Solve sqrt{3-x}=sqrt{2+x}+1

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

Solve for x

$$x=-1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Add 2 and 1 to get 3.

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

Subtract 3+x from both sides of the equation.

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

To find the opposite of 3+x, find the opposite of each term.

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

Subtract 3 from 3 to get 0.

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

Combine -x and -x to get -2x.

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

Square both sides of the equation.

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

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

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

Calculate -2 to the power of 2 and get 4.

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

Expand \left(2\sqrt{2+x}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by 2+x.

$$4x^{2}=8+4x$$

Subtract 8 from both sides.

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

Subtract 4x from both sides.

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

Divide both sides by 4.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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

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

Simplify. The value x=2 does not satisfy the equation.

$$1=3$$

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

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

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

$$2=2$$

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

$$x=-1$$