Solve sqrt{x+2}+sqrt{x-2}=4

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

Solve for x

$$x = \frac{17}{4} = 4\frac{1}{4} = 4.25$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-\sqrt{x-2}\right)^{2}.

$$x+2=16-8\sqrt{x-2}+\left(\sqrt{x-2}\right)^{2}$$

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

$$x+2=16-8\sqrt{x-2}+x-2$$

Subtract 2 from 16 to get 14.

$$x+2=14-8\sqrt{x-2}+x$$

Add 8\sqrt{x-2} to both sides.

$$x+2+8\sqrt{x-2}=14+x$$

Subtract x from both sides.

$$x+2+8\sqrt{x-2}-x=14$$

Combine x and -x to get 0.

$$2+8\sqrt{x-2}=14$$

Subtract 2 from both sides.

$$8\sqrt{x-2}=14-2$$

Subtract 2 from 14 to get 12.

$$8\sqrt{x-2}=12$$

Divide both sides by 8.

$$\sqrt{x-2}=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

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

Square both sides of the equation.

$$x-2=\frac{9}{4}$$

Add 2 to both sides of the equation.

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

Subtracting -2 from itself leaves 0.

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

Subtract -2 from \frac{9}{4}.

$$x=\frac{17}{4}$$

Substitute \frac{17}{4} for x in the equation \sqrt{x+2}+\sqrt{x-2}=4.

$$\sqrt{\frac{17}{4}+2}+\sqrt{\frac{17}{4}-2}=4$$

Simplify. The value x=\frac{17}{4} satisfies the equation.

$$4=4$$

Equation \sqrt{x+2}=-\sqrt{x-2}+4 has a unique solution.

$$x=\frac{17}{4}$$