Solve by Completing the Square 3z^2-8z+1=0

$3 z^{2} - 8 z + 1 = 0$

Step 1

Subtract $1$ from both sides of the equation.

$3 z^{2} - 8 z = - 1$

Step 2

Divide each term in $3 z^{2} - 8 z = - 1$ by $3$ and simplify.

Divide each term in $3 z^{2} - 8 z = - 1$ by $3$ .

$\frac{3 z^{2}}{3} + \frac{- 8 z}{3} = \frac{- 1}{3}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 z^{2}}{3} + \frac{- 8 z}{3} = \frac{- 1}{3}$

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

$z^{2} + \frac{- 8 z}{3} = \frac{- 1}{3}$

Move the negative in front of the fraction.

$z^{2} - \frac{8 z}{3} = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$z^{2} - \frac{8 z}{3} = - \frac{1}{3}$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{4}{3})}^{2}$

Step 4

Add the term to each side of the equation.

$z^{2} - \frac{8 z}{3} + {(- \frac{4}{3})}^{2} = - \frac{1}{3} + {(- \frac{4}{3})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{4}{3}$ .

$z^{2} - \frac{8 z}{3} + {(- 1)}^{2} {(\frac{4}{3})}^{2} = - \frac{1}{3} + {(- \frac{4}{3})}^{2}$

Apply the product rule to $\frac{4}{3}$ .

$z^{2} - \frac{8 z}{3} + {(- 1)}^{2} \frac{4^{2}}{3^{2}} = - \frac{1}{3} + {(- \frac{4}{3})}^{2}$

Raise $- 1$ to the power of $2$ .

$z^{2} - \frac{8 z}{3} + 1 \frac{4^{2}}{3^{2}} = - \frac{1}{3} + {(- \frac{4}{3})}^{2}$

Multiply $\frac{4^{2}}{3^{2}}$ by $1$ .

$z^{2} - \frac{8 z}{3} + \frac{4^{2}}{3^{2}} = - \frac{1}{3} + {(- \frac{4}{3})}^{2}$

Raise $4$ to the power of $2$ .

$z^{2} - \frac{8 z}{3} + \frac{16}{3^{2}} = - \frac{1}{3} + {(- \frac{4}{3})}^{2}$

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

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = - \frac{1}{3} + {(- \frac{4}{3})}^{2}$

Simplify the right side.

Simplify $- \frac{1}{3} + {(- \frac{4}{3})}^{2}$ .

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{4}{3}$ .

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = - \frac{1}{3} + {(- 1)}^{2} {(\frac{4}{3})}^{2}$

Apply the product rule to $\frac{4}{3}$ .

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = - \frac{1}{3} + {(- 1)}^{2} \frac{4^{2}}{3^{2}}$

Raise $- 1$ to the power of $2$ .

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = - \frac{1}{3} + 1 \frac{4^{2}}{3^{2}}$

Multiply $\frac{4^{2}}{3^{2}}$ by $1$ .

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = - \frac{1}{3} + \frac{4^{2}}{3^{2}}$

Raise $4$ to the power of $2$ .

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = - \frac{1}{3} + \frac{16}{3^{2}}$

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

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = - \frac{1}{3} + \frac{16}{9}$

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

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = - \frac{1}{3} \cdot \frac{3}{3} + \frac{16}{9}$

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

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

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = - \frac{3}{3 \cdot 3} + \frac{16}{9}$

Multiply $3$ by $3$ .

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = - \frac{3}{9} + \frac{16}{9}$

Combine the numerators over the common denominator.

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = \frac{- 3 + 16}{9}$

Add $- 3$ and $16$ .

$z^{2} - \frac{8 z}{3} + \frac{16}{9} = \frac{13}{9}$

Step 6

Factor the perfect trinomial square into ${(z - \frac{4}{3})}^{2}$ .

${(z - \frac{4}{3})}^{2} = \frac{13}{9}$

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.

$z - \frac{4}{3} = \pm \sqrt{\frac{13}{9}}$

Simplify $\pm \sqrt{\frac{13}{9}}$ .

Rewrite $\sqrt{\frac{13}{9}}$ as $\frac{\sqrt{13}}{\sqrt{9}}$ .

$z - \frac{4}{3} = \pm \frac{\sqrt{13}}{\sqrt{9}}$

Simplify the denominator.

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

$z - \frac{4}{3} = \pm \frac{\sqrt{13}}{\sqrt{3^{2}}}$

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

$z - \frac{4}{3} = \pm \frac{\sqrt{13}}{3}$

Add $\frac{4}{3}$ to both sides of the equation.

$z = \pm \frac{\sqrt{13}}{3} + \frac{4}{3}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$z = \pm \frac{\sqrt{13}}{3} + \frac{4}{3}$

Decimal Form:

$z = 2.53518375 \dots, 0.13148290 \dots$

Solve by Completing the Square 3z^2-8z+1=0

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