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

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

Solve for x

$$x=2$$

$$x=-1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

$$3x+3=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$3x+3-x^{2}=2x+1$$

Subtract 2x from both sides.

$$3x+3-x^{2}-2x=1$$

Combine 3x and -2x to get x.

$$x+3-x^{2}=1$$

Subtract 1 from both sides.

$$x+3-x^{2}-1=0$$

Subtract 1 from 3 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 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{3x+3}=x+1.

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

Simplify. The value x=2 satisfies the equation.

$$3=3$$

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

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

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

$$0=0$$

List all solutions of \sqrt{3x+3}=x+1.

$$x=2 x=-1$$