Solve by Completing the Square 9x^2+6x=8

$9 x^{2} + 6 x = 8$

Step 1

Divide each term in $9 x^{2} + 6 x = 8$ by $9$ and simplify.

Divide each term in $9 x^{2} + 6 x = 8$ by $9$ .

$\frac{9 x^{2}}{9} + \frac{6 x}{9} = \frac{8}{9}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x^{2}}{9} + \frac{6 x}{9} = \frac{8}{9}$

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

$x^{2} + \frac{6 x}{9} = \frac{8}{9}$

Cancel the common factor of $6$ and $9$ .

Factor $3$ out of $6 x$ .

$x^{2} + \frac{3 (2 x)}{9} = \frac{8}{9}$

Cancel the common factors.

Factor $3$ out of $9$ .

$x^{2} + \frac{3 (2 x)}{3 (3)} = \frac{8}{9}$

Cancel the common factor.

$x^{2} + \frac{3 (2 x)}{3 \cdot 3} = \frac{8}{9}$

Rewrite the expression.

$x^{2} + \frac{2 x}{3} = \frac{8}{9}$

Step 2

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

${(\frac{b}{2})}^{2} = {(\frac{1}{3})}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + \frac{2 x}{3} + {(\frac{1}{3})}^{2} = \frac{8}{9} + {(\frac{1}{3})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{1}{3}$ .

$x^{2} + \frac{2 x}{3} + \frac{1^{2}}{3^{2}} = \frac{8}{9} + {(\frac{1}{3})}^{2}$

One to any power is one.

$x^{2} + \frac{2 x}{3} + \frac{1}{3^{2}} = \frac{8}{9} + {(\frac{1}{3})}^{2}$

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

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + {(\frac{1}{3})}^{2}$

Simplify the right side.

Simplify $\frac{8}{9} + {(\frac{1}{3})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{1}{3}$ .

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + \frac{1^{2}}{3^{2}}$

One to any power is one.

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + \frac{1}{3^{2}}$

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

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + \frac{1}{9}$

Combine fractions.

Combine the numerators over the common denominator.

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = \frac{8 + 1}{9}$

Simplify the expression.

Add $8$ and $1$ .

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

Divide $9$ by $9$ .

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

Step 5

Factor the perfect trinomial square into ${(x + \frac{1}{3})}^{2}$ .

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

Step 6

Solve the equation for $x$ .

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

$x + \frac{1}{3} = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x + \frac{1}{3} = \pm 1$

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.

$x + \frac{1}{3} = 1$

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

Subtract $\frac{1}{3}$ from both sides of the equation.

$x = 1 - \frac{1}{3}$

Write $1$ as a fraction with a common denominator.

$x = \frac{3}{3} - \frac{1}{3}$

Combine the numerators over the common denominator.

$x = \frac{3 - 1}{3}$

Subtract $1$ from $3$ .

$x = \frac{2}{3}$

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

$x + \frac{1}{3} = - 1$

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

Subtract $\frac{1}{3}$ from both sides of the equation.

$x = - 1 - \frac{1}{3}$

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

$x = - 1 \cdot \frac{3}{3} - \frac{1}{3}$

Combine $- 1$ and $\frac{3}{3}$ .

$x = \frac{- 1 \cdot 3}{3} - \frac{1}{3}$

Combine the numerators over the common denominator.

$x = \frac{- 1 \cdot 3 - 1}{3}$

Simplify the numerator.

Multiply $- 1$ by $3$ .

$x = \frac{- 3 - 1}{3}$

Subtract $1$ from $- 3$ .

$x = \frac{- 4}{3}$

Move the negative in front of the fraction.

$x = - \frac{4}{3}$

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

$x = \frac{2}{3}, - \frac{4}{3}$

Solve by Completing the Square 9x^2+6x=8

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