Solve by Completing the Square 3v^2+4v-1=0

$3 v^{2} + 4 v - 1 = 0$

Step 1

Add $1$ to both sides of the equation.

$3 v^{2} + 4 v = 1$

Step 2

Divide each term in $3 v^{2} + 4 v = 1$ by $3$ and simplify.

Divide each term in $3 v^{2} + 4 v = 1$ by $3$ .

$\frac{3 v^{2}}{3} + \frac{4 v}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 v^{2}}{3} + \frac{4 v}{3} = \frac{1}{3}$

Divide $v^{2}$ by $1$ .

$v^{2} + \frac{4 v}{3} = \frac{1}{3}$

Step 3

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(\frac{2}{3})}^{2}$

Step 4

Add the term to each side of the equation.

$v^{2} + \frac{4 v}{3} + {(\frac{2}{3})}^{2} = \frac{1}{3} + {(\frac{2}{3})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{2}{3}$ .

$v^{2} + \frac{4 v}{3} + \frac{2^{2}}{3^{2}} = \frac{1}{3} + {(\frac{2}{3})}^{2}$

Raise $2$ to the power of $2$ .

$v^{2} + \frac{4 v}{3} + \frac{4}{3^{2}} = \frac{1}{3} + {(\frac{2}{3})}^{2}$

Raise $3$ to the power of $2$ .

$v^{2} + \frac{4 v}{3} + \frac{4}{9} = \frac{1}{3} + {(\frac{2}{3})}^{2}$

Simplify the right side.

Simplify $\frac{1}{3} + {(\frac{2}{3})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{2}{3}$ .

$v^{2} + \frac{4 v}{3} + \frac{4}{9} = \frac{1}{3} + \frac{2^{2}}{3^{2}}$

Raise $2$ to the power of $2$ .

$v^{2} + \frac{4 v}{3} + \frac{4}{9} = \frac{1}{3} + \frac{4}{3^{2}}$

Raise $3$ to the power of $2$ .

$v^{2} + \frac{4 v}{3} + \frac{4}{9} = \frac{1}{3} + \frac{4}{9}$

To write $\frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$v^{2} + \frac{4 v}{3} + \frac{4}{9} = \frac{1}{3} \cdot \frac{3}{3} + \frac{4}{9}$

Write each expression with a common denominator of $9$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{3}$ by $\frac{3}{3}$ .

$v^{2} + \frac{4 v}{3} + \frac{4}{9} = \frac{3}{3 \cdot 3} + \frac{4}{9}$

Multiply $3$ by $3$ .

$v^{2} + \frac{4 v}{3} + \frac{4}{9} = \frac{3}{9} + \frac{4}{9}$

Combine the numerators over the common denominator.

$v^{2} + \frac{4 v}{3} + \frac{4}{9} = \frac{3 + 4}{9}$

Add $3$ and $4$ .

$v^{2} + \frac{4 v}{3} + \frac{4}{9} = \frac{7}{9}$

Step 6

Factor the perfect trinomial square into ${(v + \frac{2}{3})}^{2}$ .

${(v + \frac{2}{3})}^{2} = \frac{7}{9}$

Step 7

Solve the equation for $v$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$v + \frac{2}{3} = \pm \sqrt{\frac{7}{9}}$

Simplify $\pm \sqrt{\frac{7}{9}}$ .

Rewrite $\sqrt{\frac{7}{9}}$ as $\frac{\sqrt{7}}{\sqrt{9}}$ .

$v + \frac{2}{3} = \pm \frac{\sqrt{7}}{\sqrt{9}}$

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$v + \frac{2}{3} = \pm \frac{\sqrt{7}}{\sqrt{3^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$v + \frac{2}{3} = \pm \frac{\sqrt{7}}{3}$

Subtract $\frac{2}{3}$ from both sides of the equation.

$v = \pm \frac{\sqrt{7}}{3} - \frac{2}{3}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$v = \pm \frac{\sqrt{7}}{3} - \frac{2}{3}$

Decimal Form:

$v = 0.21525043 \dots, - 1.54858377 \dots$

Solve by Completing the Square 3v^2+4v-1=0

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