Solve sqrt{x+2}+sqrt{3-x}=3

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

Solve for x

$$x=2$$

$$x=-1$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

$$x+2=9-6\sqrt{3-x}+3-x$$

Add 9 and 3 to get 12.

$$x+2=12-6\sqrt{3-x}-x$$

Subtract 12-x from both sides of the equation.

$$x+2-\left(12-x\right)=-6\sqrt{3-x}$$

To find the opposite of 12-x, find the opposite of each term.

$$x+2-12+x=-6\sqrt{3-x}$$

Subtract 12 from 2 to get -10.

$$x-10+x=-6\sqrt{3-x}$$

Combine x and x to get 2x.

$$2x-10=-6\sqrt{3-x}$$

Square both sides of the equation.

$$\left(2x-10\right)^{2}=\left(-6\sqrt{3-x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-10\right)^{2}.

$$4x^{2}-40x+100=\left(-6\sqrt{3-x}\right)^{2}$$

Expand \left(-6\sqrt{3-x}\right)^{2}.

$$4x^{2}-40x+100=\left(-6\right)^{2}\left(\sqrt{3-x}\right)^{2}$$

Calculate -6 to the power of 2 and get 36.

$$4x^{2}-40x+100=36\left(\sqrt{3-x}\right)^{2}$$

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

$$4x^{2}-40x+100=36\left(3-x\right)$$

Use the distributive property to multiply 36 by 3-x.

$$4x^{2}-40x+100=108-36x$$

Subtract 108 from both sides.

$$4x^{2}-40x+100-108=-36x$$

Subtract 108 from 100 to get -8.

$$4x^{2}-40x-8=-36x$$

Add 36x to both sides.

$$4x^{2}-40x-8+36x=0$$

Combine -40x and 36x to get -4x.

$$4x^{2}-4x-8=0$$

Divide both sides by 4.

$$x^{2}-x-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

Rewrite x^{2}-x-2 as \left(x^{2}-2x\right)+\left(x-2\right).

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

Factor out x in x^{2}-2x.

$$x\left(x-2\right)+x-2$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+1\right)$$

To find equation solutions, solve x-2=0 and x+1=0.

$$x=2 x=-1$$

Substitute 2 for x in the equation \sqrt{x+2}+\sqrt{3-x}=3.

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

Simplify. The value x=2 satisfies the equation.

$$3=3$$

Substitute -1 for x in the equation \sqrt{x+2}+\sqrt{3-x}=3.

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

Simplify. The value x=-1 satisfies the equation.

$$3=3$$

List all solutions of \sqrt{x+2}=-\sqrt{3-x}+3.

$$x=2 x=-1$$