

$$x=\frac{\sqrt{3}}{2}\approx 0.866025404$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Add 1 and 1 to get 2.

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

Subtract 2-x from both sides of the equation.

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

To find the opposite of 2-x, find the opposite of each term.

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

Subtract 2 from 1 to get -1.

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

Combine x and x to get 2x.

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

Square both sides of the equation.

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

$$1-4x+4x^{2}=4-4x$$

Add 4x to both sides.

$$1-4x+4x^{2}+4x=4$$

Combine -4x and 4x to get 0.

$$1+4x^{2}=4$$

Subtract 1 from both sides.

$$4x^{2}=4-1$$

Subtract 1 from 4 to get 3.

$$4x^{2}=3$$

Divide both sides by 4.

$$x^{2}=\frac{3}{4}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{3}}{2} x=-\frac{\sqrt{3}}{2}$$

Substitute \frac{\sqrt{3}}{2} for x in the equation \sqrt{1+x}=1+\sqrt{1-x}.

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

Simplify. The value x=\frac{\sqrt{3}}{2} satisfies the equation.

$$\frac{1}{2}+\frac{1}{2}\times 3^{\frac{1}{2}}=\frac{1}{2}+\frac{1}{2}\times 3^{\frac{1}{2}}$$

Substitute -\frac{\sqrt{3}}{2} for x in the equation \sqrt{1+x}=1+\sqrt{1-x}.

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

Simplify. The value x=-\frac{\sqrt{3}}{2} does not satisfy the equation.

$$-\left(\frac{1}{2}-\frac{1}{2}\times 3^{\frac{1}{2}}\right)=\frac{3}{2}+\frac{1}{2}\times 3^{\frac{1}{2}}$$

Substitute \frac{\sqrt{3}}{2} for x in the equation \sqrt{1+x}=1+\sqrt{1-x}.

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

Simplify. The value x=\frac{\sqrt{3}}{2} satisfies the equation.

$$\frac{1}{2}+\frac{1}{2}\times 3^{\frac{1}{2}}=\frac{1}{2}+\frac{1}{2}\times 3^{\frac{1}{2}}$$

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

$$x=\frac{\sqrt{3}}{2}$$