Trigonometry Examples

$- \frac{41}{5 x^{2}} \cdot 1 + \frac{2}{5 x} - 2 \frac{4}{5} = 0$

Step 1

Simplify the left side.

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Step 1.1

Simplify $- \frac{41}{5 x^{2}} \cdot 1 + \frac{2}{5 x} - 2 \frac{4}{5}$ .

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Step 1.1.1

Convert $2 \frac{4}{5}$ to an improper fraction.

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Step 1.1.1.1

A mixed number is an addition of its whole and fractional parts.

$- \frac{41}{5 x^{2}} \cdot 1 + \frac{2}{5 x} - (2 + \frac{4}{5}) = 0$

Step 1.1.1.2

Add $2$ and $\frac{4}{5}$ .

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Step 1.1.1.2.1

To write $2$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$- \frac{41}{5 x^{2}} \cdot 1 + \frac{2}{5 x} - (2 \cdot \frac{5}{5} + \frac{4}{5}) = 0$

Step 1.1.1.2.2

Combine $2$ and $\frac{5}{5}$ .

$- \frac{41}{5 x^{2}} \cdot 1 + \frac{2}{5 x} - (\frac{2 \cdot 5}{5} + \frac{4}{5}) = 0$

Step 1.1.1.2.3

Combine the numerators over the common denominator.

$- \frac{41}{5 x^{2}} \cdot 1 + \frac{2}{5 x} - \frac{2 \cdot 5 + 4}{5} = 0$

Step 1.1.1.2.4

Simplify the numerator.

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Step 1.1.1.2.4.1

Multiply $2$ by $5$ .

$- \frac{41}{5 x^{2}} \cdot 1 + \frac{2}{5 x} - \frac{10 + 4}{5} = 0$

Step 1.1.1.2.4.2

Add $10$ and $4$ .

$- \frac{41}{5 x^{2}} \cdot 1 + \frac{2}{5 x} - \frac{14}{5} = 0$

Step 1.1.2

Multiply $- 1$ by $1$ .

$- \frac{41}{5 x^{2}} + \frac{2}{5 x} - \frac{14}{5} = 0$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$5 x^{2}, 5 x, 5, 1$

Step 2.2

Since $5 x^{2}, 5 x, 5, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $5, 5, 5, 1$ then find LCM for the variable part $x^{2}, x^{1}$ .

Step 2.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 2.4

Since $5$ has no factors besides $1$ and $5$ .

$5$ is a prime number

Step 2.5

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 2.6

The LCM of $5, 5, 5, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$5$

Step 2.7

The factors for $x^{2}$ are $x \cdot x$ , which is $x$ multiplied by each other $2$ times.

$x^{2} = x \cdot x$

$x$ occurs $2$ times.

Step 2.8

The factor for $x^{1}$ is $x$ itself.

$x^{1} = x$

$x$ occurs $1$ time.

Step 2.9

The LCM of $x^{2}, x^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x \cdot x$

Step 2.10

Multiply $x$ by $x$ .

$x^{2}$

Step 2.11

The LCM for $5 x^{2}, 5 x, 5, 1$ is the numeric part $5$ multiplied by the variable part.

$5 x^{2}$

Step 3

Multiply each term in $- \frac{41}{5 x^{2}} + \frac{2}{5 x} - \frac{14}{5} = 0$ by $5 x^{2}$ to eliminate the fractions.

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Step 3.1

Multiply each term in $- \frac{41}{5 x^{2}} + \frac{2}{5 x} - \frac{14}{5} = 0$ by $5 x^{2}$ .

$- \frac{41}{5 x^{2}} (5 x^{2}) + \frac{2}{5 x} (5 x^{2}) - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Cancel the common factor of $5 x^{2}$ .

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Step 3.2.1.1.1

Move the leading negative in $- \frac{41}{5 x^{2}}$ into the numerator.

$\frac{- 41}{5 x^{2}} (5 x^{2}) + \frac{2}{5 x} (5 x^{2}) - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.1.2

Cancel the common factor.

$\frac{- 41}{5 x^{2}} (5 x^{2}) + \frac{2}{5 x} (5 x^{2}) - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.1.3

Rewrite the expression.

