Precalculus Examples

$4 x^{2} = - 16 x + 9$

Step 1

Add $16 x$ to both sides of the equation.

$4 x^{2} + 16 x = 9$

Step 2

Divide each term in $4 x^{2} + 16 x = 9$ by $4$ and simplify.

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Step 2.1

Divide each term in $4 x^{2} + 16 x = 9$ by $4$ .

$\frac{4 x^{2}}{4} + \frac{16 x}{4} = \frac{9}{4}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Cancel the common factor of $4$ .

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Step 2.2.1.1.1

Cancel the common factor.

$\frac{4 x^{2}}{4} + \frac{16 x}{4} = \frac{9}{4}$

Step 2.2.1.1.2

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

$x^{2} + \frac{16 x}{4} = \frac{9}{4}$

Step 2.2.1.2

Cancel the common factor of $16$ and $4$ .

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Step 2.2.1.2.1

Factor $4$ out of $16 x$ .

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

Step 2.2.1.2.2

Cancel the common factors.

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Step 2.2.1.2.2.1

Factor $4$ out of $4$ .

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

Step 2.2.1.2.2.2

Cancel the common factor.

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

Step 2.2.1.2.2.3

Rewrite the expression.

$x^{2} + \frac{4 x}{1} = \frac{9}{4}$

Step 2.2.1.2.2.4

Divide $4 x$ by $1$ .

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

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 $b$ .

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

Step 4

Add the term to each side of the equation.

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

Step 5

Simplify the equation.

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Step 5.1

Simplify the left side.

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Step 5.1.1

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

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

Step 5.2

Simplify the right side.

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Step 5.2.1

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

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Step 5.2.1.1

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

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

Step 5.2.1.2

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

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

Step 5.2.1.3

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

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

Step 5.2.1.4

Combine the numerators over the common denominator.

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

Step 5.2.1.5

Simplify the numerator.

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Step 5.2.1.5.1

Multiply $4$ by $4$ .

$x^{2} + 4 x + 4 = \frac{9 + 16}{4}$

Step 5.2.1.5.2

Add $9$ and $16$ .

$x^{2} + 4 x + 4 = \frac{25}{4}$

Step 6

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

${(x + 2)}^{2} = \frac{25}{4}$

Step 7

Solve the equation for $x$ .

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Step 7.1

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

$x + 2 = \pm \sqrt{\frac{25}{4}}$

Step 7.2

Simplify $\pm \sqrt{\frac{25}{4}}$ .

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Step 7.2.1

Rewrite $\sqrt{\frac{25}{4}}$ as $\frac{\sqrt{25}}{\sqrt{4}}$ .

$x + 2 = \pm \frac{\sqrt{25}}{\sqrt{4}}$

Step 7.2.2

Simplify the numerator.

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Step 7.2.2.1

Rewrite $25$ as $5^{2}$ .

$x + 2 = \pm \frac{\sqrt{5^{2}}}{\sqrt{4}}$

Step 7.2.2.2

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

$x + 2 = \pm \frac{5}{\sqrt{4}}$

Step 7.2.3

Simplify the denominator.

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Step 7.2.3.1

Rewrite $4$ as $2^{2}$ .

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

Step 7.2.3.2

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

$x + 2 = \pm \frac{5}{2}$

Step 7.3

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

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Step 7.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x + 2 = \frac{5}{2}$

Step 7.3.2

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

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Step 7.3.2.1

Subtract $2$ from both sides of the equation.

$x = \frac{5}{2} - 2$

Step 7.3.2.2

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

$x = \frac{5}{2} - 2 \cdot \frac{2}{2}$

Step 7.3.2.3

Combine $- 2$ and $\frac{2}{2}$ .

$x = \frac{5}{2} + \frac{- 2 \cdot 2}{2}$

Step 7.3.2.4

Combine the numerators over the common denominator.

$x = \frac{5 - 2 \cdot 2}{2}$

Step 7.3.2.5

Simplify the numerator.

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Step 7.3.2.5.1

Multiply $- 2$ by $2$ .

$x = \frac{5 - 4}{2}$

Step 7.3.2.5.2

Subtract $4$ from $5$ .

$x = \frac{1}{2}$

Step 7.3.3

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

$x + 2 = - \frac{5}{2}$

Step 7.3.4

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

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Step 7.3.4.1

Subtract $2$ from both sides of the equation.

$x = - \frac{5}{2} - 2$

Step 7.3.4.2

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

$x = - \frac{5}{2} - 2 \cdot \frac{2}{2}$

Step 7.3.4.3

Combine $- 2$ and $\frac{2}{2}$ .

$x = - \frac{5}{2} + \frac{- 2 \cdot 2}{2}$

Step 7.3.4.4

Combine the numerators over the common denominator.

$x = \frac{- 5 - 2 \cdot 2}{2}$

Step 7.3.4.5

Simplify the numerator.

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Step 7.3.4.5.1

Multiply $- 2$ by $2$ .

$x = \frac{- 5 - 4}{2}$

Step 7.3.4.5.2

Subtract $4$ from $- 5$ .

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

Step 7.3.4.6

Move the negative in front of the fraction.

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

Step 7.3.5

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

$x = \frac{1}{2}, - \frac{9}{2}$