Solve by Completing the Square 3z^2+11z-4=0

$3 z^{2} + 11 z - 4 = 0$

Step 1

Add $4$ to both sides of the equation.

$3 z^{2} + 11 z = 4$

Step 2

Divide each term in $3 z^{2} + 11 z = 4$ by $3$ and simplify.

Divide each term in $3 z^{2} + 11 z = 4$ by $3$ .

$\frac{3 z^{2}}{3} + \frac{11 z}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 z^{2}}{3} + \frac{11 z}{3} = \frac{4}{3}$

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

$z^{2} + \frac{11 z}{3} = \frac{4}{3}$

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 $b$ .

${(\frac{b}{2})}^{2} = {(\frac{11}{6})}^{2}$

Step 4

Add the term to each side of the equation.

$z^{2} + \frac{11 z}{3} + {(\frac{11}{6})}^{2} = \frac{4}{3} + {(\frac{11}{6})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{11}{6}$ .

$z^{2} + \frac{11 z}{3} + \frac{11^{2}}{6^{2}} = \frac{4}{3} + {(\frac{11}{6})}^{2}$

Raise $11$ to the power of $2$ .

$z^{2} + \frac{11 z}{3} + \frac{121}{6^{2}} = \frac{4}{3} + {(\frac{11}{6})}^{2}$

Raise $6$ to the power of $2$ .

$z^{2} + \frac{11 z}{3} + \frac{121}{36} = \frac{4}{3} + {(\frac{11}{6})}^{2}$

Simplify the right side.

Simplify $\frac{4}{3} + {(\frac{11}{6})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{11}{6}$ .

$z^{2} + \frac{11 z}{3} + \frac{121}{36} = \frac{4}{3} + \frac{11^{2}}{6^{2}}$

Raise $11$ to the power of $2$ .

$z^{2} + \frac{11 z}{3} + \frac{121}{36} = \frac{4}{3} + \frac{121}{6^{2}}$

Raise $6$ to the power of $2$ .

$z^{2} + \frac{11 z}{3} + \frac{121}{36} = \frac{4}{3} + \frac{121}{36}$

To write $\frac{4}{3}$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$z^{2} + \frac{11 z}{3} + \frac{121}{36} = \frac{4}{3} \cdot \frac{12}{12} + \frac{121}{36}$

Write each expression with a common denominator of $36$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4}{3}$ by $\frac{12}{12}$ .

$z^{2} + \frac{11 z}{3} + \frac{121}{36} = \frac{4 \cdot 12}{3 \cdot 12} + \frac{121}{36}$

Multiply $3$ by $12$ .

$z^{2} + \frac{11 z}{3} + \frac{121}{36} = \frac{4 \cdot 12}{36} + \frac{121}{36}$

Combine the numerators over the common denominator.

$z^{2} + \frac{11 z}{3} + \frac{121}{36} = \frac{4 \cdot 12 + 121}{36}$

Simplify the numerator.

Multiply $4$ by $12$ .

$z^{2} + \frac{11 z}{3} + \frac{121}{36} = \frac{48 + 121}{36}$

Add $48$ and $121$ .

$z^{2} + \frac{11 z}{3} + \frac{121}{36} = \frac{169}{36}$

Step 6

Factor the perfect trinomial square into ${(z + \frac{11}{6})}^{2}$ .

${(z + \frac{11}{6})}^{2} = \frac{169}{36}$

Step 7

Solve the equation for $z$ .

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

$z + \frac{11}{6} = \pm \sqrt{\frac{169}{36}}$

Simplify $\pm \sqrt{\frac{169}{36}}$ .

Rewrite $\sqrt{\frac{169}{36}}$ as $\frac{\sqrt{169}}{\sqrt{36}}$ .

$z + \frac{11}{6} = \pm \frac{\sqrt{169}}{\sqrt{36}}$

Simplify the numerator.

Rewrite $169$ as $13^{2}$ .

$z + \frac{11}{6} = \pm \frac{\sqrt{13^{2}}}{\sqrt{36}}$

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

$z + \frac{11}{6} = \pm \frac{13}{\sqrt{36}}$

Simplify the denominator.

Rewrite $36$ as $6^{2}$ .

$z + \frac{11}{6} = \pm \frac{13}{\sqrt{6^{2}}}$

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

$z + \frac{11}{6} = \pm \frac{13}{6}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$z + \frac{11}{6} = \frac{13}{6}$

Move all terms not containing $z$ to the right side of the equation.

Subtract $\frac{11}{6}$ from both sides of the equation.

$z = \frac{13}{6} - \frac{11}{6}$

Combine the numerators over the common denominator.

$z = \frac{13 - 11}{6}$

Subtract $11$ from $13$ .

$z = \frac{2}{6}$

Cancel the common factor of $2$ and $6$ .

Factor $2$ out of $2$ .

$z = \frac{2 (1)}{6}$

Cancel the common factors.

Factor $2$ out of $6$ .

$z = \frac{2 \cdot 1}{2 \cdot 3}$

Cancel the common factor.

$z = \frac{2 \cdot 1}{2 \cdot 3}$

Rewrite the expression.

$z = \frac{1}{3}$

Next, use the negative value of the $\pm$ to find the second solution.

$z + \frac{11}{6} = - \frac{13}{6}$

Move all terms not containing $z$ to the right side of the equation.

Subtract $\frac{11}{6}$ from both sides of the equation.

$z = - \frac{13}{6} - \frac{11}{6}$

Combine the numerators over the common denominator.

$z = \frac{- 13 - 11}{6}$

Subtract $11$ from $- 13$ .

$z = \frac{- 24}{6}$

Divide $- 24$ by $6$ .

$z = - 4$

The complete solution is the result of both the positive and negative portions of the solution.

$z = \frac{1}{3}, - 4$

Solve by Completing the Square 3z^2+11z-4=0

Related Questions

Support LCMGCF.com by disabling your adblocker.