Solve sqrt{x+3}+sqrt{x-3}=6

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

Solve for x

$$x = \frac{37}{4} = 9\frac{1}{4} = 9.25$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-\sqrt{x-3}\right)^{2}.

$$x+3=36-12\sqrt{x-3}+\left(\sqrt{x-3}\right)^{2}$$

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

$$x+3=36-12\sqrt{x-3}+x-3$$

Subtract 3 from 36 to get 33.

$$x+3=33-12\sqrt{x-3}+x$$

Add 12\sqrt{x-3} to both sides.

$$x+3+12\sqrt{x-3}=33+x$$

Subtract x from both sides.

$$x+3+12\sqrt{x-3}-x=33$$

Combine x and -x to get 0.

$$3+12\sqrt{x-3}=33$$

Subtract 3 from both sides.

$$12\sqrt{x-3}=33-3$$

Subtract 3 from 33 to get 30.

$$12\sqrt{x-3}=30$$

Divide both sides by 12.

$$\sqrt{x-3}=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

$$\sqrt{x-3}=\frac{5}{2}$$

Square both sides of the equation.

$$x-3=\frac{25}{4}$$

Add 3 to both sides of the equation.

$$x-3-\left(-3\right)=\frac{25}{4}-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$x=\frac{25}{4}-\left(-3\right)$$

Subtract -3 from \frac{25}{4}.

$$x=\frac{37}{4}$$

Substitute \frac{37}{4} for x in the equation \sqrt{x+3}+\sqrt{x-3}=6.

$$\sqrt{\frac{37}{4}+3}+\sqrt{\frac{37}{4}-3}=6$$

Simplify. The value x=\frac{37}{4} satisfies the equation.

$$6=6$$

Equation \sqrt{x+3}=-\sqrt{x-3}+6 has a unique solution.

$$x=\frac{37}{4}$$