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

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

Solve for v

$$v=-1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract v^{2} from both sides.

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

Subtract 4v from both sides.

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

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

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

Subtract 4 from both sides.

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

Subtract 4 from 3 to get -1.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -v^{2}+av+bv-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 -v^{2}-2v-1 as \left(-v^{2}-v\right)+\left(-v-1\right).

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

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

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

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

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

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

$$v=-1 v=-1$$

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

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

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

$$1=1$$

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

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

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

$$1=1$$

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

$$v=-1 v=-1$$