Precalculus Examples

$x^{2} - 7 x = 3$

Step 1

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

${(\frac{b}{2})}^{2} = {(- \frac{7}{2})}^{2}$

Step 2

Add the term to each side of the equation.

$x^{2} - 7 x + {(- \frac{7}{2})}^{2} = 3 + {(- \frac{7}{2})}^{2}$

Step 3

Simplify the equation.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Simplify each term.

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Step 3.1.1.1

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

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Step 3.1.1.1.1

Apply the product rule to $- \frac{7}{2}$ .

$x^{2} - 7 x + {(- 1)}^{2} {(\frac{7}{2})}^{2} = 3 + {(- \frac{7}{2})}^{2}$

Step 3.1.1.1.2

Apply the product rule to $\frac{7}{2}$ .

$x^{2} - 7 x + {(- 1)}^{2} \frac{7^{2}}{2^{2}} = 3 + {(- \frac{7}{2})}^{2}$

Step 3.1.1.2

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

$x^{2} - 7 x + 1 \frac{7^{2}}{2^{2}} = 3 + {(- \frac{7}{2})}^{2}$

Step 3.1.1.3

Multiply $\frac{7^{2}}{2^{2}}$ by $1$ .

$x^{2} - 7 x + \frac{7^{2}}{2^{2}} = 3 + {(- \frac{7}{2})}^{2}$

Step 3.1.1.4

Raise $7$ to the power of $2$ .

$x^{2} - 7 x + \frac{49}{2^{2}} = 3 + {(- \frac{7}{2})}^{2}$

Step 3.1.1.5

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

$x^{2} - 7 x + \frac{49}{4} = 3 + {(- \frac{7}{2})}^{2}$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $3 + {(- \frac{7}{2})}^{2}$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

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

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Step 3.2.1.1.1.1

Apply the product rule to $- \frac{7}{2}$ .

$x^{2} - 7 x + \frac{49}{4} = 3 + {(- 1)}^{2} {(\frac{7}{2})}^{2}$

Step 3.2.1.1.1.2

Apply the product rule to $\frac{7}{2}$ .

$x^{2} - 7 x + \frac{49}{4} = 3 + {(- 1)}^{2} \frac{7^{2}}{2^{2}}$

Step 3.2.1.1.2

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

$x^{2} - 7 x + \frac{49}{4} = 3 + 1 \frac{7^{2}}{2^{2}}$

Step 3.2.1.1.3

Multiply $\frac{7^{2}}{2^{2}}$ by $1$ .

$x^{2} - 7 x + \frac{49}{4} = 3 + \frac{7^{2}}{2^{2}}$

Step 3.2.1.1.4

Raise $7$ to the power of $2$ .

$x^{2} - 7 x + \frac{49}{4} = 3 + \frac{49}{2^{2}}$

Step 3.2.1.1.5

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

$x^{2} - 7 x + \frac{49}{4} = 3 + \frac{49}{4}$

Step 3.2.1.2

To write $3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x^{2} - 7 x + \frac{49}{4} = 3 \cdot \frac{4}{4} + \frac{49}{4}$

Step 3.2.1.3

Combine $3$ and $\frac{4}{4}$ .

$x^{2} - 7 x + \frac{49}{4} = \frac{3 \cdot 4}{4} + \frac{49}{4}$

Step 3.2.1.4

Combine the numerators over the common denominator.

$x^{2} - 7 x + \frac{49}{4} = \frac{3 \cdot 4 + 49}{4}$

Step 3.2.1.5

Simplify the numerator.

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Step 3.2.1.5.1

Multiply $3$ by $4$ .

$x^{2} - 7 x + \frac{49}{4} = \frac{12 + 49}{4}$

Step 3.2.1.5.2

Add $12$ and $49$ .

$x^{2} - 7 x + \frac{49}{4} = \frac{61}{4}$

Step 4

Factor the perfect trinomial square into ${(x - \frac{7}{2})}^{2}$ .

${(x - \frac{7}{2})}^{2} = \frac{61}{4}$

Step 5

Solve the equation for $x$ .

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Step 5.1

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

$x - \frac{7}{2} = \pm \sqrt{\frac{61}{4}}$

Step 5.2

Simplify $\pm \sqrt{\frac{61}{4}}$ .

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Step 5.2.1

Rewrite $\sqrt{\frac{61}{4}}$ as $\frac{\sqrt{61}}{\sqrt{4}}$ .

$x - \frac{7}{2} = \pm \frac{\sqrt{61}}{\sqrt{4}}$

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

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

$x - \frac{7}{2} = \pm \frac{\sqrt{61}}{\sqrt{2^{2}}}$

Step 5.2.2.2

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

$x - \frac{7}{2} = \pm \frac{\sqrt{61}}{2}$

Step 5.3

Add $\frac{7}{2}$ to both sides of the equation.

$x = \pm \frac{\sqrt{61}}{2} + \frac{7}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{61}}{2} + \frac{7}{2}$

Decimal Form:

$x = 7.40512483 \dots, - 0.40512483 \dots$