Solve by Completing the Square -9x^2+36x=41

$- 9 x^{2} + 36 x = 41$

Step 1

Divide each term in $- 9 x^{2} + 36 x = 41$ by $- 9$ and simplify.

Divide each term in $- 9 x^{2} + 36 x = 41$ by $- 9$ .

$\frac{- 9 x^{2}}{- 9} + \frac{36 x}{- 9} = \frac{41}{- 9}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $- 9$ .

Cancel the common factor.

$\frac{- 9 x^{2}}{- 9} + \frac{36 x}{- 9} = \frac{41}{- 9}$

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

$x^{2} + \frac{36 x}{- 9} = \frac{41}{- 9}$

Cancel the common factor of $36$ and $- 9$ .

Factor $9$ out of $36 x$ .

$x^{2} + \frac{9 (4 x)}{- 9} = \frac{41}{- 9}$

Move the negative one from the denominator of $\frac{4 x}{- 1}$ .

$x^{2} - 1 \cdot (4 x) = \frac{41}{- 9}$

Rewrite $- 1 \cdot (4 x)$ as $- (4 x)$ .

$x^{2} - (4 x) = \frac{41}{- 9}$

Multiply $4$ by $- 1$ .

$x^{2} - 4 x = \frac{41}{- 9}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} - 4 x = - \frac{41}{9}$

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 $b$ .

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

Step 3

Add the term to each side of the equation.

$x^{2} - 4 x + {(- 2)}^{2} = - \frac{41}{9} + {(- 2)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} - 4 x + 4 = - \frac{41}{9} + {(- 2)}^{2}$

Simplify the right side.

Simplify $- \frac{41}{9} + {(- 2)}^{2}$ .

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

$x^{2} - 4 x + 4 = - \frac{41}{9} + 4$

To write $4$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$x^{2} - 4 x + 4 = - \frac{41}{9} + 4 \cdot \frac{9}{9}$

Combine $4$ and $\frac{9}{9}$ .

$x^{2} - 4 x + 4 = - \frac{41}{9} + \frac{4 \cdot 9}{9}$

Combine the numerators over the common denominator.

$x^{2} - 4 x + 4 = \frac{- 41 + 4 \cdot 9}{9}$

Simplify the numerator.

Multiply $4$ by $9$ .

$x^{2} - 4 x + 4 = \frac{- 41 + 36}{9}$

Add $- 41$ and $36$ .

$x^{2} - 4 x + 4 = \frac{- 5}{9}$

Move the negative in front of the fraction.

$x^{2} - 4 x + 4 = - \frac{5}{9}$

Step 5

Factor the perfect trinomial square into ${(x - 2)}^{2}$ .

${(x - 2)}^{2} = - \frac{5}{9}$

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.

$x - 2 = \pm \sqrt{- \frac{5}{9}}$

Simplify $\pm \sqrt{- \frac{5}{9}}$ .

Rewrite $- \frac{5}{9}$ as ${(\frac{i}{3})}^{2} \cdot 5$ .

Rewrite $- 1$ as $i^{2}$ .

$x - 2 = \pm \sqrt{i^{2} \frac{5}{9}}$

Factor the perfect power $1^{2}$ out of $5$ .

$x - 2 = \pm \sqrt{i^{2} \frac{1^{2} \cdot 5}{9}}$

Factor the perfect power $3^{2}$ out of $9$ .

$x - 2 = \pm \sqrt{i^{2} \frac{1^{2} \cdot 5}{3^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} \cdot 5}{3^{2} \cdot 1}$ .

$x - 2 = \pm \sqrt{i^{2} ({(\frac{1}{3})}^{2} \cdot 5)}$

Rewrite $i^{2} {(\frac{1}{3})}^{2}$ as ${(\frac{i}{3})}^{2}$ .

$x - 2 = \pm \sqrt{{(\frac{i}{3})}^{2} \cdot 5}$

Pull terms out from under the radical.

$x - 2 = \pm \frac{i}{3} \sqrt{5}$

Combine $\frac{i}{3}$ and $\sqrt{5}$ .

$x - 2 = \pm \frac{i \sqrt{5}}{3}$

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

Add $2$ to both sides of the equation.

$x = \pm \frac{i \sqrt{5}}{3} + 2$

Reorder $\pm \frac{i \sqrt{5}}{3}$ and $2$ .

$x = 2 \pm \frac{i \sqrt{5}}{3}$

Solve by Completing the Square -9x^2+36x=41

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