$- 41 + \frac{2}{5 x} (5 x^{2}) - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.2

Rewrite using the commutative property of multiplication.

$- 41 + 5 \frac{2}{5 x} x^{2} - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.3

Cancel the common factor of $5$ .

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Step 3.2.1.3.1

Factor $5$ out of $5 x$ .

$- 41 + 5 \frac{2}{5 (x)} x^{2} - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.3.2

Cancel the common factor.

$- 41 + 5 \frac{2}{5 x} x^{2} - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.3.3

Rewrite the expression.

$- 41 + \frac{2}{x} x^{2} - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.4

Cancel the common factor of $x$ .

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Step 3.2.1.4.1

Factor $x$ out of $x^{2}$ .

$- 41 + \frac{2}{x} (x \cdot x) - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.4.2

Cancel the common factor.

$- 41 + \frac{2}{x} (x \cdot x) - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.4.3

Rewrite the expression.

$- 41 + 2 x - \frac{14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.5

Cancel the common factor of $5$ .

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Step 3.2.1.5.1

Move the leading negative in $- \frac{14}{5}$ into the numerator.

$- 41 + 2 x + \frac{- 14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.5.2

Factor $5$ out of $5 x^{2}$ .

$- 41 + 2 x + \frac{- 14}{5} (5 (x^{2})) = 0 (5 x^{2})$

Step 3.2.1.5.3

Cancel the common factor.

$- 41 + 2 x + \frac{- 14}{5} (5 x^{2}) = 0 (5 x^{2})$

Step 3.2.1.5.4

Rewrite the expression.

$- 41 + 2 x - 14 x^{2} = 0 (5 x^{2})$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply $0 (5 x^{2})$ .

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Step 3.3.1.1

Multiply $5$ by $0$ .

$- 41 + 2 x - 14 x^{2} = 0 x^{2}$

Step 3.3.1.2

Multiply $0$ by $x^{2}$ .

$- 41 + 2 x - 14 x^{2} = 0$

Step 4

Solve the equation.

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Step 4.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.2

Substitute the values $a = - 14$ , $b = 2$ , and $c = - 41$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (- 14 \cdot - 41)}}{2 \cdot - 14}$

Step 4.3

Simplify.

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Step 4.3.1

Simplify the numerator.

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Step 4.3.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 14 \cdot - 41}}{2 \cdot - 14}$

Step 4.3.1.2

Multiply $- 4 \cdot - 14 \cdot - 41$ .

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Step 4.3.1.2.1

Multiply $- 4$ by $- 14$ .

$x = \frac{- 2 \pm \sqrt{4 + 56 \cdot - 41}}{2 \cdot - 14}$

Step 4.3.1.2.2

Multiply $56$ by $- 41$ .

$x = \frac{- 2 \pm \sqrt{4 - 2296}}{2 \cdot - 14}$

Step 4.3.1.3

Subtract $2296$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 2292}}{2 \cdot - 14}$

Step 4.3.1.4

Rewrite $- 2292$ as $- 1 (2292)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 2292}}{2 \cdot - 14}$

Step 4.3.1.5

Rewrite $\sqrt{- 1 (2292)}$ as $\sqrt{- 1} \cdot \sqrt{2292}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{2292}}{2 \cdot - 14}$

Step 4.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{2292}}{2 \cdot - 14}$

Step 4.3.1.7

Rewrite $2292$ as $2^{2} \cdot 573$ .

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Step 4.3.1.7.1

Factor $4$ out of $2292$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (573)}}{2 \cdot - 14}$

Step 4.3.1.7.2

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

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 573}}{2 \cdot - 14}$

Step 4.3.1.8

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{573})}{2 \cdot - 14}$

Step 4.3.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{573}}{2 \cdot - 14}$

Step 4.3.2

Multiply $2$ by $- 14$ .

$x = \frac{- 2 \pm 2 i \sqrt{573}}{- 28}$

Step 4.3.3

Simplify $\frac{- 2 \pm 2 i \sqrt{573}}{- 28}$ .

$x = \frac{1 \pm i \sqrt{573}}{14}$

Step 4.4

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{573}}{14}, \frac{1 - i \sqrt{573}}{14}$