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

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

Step 1

Add $3$ to both sides of the equation.

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

Step 2

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

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

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

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

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

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 $b$ .

${(\frac{b}{2})}^{2} = {(\frac{11}{8})}^{2}$

Step 4

Add the term to each side of the equation.

$z^{2} + \frac{11 z}{4} + {(\frac{11}{8})}^{2} = \frac{3}{4} + {(\frac{11}{8})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{11}{8}$ .

$z^{2} + \frac{11 z}{4} + \frac{11^{2}}{8^{2}} = \frac{3}{4} + {(\frac{11}{8})}^{2}$

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

$z^{2} + \frac{11 z}{4} + \frac{121}{8^{2}} = \frac{3}{4} + {(\frac{11}{8})}^{2}$

Raise $8$ to the power of $2$ .

$z^{2} + \frac{11 z}{4} + \frac{121}{64} = \frac{3}{4} + {(\frac{11}{8})}^{2}$

Simplify the right side.

Simplify $\frac{3}{4} + {(\frac{11}{8})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{11}{8}$ .

$z^{2} + \frac{11 z}{4} + \frac{121}{64} = \frac{3}{4} + \frac{11^{2}}{8^{2}}$

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

$z^{2} + \frac{11 z}{4} + \frac{121}{64} = \frac{3}{4} + \frac{121}{8^{2}}$

Raise $8$ to the power of $2$ .

$z^{2} + \frac{11 z}{4} + \frac{121}{64} = \frac{3}{4} + \frac{121}{64}$

To write $\frac{3}{4}$ as a fraction with a common denominator, multiply by $\frac{16}{16}$ .

$z^{2} + \frac{11 z}{4} + \frac{121}{64} = \frac{3}{4} \cdot \frac{16}{16} + \frac{121}{64}$

Write each expression with a common denominator of $64$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3}{4}$ by $\frac{16}{16}$ .

$z^{2} + \frac{11 z}{4} + \frac{121}{64} = \frac{3 \cdot 16}{4 \cdot 16} + \frac{121}{64}$

Multiply $4$ by $16$ .

$z^{2} + \frac{11 z}{4} + \frac{121}{64} = \frac{3 \cdot 16}{64} + \frac{121}{64}$

Combine the numerators over the common denominator.

$z^{2} + \frac{11 z}{4} + \frac{121}{64} = \frac{3 \cdot 16 + 121}{64}$

Simplify the numerator.

Multiply $3$ by $16$ .

$z^{2} + \frac{11 z}{4} + \frac{121}{64} = \frac{48 + 121}{64}$

Add $48$ and $121$ .

$z^{2} + \frac{11 z}{4} + \frac{121}{64} = \frac{169}{64}$

Step 6

Factor the perfect trinomial square into ${(z + \frac{11}{8})}^{2}$ .

${(z + \frac{11}{8})}^{2} = \frac{169}{64}$

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.

$z + \frac{11}{8} = \pm \sqrt{\frac{169}{64}}$

Simplify $\pm \sqrt{\frac{169}{64}}$ .

Rewrite $\sqrt{\frac{169}{64}}$ as $\frac{\sqrt{169}}{\sqrt{64}}$ .

$z + \frac{11}{8} = \pm \frac{\sqrt{169}}{\sqrt{64}}$

Simplify the numerator.

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

$z + \frac{11}{8} = \pm \frac{\sqrt{13^{2}}}{\sqrt{64}}$

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

$z + \frac{11}{8} = \pm \frac{13}{\sqrt{64}}$

Simplify the denominator.

Rewrite $64$ as $8^{2}$ .

$z + \frac{11}{8} = \pm \frac{13}{\sqrt{8^{2}}}$

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

$z + \frac{11}{8} = \pm \frac{13}{8}$

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.

$z + \frac{11}{8} = \frac{13}{8}$

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

Subtract $\frac{11}{8}$ from both sides of the equation.

$z = \frac{13}{8} - \frac{11}{8}$

Combine the numerators over the common denominator.

$z = \frac{13 - 11}{8}$

Subtract $11$ from $13$ .

$z = \frac{2}{8}$

Cancel the common factor of $2$ and $8$ .

Factor $2$ out of $2$ .

$z = \frac{2 (1)}{8}$

Cancel the common factors.

Factor $2$ out of $8$ .

$z = \frac{2 \cdot 1}{2 \cdot 4}$

Cancel the common factor.

$z = \frac{2 \cdot 1}{2 \cdot 4}$

Rewrite the expression.

$z = \frac{1}{4}$

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

$z + \frac{11}{8} = - \frac{13}{8}$

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

Subtract $\frac{11}{8}$ from both sides of the equation.

$z = - \frac{13}{8} - \frac{11}{8}$

Combine the numerators over the common denominator.

$z = \frac{- 13 - 11}{8}$

Subtract $11$ from $- 13$ .

$z = \frac{- 24}{8}$

Divide $- 24$ by $8$ .

$z = - 3$

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

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

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

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