Solve by Completing the Square 4x^2-9x-9=0

$4 x^{2} - 9 x - 9 = 0$

Step 1

Add $9$ to both sides of the equation.

$4 x^{2} - 9 x = 9$

Step 2

Divide each term in $4 x^{2} - 9 x = 9$ by $4$ and simplify.

Divide each term in $4 x^{2} - 9 x = 9$ by $4$ .

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

Simplify the left side.

Simplify each term.

Cancel the common factor of $4$ .

Cancel the common factor.

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

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

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

Move the negative in front of the fraction.

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

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 $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{9}{8})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} - \frac{9 x}{4} + {(- \frac{9}{8})}^{2} = \frac{9}{4} + {(- \frac{9}{8})}^{2}$

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 $- \frac{9}{8}$ .

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

Apply the product rule to $\frac{9}{8}$ .

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

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

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

Multiply $\frac{9^{2}}{8^{2}}$ by $1$ .

$x^{2} - \frac{9 x}{4} + \frac{9^{2}}{8^{2}} = \frac{9}{4} + {(- \frac{9}{8})}^{2}$

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

$x^{2} - \frac{9 x}{4} + \frac{81}{8^{2}} = \frac{9}{4} + {(- \frac{9}{8})}^{2}$

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

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9}{4} + {(- \frac{9}{8})}^{2}$

Simplify the right side.

Simplify $\frac{9}{4} + {(- \frac{9}{8})}^{2}$ .

Simplify each term.

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

Apply the product rule to $- \frac{9}{8}$ .

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9}{4} + {(- 1)}^{2} {(\frac{9}{8})}^{2}$

Apply the product rule to $\frac{9}{8}$ .

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9}{4} + {(- 1)}^{2} \frac{9^{2}}{8^{2}}$

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

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9}{4} + 1 \frac{9^{2}}{8^{2}}$

Multiply $\frac{9^{2}}{8^{2}}$ by $1$ .

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9}{4} + \frac{9^{2}}{8^{2}}$

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

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9}{4} + \frac{81}{8^{2}}$

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

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9}{4} + \frac{81}{64}$

To write $\frac{9}{4}$ as a fraction with a common denominator, multiply by $\frac{16}{16}$ .

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9}{4} \cdot \frac{16}{16} + \frac{81}{64}$

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

Multiply $\frac{9}{4}$ by $\frac{16}{16}$ .

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9 \cdot 16}{4 \cdot 16} + \frac{81}{64}$

Multiply $4$ by $16$ .

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9 \cdot 16}{64} + \frac{81}{64}$

Combine the numerators over the common denominator.

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{9 \cdot 16 + 81}{64}$

Simplify the numerator.

Multiply $9$ by $16$ .

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{144 + 81}{64}$

Add $144$ and $81$ .

$x^{2} - \frac{9 x}{4} + \frac{81}{64} = \frac{225}{64}$

Step 6

Factor the perfect trinomial square into ${(x - \frac{9}{8})}^{2}$ .

${(x - \frac{9}{8})}^{2} = \frac{225}{64}$

Step 7

Solve the equation for $x$ .

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

$x - \frac{9}{8} = \pm \sqrt{\frac{225}{64}}$

Simplify $\pm \sqrt{\frac{225}{64}}$ .

Rewrite $\sqrt{\frac{225}{64}}$ as $\frac{\sqrt{225}}{\sqrt{64}}$ .

$x - \frac{9}{8} = \pm \frac{\sqrt{225}}{\sqrt{64}}$

Simplify the numerator.

Rewrite $225$ as $15^{2}$ .

$x - \frac{9}{8} = \pm \frac{\sqrt{15^{2}}}{\sqrt{64}}$

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

$x - \frac{9}{8} = \pm \frac{15}{\sqrt{64}}$

Simplify the denominator.

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

$x - \frac{9}{8} = \pm \frac{15}{\sqrt{8^{2}}}$

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

$x - \frac{9}{8} = \pm \frac{15}{8}$

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.

$x - \frac{9}{8} = \frac{15}{8}$

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

Add $\frac{9}{8}$ to both sides of the equation.

$x = \frac{15}{8} + \frac{9}{8}$

Combine the numerators over the common denominator.

$x = \frac{15 + 9}{8}$

Add $15$ and $9$ .

$x = \frac{24}{8}$

Divide $24$ by $8$ .

$x = 3$

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

$x - \frac{9}{8} = - \frac{15}{8}$

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

Add $\frac{9}{8}$ to both sides of the equation.

$x = - \frac{15}{8} + \frac{9}{8}$

Combine the numerators over the common denominator.

$x = \frac{- 15 + 9}{8}$

Add $- 15$ and $9$ .

$x = \frac{- 6}{8}$

Cancel the common factor of $- 6$ and $8$ .

Factor $2$ out of $- 6$ .

$x = \frac{2 (- 3)}{8}$

Cancel the common factors.

Factor $2$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

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

Solve by Completing the Square 4x^2-9x-9=0

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