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

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

Solve for x

$$x = \frac{\sqrt{5} + 3}{2} \approx 2.618033989$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract 1 from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

Subtract x from both sides.

$$x^{2}-2x+1-x=0$$

Combine -2x and -x to get -3x.

$$x^{2}-3x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4}}{2}$$

Add 9 to -4.

$$x=\frac{-\left(-3\right)±\sqrt{5}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{5}}{2}$$

Now solve the equation x=\frac{3±\sqrt{5}}{2} when ± is plus. Add 3 to \sqrt{5}.

$$x=\frac{\sqrt{5}+3}{2}$$

Now solve the equation x=\frac{3±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from 3.

$$x=\frac{3-\sqrt{5}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}+3}{2} x=\frac{3-\sqrt{5}}{2}$$

Substitute \frac{\sqrt{5}+3}{2} for x in the equation \sqrt{x}=\sqrt{1+\sqrt{x}}.

$$\sqrt{\frac{\sqrt{5}+3}{2}}=\sqrt{1+\sqrt{\frac{\sqrt{5}+3}{2}}}$$

Simplify. The value x=\frac{\sqrt{5}+3}{2} satisfies the equation.

$$\frac{1}{2}+\frac{1}{2}\times 5^{\frac{1}{2}}=\frac{1}{2}+\frac{1}{2}\times 5^{\frac{1}{2}}$$

Substitute \frac{3-\sqrt{5}}{2} for x in the equation \sqrt{x}=\sqrt{1+\sqrt{x}}.

$$\sqrt{\frac{3-\sqrt{5}}{2}}=\sqrt{1+\sqrt{\frac{3-\sqrt{5}}{2}}}$$

Simplify. The value x=\frac{3-\sqrt{5}}{2} does not satisfy the equation.

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

Substitute \frac{\sqrt{5}+3}{2} for x in the equation \sqrt{x}=\sqrt{1+\sqrt{x}}.

$$\sqrt{\frac{\sqrt{5}+3}{2}}=\sqrt{1+\sqrt{\frac{\sqrt{5}+3}{2}}}$$

Simplify. The value x=\frac{\sqrt{5}+3}{2} satisfies the equation.

$$\frac{1}{2}+\frac{1}{2}\times 5^{\frac{1}{2}}=\frac{1}{2}+\frac{1}{2}\times 5^{\frac{1}{2}}$$

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

$$x=\frac{\sqrt{5}+3}{2}$$