Solve sqrt{2x}-sqrt{x}=1

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

Solve for x

$$x=2\sqrt{2}+3\approx 5.828427125$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 1+x from both sides of the equation.

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

To find the opposite of 1+x, find the opposite of each term.

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

Combine 2x and -x to get x.

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

Square both sides of the equation.

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Subtract 4x from both sides.

$$x^{2}-2x+1-4x=0$$

Combine -2x and -4x to get -6x.

$$x^{2}-6x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4}}{2}$$

Add 36 to -4.

$$x=\frac{-\left(-6\right)±\sqrt{32}}{2}$$

Take the square root of 32.

$$x=\frac{-\left(-6\right)±4\sqrt{2}}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±4\sqrt{2}}{2}$$

Now solve the equation x=\frac{6±4\sqrt{2}}{2} when ± is plus. Add 6 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}+6}{2}$$

Divide 6+4\sqrt{2} by 2.

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

Now solve the equation x=\frac{6±4\sqrt{2}}{2} when ± is minus. Subtract 4\sqrt{2} from 6.

$$x=\frac{6-4\sqrt{2}}{2}$$

Divide 6-4\sqrt{2} by 2.

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

The equation is now solved.

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

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

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

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

$$1=1$$

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

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

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

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

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

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

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

$$1=1$$

Equation \sqrt{2x}=\sqrt{x}+1 has a unique solution.

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