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

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

Solve for x

$$x=0$$

$$x=1$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

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

Add 1 and 1 to get 2.

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

Subtract 2-x from both sides of the equation.

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

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

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

Combine x and x to get 2x.

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

Square both sides of the equation.

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

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

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

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

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

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

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

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

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

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

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Add 4x to both sides.

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

Combine -8x and 4x to get -4x.

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

Factor out x.

$$x\left(4x-4\right)=0$$

To find equation solutions, solve x=0 and 4x-4=0.

$$x=0 x=1$$

Substitute 0 for x in the equation \sqrt{x}+\sqrt{1-x}=1.

$$\sqrt{0}+\sqrt{1-0}=1$$

Simplify. The value x=0 satisfies the equation.

$$1=1$$

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

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

Simplify. The value x=1 satisfies the equation.

$$1=1$$

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

$$x=0 x=1$$