Solve v-2/v+3=2v

$$\frac { v - 2 } { v + 3 } = 2 v$$

Solve for v

$$v=-\frac{1}{2}=-0.5$$

$$v=-2$$

Steps Using Factoring By Grouping

Variable v cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by v+3.

$$v-2=2v\left(v+3\right)$$

Use the distributive property to multiply 2v by v+3.

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

Subtract 2v^{2} from both sides.

$$v-2-2v^{2}=6v$$

Subtract 6v from both sides.

$$v-2-2v^{2}-6v=0$$

Combine v and -6v to get -5v.

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

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

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

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

$$a+b=-5 ab=-2\left(-2\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-1 b=-4$$

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

$$\left(-2v^{2}-v\right)+\left(-4v-2\right)$$

Factor out -v in the first and -2 in the second group.

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

Factor out common term 2v+1 by using distributive property.

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

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

$$v=-\frac{1}{2} v=-2$$

Steps Using the Quadratic Formula

Variable v cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by v+3.

$$v-2=2v\left(v+3\right)$$

Use the distributive property to multiply 2v by v+3.

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

Subtract 2v^{2} from both sides.

$$v-2-2v^{2}=6v$$

Subtract 6v from both sides.

$$v-2-2v^{2}-6v=0$$

Combine v and -6v to get -5v.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -5 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)\left(-2\right)}}{2\left(-2\right)}$$

Square -5.

$$v=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\left(-2\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$v=\frac{-\left(-5\right)±\sqrt{25+8\left(-2\right)}}{2\left(-2\right)}$$

Multiply 8 times -2.

$$v=\frac{-\left(-5\right)±\sqrt{25-16}}{2\left(-2\right)}$$

Add 25 to -16.

$$v=\frac{-\left(-5\right)±\sqrt{9}}{2\left(-2\right)}$$

Take the square root of 9.

$$v=\frac{-\left(-5\right)±3}{2\left(-2\right)}$$

The opposite of -5 is 5.

$$v=\frac{5±3}{2\left(-2\right)}$$

Multiply 2 times -2.

$$v=\frac{5±3}{-4}$$

Now solve the equation v=\frac{5±3}{-4} when ± is plus. Add 5 to 3.

$$v=\frac{8}{-4}$$

Divide 8 by -4.

$$v=-2$$

Now solve the equation v=\frac{5±3}{-4} when ± is minus. Subtract 3 from 5.

$$v=\frac{2}{-4}$$

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$v=-\frac{1}{2}$$

The equation is now solved.

$$v=-2 v=-\frac{1}{2}$$

Steps for Completing the Square

Variable v cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by v+3.

$$v-2=2v\left(v+3\right)$$

Use the distributive property to multiply 2v by v+3.

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

Subtract 2v^{2} from both sides.

$$v-2-2v^{2}=6v$$

Subtract 6v from both sides.

$$v-2-2v^{2}-6v=0$$

Combine v and -6v to get -5v.

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

Add 2 to both sides. Anything plus zero gives itself.

$$-5v-2v^{2}=2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2v^{2}-5v=2$$

Divide both sides by -2.

$$\frac{-2v^{2}-5v}{-2}=\frac{2}{-2}$$

Dividing by -2 undoes the multiplication by -2.

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

Divide -5 by -2.

$$v^{2}+\frac{5}{2}v=\frac{2}{-2}$$

Divide 2 by -2.

$$v^{2}+\frac{5}{2}v=-1$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$v^{2}+\frac{5}{2}v+\left(\frac{5}{4}\right)^{2}=-1+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$v^{2}+\frac{5}{2}v+\frac{25}{16}=-1+\frac{25}{16}$$

Add -1 to \frac{25}{16}.

$$v^{2}+\frac{5}{2}v+\frac{25}{16}=\frac{9}{16}$$

Factor v^{2}+\frac{5}{2}v+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(v+\frac{5}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v+\frac{5}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$v+\frac{5}{4}=\frac{3}{4} v+\frac{5}{4}=-\frac{3}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$v=-\frac{1}{2} v=-2$$