Precalculus Examples

$x (2 x - 3) = 93$

Step 1

Simplify $x (2 x - 3)$ .

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Step 1.1

Simplify by multiplying through.

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Step 1.1.1

Apply the distributive property.

$x (2 x) + x \cdot - 3 = 93$

Step 1.1.2

Reorder.

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Step 1.1.2.1

Rewrite using the commutative property of multiplication.

$2 x \cdot x + x \cdot - 3 = 93$

Step 1.1.2.2

Move $- 3$ to the left of $x$ .

$2 x \cdot x - 3 \cdot x = 93$

Step 1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 1.2.1

Move $x$ .

$2 (x \cdot x) - 3 \cdot x = 93$

Step 1.2.2

Multiply $x$ by $x$ .

$2 x^{2} - 3 \cdot x = 93$

$2 x^{2} - 3 x = 93$

Step 2

Divide each term in $2 x^{2} - 3 x = 93$ by $2$ and simplify.

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Step 2.1

Divide each term in $2 x^{2} - 3 x = 93$ by $2$ .

$\frac{2 x^{2}}{2} + \frac{- 3 x}{2} = \frac{93}{2}$

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 $2$ .

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Step 2.2.1.1.1

Cancel the common factor.

$\frac{2 x^{2}}{2} + \frac{- 3 x}{2} = \frac{93}{2}$

Step 2.2.1.1.2

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

$x^{2} + \frac{- 3 x}{2} = \frac{93}{2}$

Step 2.2.1.2

Move the negative in front of the fraction.

$x^{2} - \frac{3 x}{2} = \frac{93}{2}$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{3}{4})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} - \frac{3 x}{2} + {(- \frac{3}{4})}^{2} = \frac{93}{2} + {(- \frac{3}{4})}^{2}$

Step 5

Simplify the equation.

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Step 5.1

Simplify the left side.

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Step 5.1.1

Simplify each term.

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Step 5.1.1.1

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

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Step 5.1.1.1.1

Apply the product rule to $- \frac{3}{4}$ .

$x^{2} - \frac{3 x}{2} + {(- 1)}^{2} {(\frac{3}{4})}^{2} = \frac{93}{2} + {(- \frac{3}{4})}^{2}$

Step 5.1.1.1.2

Apply the product rule to $\frac{3}{4}$ .

$x^{2} - \frac{3 x}{2} + {(- 1)}^{2} \frac{3^{2}}{4^{2}} = \frac{93}{2} + {(- \frac{3}{4})}^{2}$

Step 5.1.1.2

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

$x^{2} - \frac{3 x}{2} + 1 \frac{3^{2}}{4^{2}} = \frac{93}{2} + {(- \frac{3}{4})}^{2}$

Step 5.1.1.3

Multiply $\frac{3^{2}}{4^{2}}$ by $1$ .

$x^{2} - \frac{3 x}{2} + \frac{3^{2}}{4^{2}} = \frac{93}{2} + {(- \frac{3}{4})}^{2}$

Step 5.1.1.4

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

$x^{2} - \frac{3 x}{2} + \frac{9}{4^{2}} = \frac{93}{2} + {(- \frac{3}{4})}^{2}$

Step 5.1.1.5

Raise $4$ to the power of $2$ .

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93}{2} + {(- \frac{3}{4})}^{2}$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $\frac{93}{2} + {(- \frac{3}{4})}^{2}$ .

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Step 5.2.1.1

Simplify each term.

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Step 5.2.1.1.1

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

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Step 5.2.1.1.1.1

Apply the product rule to $- \frac{3}{4}$ .

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93}{2} + {(- 1)}^{2} {(\frac{3}{4})}^{2}$

Step 5.2.1.1.1.2

Apply the product rule to $\frac{3}{4}$ .

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93}{2} + {(- 1)}^{2} \frac{3^{2}}{4^{2}}$

Step 5.2.1.1.2

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

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93}{2} + 1 \frac{3^{2}}{4^{2}}$

Step 5.2.1.1.3

Multiply $\frac{3^{2}}{4^{2}}$ by $1$ .

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93}{2} + \frac{3^{2}}{4^{2}}$

Step 5.2.1.1.4

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

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93}{2} + \frac{9}{4^{2}}$

Step 5.2.1.1.5

Raise $4$ to the power of $2$ .

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93}{2} + \frac{9}{16}$

Step 5.2.1.2

To write $\frac{93}{2}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93}{2} \cdot \frac{8}{8} + \frac{9}{16}$

Step 5.2.1.3

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.2.1.3.1

Multiply $\frac{93}{2}$ by $\frac{8}{8}$ .

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93 \cdot 8}{2 \cdot 8} + \frac{9}{16}$

Step 5.2.1.3.2

Multiply $2$ by $8$ .

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93 \cdot 8}{16} + \frac{9}{16}$

Step 5.2.1.4

Combine the numerators over the common denominator.

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{93 \cdot 8 + 9}{16}$

Step 5.2.1.5

Simplify the numerator.

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Step 5.2.1.5.1

Multiply $93$ by $8$ .

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{744 + 9}{16}$

Step 5.2.1.5.2

Add $744$ and $9$ .

$x^{2} - \frac{3 x}{2} + \frac{9}{16} = \frac{753}{16}$

Step 6

Factor the perfect trinomial square into ${(x - \frac{3}{4})}^{2}$ .

${(x - \frac{3}{4})}^{2} = \frac{753}{16}$

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.

$x - \frac{3}{4} = \pm \sqrt{\frac{753}{16}}$

Step 7.2

Simplify $\pm \sqrt{\frac{753}{16}}$ .

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Step 7.2.1

Rewrite $\sqrt{\frac{753}{16}}$ as $\frac{\sqrt{753}}{\sqrt{16}}$ .

$x - \frac{3}{4} = \pm \frac{\sqrt{753}}{\sqrt{16}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

Rewrite $16$ as $4^{2}$ .

$x - \frac{3}{4} = \pm \frac{\sqrt{753}}{\sqrt{4^{2}}}$

Step 7.2.2.2

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

$x - \frac{3}{4} = \pm \frac{\sqrt{753}}{4}$

Step 7.3

Add $\frac{3}{4}$ to both sides of the equation.

$x = \pm \frac{\sqrt{753}}{4} + \frac{3}{4}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{753}}{4} + \frac{3}{4}$

Decimal Form:

$x = 7.61021136 \dots, - 6.11021136 \dots$