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

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

Solve for x

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

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 1 from 4 to get 3.

$$2x+1=3-4\sqrt{2x-1}+2x$$

Add 4\sqrt{2x-1} to both sides.

$$2x+1+4\sqrt{2x-1}=3+2x$$

Subtract 2x from both sides.

$$2x+1+4\sqrt{2x-1}-2x=3$$

Combine 2x and -2x to get 0.

$$1+4\sqrt{2x-1}=3$$

Subtract 1 from both sides.

$$4\sqrt{2x-1}=3-1$$

Subtract 1 from 3 to get 2.

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

Divide both sides by 4.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

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}=2.

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

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

$$2=2$$

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

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