Solve sqrt{3x+6}=x+2

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

Solve for x

$$x=1$$

$$x=-2$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{3x+6} to the power of 2 and get 3x+6.

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

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

$$3x+6=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$3x+6-x^{2}=4x+4$$

Subtract 4x from both sides.

$$3x+6-x^{2}-4x=4$$

Combine 3x and -4x to get -x.

$$-x+6-x^{2}=4$$

Subtract 4 from both sides.

$$-x+6-x^{2}-4=0$$

Subtract 4 from 6 to get 2.

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

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 negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-2$$

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

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

Factor out x in the first and 2 in the second group.

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

Factor out common term -x+1 by using distributive property.

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

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

$$x=1 x=-2$$

Substitute 1 for x in the equation \sqrt{3x+6}=x+2.

$$\sqrt{3\times 1+6}=1+2$$

Simplify. The value x=1 satisfies the equation.

$$3=3$$

Substitute -2 for x in the equation \sqrt{3x+6}=x+2.

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

Simplify. The value x=-2 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{3x+6}=x+2.

$$x=1 x=-2$$