Solve by Completing the Square 5z^2+9z-2=0

$5 z^{2} + 9 z - 2 = 0$

Step 1

Add $2$ to both sides of the equation.

$5 z^{2} + 9 z = 2$

Step 2

Divide each term in $5 z^{2} + 9 z = 2$ by $5$ and simplify.

Divide each term in $5 z^{2} + 9 z = 2$ by $5$ .

$\frac{5 z^{2}}{5} + \frac{9 z}{5} = \frac{2}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 z^{2}}{5} + \frac{9 z}{5} = \frac{2}{5}$

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

$z^{2} + \frac{9 z}{5} = \frac{2}{5}$

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 $b$ .

${(\frac{b}{2})}^{2} = {(\frac{9}{10})}^{2}$

Step 4

Add the term to each side of the equation.

$z^{2} + \frac{9 z}{5} + {(\frac{9}{10})}^{2} = \frac{2}{5} + {(\frac{9}{10})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{9}{10}$ .

$z^{2} + \frac{9 z}{5} + \frac{9^{2}}{10^{2}} = \frac{2}{5} + {(\frac{9}{10})}^{2}$

Raise $9$ to the power of $2$ .

$z^{2} + \frac{9 z}{5} + \frac{81}{10^{2}} = \frac{2}{5} + {(\frac{9}{10})}^{2}$

Raise $10$ to the power of $2$ .

$z^{2} + \frac{9 z}{5} + \frac{81}{100} = \frac{2}{5} + {(\frac{9}{10})}^{2}$

Simplify the right side.

Simplify $\frac{2}{5} + {(\frac{9}{10})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{9}{10}$ .

$z^{2} + \frac{9 z}{5} + \frac{81}{100} = \frac{2}{5} + \frac{9^{2}}{10^{2}}$

Raise $9$ to the power of $2$ .

$z^{2} + \frac{9 z}{5} + \frac{81}{100} = \frac{2}{5} + \frac{81}{10^{2}}$

Raise $10$ to the power of $2$ .

$z^{2} + \frac{9 z}{5} + \frac{81}{100} = \frac{2}{5} + \frac{81}{100}$

To write $\frac{2}{5}$ as a fraction with a common denominator, multiply by $\frac{20}{20}$ .

$z^{2} + \frac{9 z}{5} + \frac{81}{100} = \frac{2}{5} \cdot \frac{20}{20} + \frac{81}{100}$

Write each expression with a common denominator of $100$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2}{5}$ by $\frac{20}{20}$ .

$z^{2} + \frac{9 z}{5} + \frac{81}{100} = \frac{2 \cdot 20}{5 \cdot 20} + \frac{81}{100}$

Multiply $5$ by $20$ .

$z^{2} + \frac{9 z}{5} + \frac{81}{100} = \frac{2 \cdot 20}{100} + \frac{81}{100}$

Combine the numerators over the common denominator.

$z^{2} + \frac{9 z}{5} + \frac{81}{100} = \frac{2 \cdot 20 + 81}{100}$

Simplify the numerator.

Multiply $2$ by $20$ .

$z^{2} + \frac{9 z}{5} + \frac{81}{100} = \frac{40 + 81}{100}$

Add $40$ and $81$ .

$z^{2} + \frac{9 z}{5} + \frac{81}{100} = \frac{121}{100}$

Step 6

Factor the perfect trinomial square into ${(z + \frac{9}{10})}^{2}$ .

${(z + \frac{9}{10})}^{2} = \frac{121}{100}$

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.

$z + \frac{9}{10} = \pm \sqrt{\frac{121}{100}}$

Simplify $\pm \sqrt{\frac{121}{100}}$ .

Rewrite $\sqrt{\frac{121}{100}}$ as $\frac{\sqrt{121}}{\sqrt{100}}$ .

$z + \frac{9}{10} = \pm \frac{\sqrt{121}}{\sqrt{100}}$

Simplify the numerator.

Rewrite $121$ as $11^{2}$ .

$z + \frac{9}{10} = \pm \frac{\sqrt{11^{2}}}{\sqrt{100}}$

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

$z + \frac{9}{10} = \pm \frac{11}{\sqrt{100}}$

Simplify the denominator.

Rewrite $100$ as $10^{2}$ .

$z + \frac{9}{10} = \pm \frac{11}{\sqrt{10^{2}}}$

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

$z + \frac{9}{10} = \pm \frac{11}{10}$

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.

$z + \frac{9}{10} = \frac{11}{10}$

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

Subtract $\frac{9}{10}$ from both sides of the equation.

$z = \frac{11}{10} - \frac{9}{10}$

Combine the numerators over the common denominator.

$z = \frac{11 - 9}{10}$

Subtract $9$ from $11$ .

$z = \frac{2}{10}$

Cancel the common factor of $2$ and $10$ .

Factor $2$ out of $2$ .

$z = \frac{2 (1)}{10}$

Cancel the common factors.

Factor $2$ out of $10$ .

$z = \frac{2 \cdot 1}{2 \cdot 5}$

Cancel the common factor.

$z = \frac{2 \cdot 1}{2 \cdot 5}$

Rewrite the expression.

$z = \frac{1}{5}$

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

$z + \frac{9}{10} = - \frac{11}{10}$

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

Subtract $\frac{9}{10}$ from both sides of the equation.

$z = - \frac{11}{10} - \frac{9}{10}$

Combine the numerators over the common denominator.

$z = \frac{- 11 - 9}{10}$

Subtract $9$ from $- 11$ .

$z = \frac{- 20}{10}$

Divide $- 20$ by $10$ .

$z = - 2$

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

$z = \frac{1}{5}, - 2$

Solve by Completing the Square 5z^2+9z-2=0

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