Solve x-1=sqrt{2x}

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

Solve for x

$$x=\sqrt{3}+2\approx 3.732050808$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract 2x from both sides.

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

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

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

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

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

Square -4.

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

Add 16 to -4.

$$x=\frac{-\left(-4\right)±\sqrt{12}}{2}$$

Take the square root of 12.

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

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{3}}{2}$$

Now solve the equation x=\frac{4±2\sqrt{3}}{2} when ± is plus. Add 4 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+4}{2}$$

Divide 4+2\sqrt{3} by 2.

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

Now solve the equation x=\frac{4±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 4.

$$x=\frac{4-2\sqrt{3}}{2}$$

Divide 4-2\sqrt{3} by 2.

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

The equation is now solved.

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

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

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

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

$$3^{\frac{1}{2}}+1=3^{\frac{1}{2}}+1$$

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

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

Simplify. The value x=2-\sqrt{3} does not satisfy the equation because the left and the right hand side have opposite signs.

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

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

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