Solve sqrt{1+4sqrt{x}}=sqrt{x}+1

$$\sqrt{ 1+4 \sqrt{ x } } = \sqrt{ x } +1$$

Solve for x

$$x=0$$

$$x=4$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{1+4\sqrt{x}} to the power of 2 and get 1+4\sqrt{x}.

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

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

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

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

$$1+4\sqrt{x}=x+2\sqrt{x}+1$$

Subtract x from both sides.

$$1+4\sqrt{x}-x=2\sqrt{x}+1$$

Subtract 2\sqrt{x} from both sides.

$$1+4\sqrt{x}-x-2\sqrt{x}=1$$

Combine 4\sqrt{x} and -2\sqrt{x} to get 2\sqrt{x}.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

$$2\sqrt{x}-x=0$$

Subtract -x from both sides of the equation.

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

Square both sides of the equation.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Subtract x^{2} from both sides.

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

Factor out x.

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

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

$$x=0 x=4$$

Substitute 0 for x in the equation \sqrt{1+4\sqrt{x}}=\sqrt{x}+1.

$$\sqrt{1+4\sqrt{0}}=\sqrt{0}+1$$

Simplify. The value x=0 satisfies the equation.

$$1=1$$

Substitute 4 for x in the equation \sqrt{1+4\sqrt{x}}=\sqrt{x}+1.

$$\sqrt{1+4\sqrt{4}}=\sqrt{4}+1$$

Simplify. The value x=4 satisfies the equation.

$$3=3$$

List all solutions of \sqrt{4\sqrt{x}+1}=\sqrt{x}+1.

$$x=0 x=4$$