Solve sqrt{x+2}+sqrt{x-1}=sqrt{9}

$$\sqrt { x + 2 } + \sqrt { x - 1 } = \sqrt { 9 }$$

Solve for x

$$x=2$$

Solution Steps

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

$$\sqrt{x+2}=\sqrt{9}-\sqrt{x-1}$$

Calculate the square root of 9 and get 3.

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

Square both sides of the equation.

$$\left(\sqrt{x+2}\right)^{2}=\left(3-\sqrt{x-1}\right)^{2}$$

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

$$x+2=\left(3-\sqrt{x-1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-\sqrt{x-1}\right)^{2}.

$$x+2=9-6\sqrt{x-1}+\left(\sqrt{x-1}\right)^{2}$$

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

$$x+2=9-6\sqrt{x-1}+x-1$$

Subtract 1 from 9 to get 8.

$$x+2=8-6\sqrt{x-1}+x$$

Add 6\sqrt{x-1} to both sides.

$$x+2+6\sqrt{x-1}=8+x$$

Subtract x from both sides.

$$x+2+6\sqrt{x-1}-x=8$$

Combine x and -x to get 0.

$$2+6\sqrt{x-1}=8$$

Subtract 2 from both sides.

$$6\sqrt{x-1}=8-2$$

Subtract 2 from 8 to get 6.

$$6\sqrt{x-1}=6$$

Divide both sides by 6.

$$\sqrt{x-1}=\frac{6}{6}$$

Divide 6 by 6 to get 1.

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

Square both sides of the equation.

$$x-1=1$$

Add 1 to both sides of the equation.

$$x-1-\left(-1\right)=1-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$x=1-\left(-1\right)$$

Subtract -1 from 1.

$$x=2$$

Substitute 2 for x in the equation \sqrt{x+2}+\sqrt{x-1}=\sqrt{9}.

$$\sqrt{2+2}+\sqrt{2-1}=\sqrt{9}$$

Simplify. The value x=2 satisfies the equation.

$$3=3$$

Equation \sqrt{x+2}=-\sqrt{x-1}+3 has a unique solution.

$$x=2$$