Solve sqrt{16-x}+sqrt{x}=4

$$\sqrt { 16 - x } + \sqrt { x } = 4$$

Solve for x

$$x=0$$

$$x=16$$

Solution Steps

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

$$\sqrt{16-x}=4-\sqrt{x}$$

Square both sides of the equation.

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

Calculate \sqrt{16-x} to the power of 2 and get 16-x.

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

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

$$16-x=16-8\sqrt{x}+\left(\sqrt{x}\right)^{2}$$

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

$$16-x=16-8\sqrt{x}+x$$

Subtract 16+x from both sides of the equation.

$$16-x-\left(16+x\right)=-8\sqrt{x}$$

To find the opposite of 16+x, find the opposite of each term.

$$16-x-16-x=-8\sqrt{x}$$

Subtract 16 from 16 to get 0.

$$-x-x=-8\sqrt{x}$$

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

$$-2x=-8\sqrt{x}$$

Square both sides of the equation.

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

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

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

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

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

Expand \left(-8\sqrt{x}\right)^{2}.

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

Calculate -8 to the power of 2 and get 64.

$$4x^{2}=64\left(\sqrt{x}\right)^{2}$$

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

$$4x^{2}=64x$$

Subtract 64x from both sides.

$$4x^{2}-64x=0$$

Factor out x.

$$x\left(4x-64\right)=0$$

To find equation solutions, solve x=0 and 4x-64=0.

$$x=0 x=16$$

Substitute 0 for x in the equation \sqrt{16-x}+\sqrt{x}=4.

$$\sqrt{16-0}+\sqrt{0}=4$$

Simplify. The value x=0 satisfies the equation.

$$4=4$$

Substitute 16 for x in the equation \sqrt{16-x}+\sqrt{x}=4.

$$\sqrt{16-16}+\sqrt{16}=4$$

Simplify. The value x=16 satisfies the equation.

$$4=4$$

List all solutions of \sqrt{16-x}=-\sqrt{x}+4.

$$x=0 x=16$$