Solve sqrt{x+3}+sqrt{x-3}=3

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

Solve for x

$$x = \frac{13}{4} = 3\frac{1}{4} = 3.25$$

Solution Steps

Subtract \sqrt{x-3} from both sides of the equation.

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

Square both sides of the equation.

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

Calculate \sqrt{x+3} to the power of 2 and get x+3.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-\sqrt{x-3}\right)^{2}.

$$x+3=9-6\sqrt{x-3}+\left(\sqrt{x-3}\right)^{2}$$

Calculate \sqrt{x-3} to the power of 2 and get x-3.

$$x+3=9-6\sqrt{x-3}+x-3$$

Subtract 3 from 9 to get 6.

$$x+3=6-6\sqrt{x-3}+x$$

Add 6\sqrt{x-3} to both sides.

$$x+3+6\sqrt{x-3}=6+x$$

Subtract x from both sides.

$$x+3+6\sqrt{x-3}-x=6$$

Combine x and -x to get 0.

$$3+6\sqrt{x-3}=6$$

Subtract 3 from both sides.

$$6\sqrt{x-3}=6-3$$

Subtract 3 from 6 to get 3.

$$6\sqrt{x-3}=3$$

Divide both sides by 6.

$$\sqrt{x-3}=\frac{3}{6}$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

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

Square both sides of the equation.

$$x-3=\frac{1}{4}$$

Add 3 to both sides of the equation.

$$x-3-\left(-3\right)=\frac{1}{4}-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$x=\frac{1}{4}-\left(-3\right)$$

Subtract -3 from \frac{1}{4}.

$$x=\frac{13}{4}$$

Substitute \frac{13}{4} for x in the equation \sqrt{x+3}+\sqrt{x-3}=3.

$$\sqrt{\frac{13}{4}+3}+\sqrt{\frac{13}{4}-3}=3$$

Simplify. The value x=\frac{13}{4} satisfies the equation.

$$3=3$$

Equation \sqrt{x+3}=-\sqrt{x-3}+3 has a unique solution.

$$x=\frac{13}{4}$$