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

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

Solve for x

$$x=\frac{5}{8}=0.625$$

Solution Steps

Subtract -\sqrt{2x-1} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 1 from 1 to get 0.

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

Subtract 2\sqrt{2x-1} from both sides.

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

Subtract 2x from both sides.

$$2x+1-2\sqrt{2x-1}-2x=0$$

Combine 2x and -2x to get 0.

$$1-2\sqrt{2x-1}=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -2.

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

Fraction \frac{-1}{-2} can be simplified to \frac{1}{2} by removing the negative sign from both the numerator and the denominator.

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

Square both sides of the equation.

$$2x-1=\frac{1}{4}$$

Add 1 to both sides of the equation.

$$2x-1-\left(-1\right)=\frac{1}{4}-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$2x=\frac{1}{4}-\left(-1\right)$$

Subtract -1 from \frac{1}{4}.

$$2x=\frac{5}{4}$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{\frac{5}{4}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{\frac{5}{4}}{2}$$

Divide \frac{5}{4} by 2.

$$x=\frac{5}{8}$$

Substitute \frac{5}{8} for x in the equation \sqrt{2x+1}-\sqrt{2x-1}=1.

$$\sqrt{2\times \frac{5}{8}+1}-\sqrt{2\times \frac{5}{8}-1}=1$$

Simplify. The value x=\frac{5}{8} satisfies the equation.

$$1=1$$

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

$$x=\frac{5}{8}$$