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

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

Solve for x

$$x=-1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Subtract 4x from both sides.

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

Combine 2x and -4x to get -2x.

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

Subtract 4 from both sides.

$$-2x+3-x^{2}-4=0$$

Subtract 4 from 3 to get -1.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

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

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-1$$

Rewrite -x^{2}-2x-1 as \left(-x^{2}-x\right)+\left(-x-1\right).

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

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

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

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

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

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

$$x=-1 x=-1$$

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

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

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

$$1=1$$

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

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

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

$$1=1$$

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

$$x=-1 x=-1$$