Precalculus Examples

$8 {(1 + \frac{1}{t})}^{2} + 22 (1 + \frac{1}{t}) + 15 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = 1 + \frac{1}{t}$ . Substitute $u$ for all occurrences of $1 + \frac{1}{t}$ .

$8 u^{2} + 22 u + 15 = 0$

Step 1.2

Factor by grouping.

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Step 1.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 8 \cdot 15 = 120$ and whose sum is $b = 22$ .

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Step 1.2.1.1

Factor $22$ out of $22 u$ .

$8 u^{2} + 22 (u) + 15 = 0$

Step 1.2.1.2

Rewrite $22$ as $10$ plus $12$

$8 u^{2} + (10 + 12) u + 15 = 0$

Step 1.2.1.3

Apply the distributive property.

$8 u^{2} + 10 u + 12 u + 15 = 0$

Step 1.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.1

Group the first two terms and the last two terms.

$(8 u^{2} + 10 u) + 12 u + 15 = 0$

Step 1.2.2.2

Factor out the greatest common factor (GCF) from each group.

$2 u (4 u + 5) + 3 (4 u + 5) = 0$

Step 1.2.3

Factor the polynomial by factoring out the greatest common factor, $4 u + 5$ .

$(4 u + 5) (2 u + 3) = 0$

Step 1.3

Replace all occurrences of $u$ with $1 + \frac{1}{t}$ .

$(4 (1 + \frac{1}{t}) + 5) (2 (1 + \frac{1}{t}) + 3) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$4 (1 + \frac{1}{t}) + 5 = 0$

$2 (1 + \frac{1}{t}) + 3 = 0$

Step 3

Set $4 (1 + \frac{1}{t}) + 5$ equal to $0$ and solve for $t$ .

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Step 3.1

Set $4 (1 + \frac{1}{t}) + 5$ equal to $0$ .

$4 (1 + \frac{1}{t}) + 5 = 0$

Step 3.2

Solve $4 (1 + \frac{1}{t}) + 5 = 0$ for $t$ .

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Step 3.2.1

Simplify $4 (1 + \frac{1}{t}) + 5$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

Apply the distributive property.

$4 \cdot 1 + 4 \frac{1}{t} + 5 = 0$

Step 3.2.1.1.2

Multiply $4$ by $1$ .

$4 + 4 \frac{1}{t} + 5 = 0$

Step 3.2.1.1.3

Combine $4$ and $\frac{1}{t}$ .

$4 + \frac{4}{t} + 5 = 0$

Step 3.2.1.2

Add $4$ and $5$ .

$\frac{4}{t} + 9 = 0$

Step 3.2.2

Subtract $9$ from both sides of the equation.

$\frac{4}{t} = - 9$

Step 3.2.3

Find the LCD of the terms in the equation.

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Step 3.2.3.1

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

$t, 1$

Step 3.2.3.2

The LCM of one and any expression is the expression.

$t$

Step 3.2.4

Multiply each term in $\frac{4}{t} = - 9$ by $t$ to eliminate the fractions.

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Step 3.2.4.1

Multiply each term in $\frac{4}{t} = - 9$ by $t$ .

$\frac{4}{t} t = - 9 t$

Step 3.2.4.2

Simplify the left side.

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Step 3.2.4.2.1

Cancel the common factor of $t$ .

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Step 3.2.4.2.1.1

Cancel the common factor.

$\frac{4}{t} t = - 9 t$

Step 3.2.4.2.1.2

Rewrite the expression.

$4 = - 9 t$

Step 3.2.5

Solve the equation.

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Step 3.2.5.1

Rewrite the equation as $- 9 t = 4$ .

$- 9 t = 4$

Step 3.2.5.2

Divide each term in $- 9 t = 4$ by $- 9$ and simplify.

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Step 3.2.5.2.1

Divide each term in $- 9 t = 4$ by $- 9$ .

$\frac{- 9 t}{- 9} = \frac{4}{- 9}$

Step 3.2.5.2.2

Simplify the left side.

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Step 3.2.5.2.2.1

Cancel the common factor of $- 9$ .

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Step 3.2.5.2.2.1.1

Cancel the common factor.

$\frac{- 9 t}{- 9} = \frac{4}{- 9}$

Step 3.2.5.2.2.1.2

Divide $t$ by $1$ .

$t = \frac{4}{- 9}$

Step 3.2.5.2.3

Simplify the right side.

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Step 3.2.5.2.3.1

Move the negative in front of the fraction.

$t = - \frac{4}{9}$

Step 4

Set $2 (1 + \frac{1}{t}) + 3$ equal to $0$ and solve for $t$ .

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Step 4.1

Set $2 (1 + \frac{1}{t}) + 3$ equal to $0$ .

$2 (1 + \frac{1}{t}) + 3 = 0$

Step 4.2

Solve $2 (1 + \frac{1}{t}) + 3 = 0$ for $t$ .

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Step 4.2.1

Simplify $2 (1 + \frac{1}{t}) + 3$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Apply the distributive property.

$2 \cdot 1 + 2 \frac{1}{t} + 3 = 0$

Step 4.2.1.1.2

Multiply $2$ by $1$ .

$2 + 2 \frac{1}{t} + 3 = 0$

Step 4.2.1.1.3

Combine $2$ and $\frac{1}{t}$ .

$2 + \frac{2}{t} + 3 = 0$

Step 4.2.1.2

Add $2$ and $3$ .

$\frac{2}{t} + 5 = 0$

Step 4.2.2

Subtract $5$ from both sides of the equation.

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

Step 4.2.3

Find the LCD of the terms in the equation.

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Step 4.2.3.1

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

$t, 1$

Step 4.2.3.2

The LCM of one and any expression is the expression.

$t$

Step 4.2.4

Multiply each term in $\frac{2}{t} = - 5$ by $t$ to eliminate the fractions.

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Step 4.2.4.1

Multiply each term in $\frac{2}{t} = - 5$ by $t$ .

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

Step 4.2.4.2

Simplify the left side.

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Step 4.2.4.2.1

Cancel the common factor of $t$ .

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Step 4.2.4.2.1.1

Cancel the common factor.

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

Step 4.2.4.2.1.2

Rewrite the expression.

$2 = - 5 t$

Step 4.2.5

Solve the equation.

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Step 4.2.5.1

Rewrite the equation as $- 5 t = 2$ .

$- 5 t = 2$

Step 4.2.5.2

Divide each term in $- 5 t = 2$ by $- 5$ and simplify.

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Step 4.2.5.2.1

Divide each term in $- 5 t = 2$ by $- 5$ .

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

Step 4.2.5.2.2

Simplify the left side.

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Step 4.2.5.2.2.1

Cancel the common factor of $- 5$ .

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Step 4.2.5.2.2.1.1

Cancel the common factor.

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

Step 4.2.5.2.2.1.2

Divide $t$ by $1$ .

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

Step 4.2.5.2.3

Simplify the right side.

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Step 4.2.5.2.3.1

Move the negative in front of the fraction.

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

Step 5

The final solution is all the values that make $(4 (1 + \frac{1}{t}) + 5) (2 (1 + \frac{1}{t}) + 3) = 0$ true.

$t = - \frac{4}{9}, - \frac{2}{5}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$t = - \frac{4}{9}, - \frac{2}{5}$

Decimal Form:

$t = - 0.\overline{4}, - 0.4$