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

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

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Add 2 and 1 to get 3.

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

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

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

Subtract x+3 from both sides of the equation.

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

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

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

Combine 3x and -x to get 2x.

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

Subtract 3 from 3 to get 0.

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

Cancel out 2 on both sides.

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

Square both sides of the equation.

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

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

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

Subtract x^{2} from both sides.

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

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.

$$a+b=1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 the first and -1 in the second group.

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

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

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

Simplify. The value x=2 satisfies the equation.

$$3=3$$

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

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

Simplify. The value x=-1 does not satisfy the equation.

$$2=0$$

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

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

Simplify. The value x=2 satisfies the equation.

$$3=3$$

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

$$x=2$$