Solve sqrt{2v+14}=v+3

$$\sqrt { 2 v + 14 } = v + 3$$

Solve for v

$$v=1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2v+14}\right)^{2}=\left(v+3\right)^{2}$$

Calculate \sqrt{2v+14} to the power of 2 and get 2v+14.

$$2v+14=\left(v+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(v+3\right)^{2}.

$$2v+14=v^{2}+6v+9$$

Subtract v^{2} from both sides.

$$2v+14-v^{2}=6v+9$$

Subtract 6v from both sides.

$$2v+14-v^{2}-6v=9$$

Combine 2v and -6v to get -4v.

$$-4v+14-v^{2}=9$$

Subtract 9 from both sides.

$$-4v+14-v^{2}-9=0$$

Subtract 9 from 14 to get 5.

$$-4v+5-v^{2}=0$$

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

$$-v^{2}-4v+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -v^{2}+av+bv+5. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-5=-5$$

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.

$$a=1 b=-5$$

Rewrite -v^{2}-4v+5 as \left(-v^{2}+v\right)+\left(-5v+5\right).

$$\left(-v^{2}+v\right)+\left(-5v+5\right)$$

Factor out v in the first and 5 in the second group.

$$v\left(-v+1\right)+5\left(-v+1\right)$$

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

$$\left(-v+1\right)\left(v+5\right)$$

To find equation solutions, solve -v+1=0 and v+5=0.

$$v=1 v=-5$$

Substitute 1 for v in the equation \sqrt{2v+14}=v+3.

$$\sqrt{2\times 1+14}=1+3$$

Simplify. The value v=1 satisfies the equation.

$$4=4$$

Substitute -5 for v in the equation \sqrt{2v+14}=v+3.

$$\sqrt{2\left(-5\right)+14}=-5+3$$

Simplify. The value v=-5 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

Equation \sqrt{2v+14}=v+3 has a unique solution.

$$v=1$$