Precalculus Examples

$20 y^{- 2} - 20 y^{- 1} + 32 = 0$

Step 1

Simplify each term.

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Step 1.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$20 \frac{1}{y^{2}} - 20 y^{- 1} + 32 = 0$

Step 1.2

Combine $20$ and $\frac{1}{y^{2}}$ .

$\frac{20}{y^{2}} - 20 y^{- 1} + 32 = 0$

Step 1.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{20}{y^{2}} - 20 \frac{1}{y} + 32 = 0$

Step 1.4

Combine $- 20$ and $\frac{1}{y}$ .

$\frac{20}{y^{2}} + \frac{- 20}{y} + 32 = 0$

Step 1.5

Move the negative in front of the fraction.

$\frac{20}{y^{2}} - \frac{20}{y} + 32 = 0$

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.

$y^{2}, y, 1, 1$

Step 2.2

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

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

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

Not prime

Step 2.5

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

$1$

Step 2.6

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

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

$y$ occurs $2$ times.

Step 2.7

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

$y^{1} = y$

$y$ occurs $1$ time.

Step 2.8

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

$y \cdot y$

Step 2.9

Multiply $y$ by $y$ .

$y^{2}$

Step 3

Multiply each term in $\frac{20}{y^{2}} - \frac{20}{y} + 32 = 0$ by $y^{2}$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{20}{y^{2}} - \frac{20}{y} + 32 = 0$ by $y^{2}$ .

$\frac{20}{y^{2}} y^{2} - \frac{20}{y} y^{2} + 32 y^{2} = 0 y^{2}$

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 $y^{2}$ .

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Step 3.2.1.1.1

Cancel the common factor.

$\frac{20}{y^{2}} y^{2} - \frac{20}{y} y^{2} + 32 y^{2} = 0 y^{2}$

Step 3.2.1.1.2

Rewrite the expression.

$20 - \frac{20}{y} y^{2} + 32 y^{2} = 0 y^{2}$

Step 3.2.1.2

Cancel the common factor of $y$ .

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Step 3.2.1.2.1

Move the leading negative in $- \frac{20}{y}$ into the numerator.

$20 + \frac{- 20}{y} y^{2} + 32 y^{2} = 0 y^{2}$

Step 3.2.1.2.2

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

$20 + \frac{- 20}{y} (y \cdot y) + 32 y^{2} = 0 y^{2}$

Step 3.2.1.2.3

Cancel the common factor.

$20 + \frac{- 20}{y} (y \cdot y) + 32 y^{2} = 0 y^{2}$

Step 3.2.1.2.4

Rewrite the expression.

$20 - 20 y + 32 y^{2} = 0 y^{2}$

Step 3.3

Simplify the right side.

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Step 3.3.1

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

$20 - 20 y + 32 y^{2} = 0$

Step 4

Solve the equation.

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Step 4.1

Factor $4$ out of $20 - 20 y + 32 y^{2}$ .

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Step 4.1.1

Factor $4$ out of $20$ .

$4 (5) - 20 y + 32 y^{2} = 0$

Step 4.1.2

Factor $4$ out of $- 20 y$ .

$4 (5) + 4 (- 5 y) + 32 y^{2} = 0$

Step 4.1.3

Factor $4$ out of $32 y^{2}$ .

$4 (5) + 4 (- 5 y) + 4 (8 y^{2}) = 0$

Step 4.1.4

Factor $4$ out of $4 (5) + 4 (- 5 y)$ .

$4 (5 - 5 y) + 4 (8 y^{2}) = 0$

Step 4.1.5

Factor $4$ out of $4 (5 - 5 y) + 4 (8 y^{2})$ .

$4 (5 - 5 y + 8 y^{2}) = 0$

Step 4.2

Divide each term in $4 (5 - 5 y + 8 y^{2}) = 0$ by $4$ and simplify.

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Step 4.2.1

Divide each term in $4 (5 - 5 y + 8 y^{2}) = 0$ by $4$ .

$\frac{4 (5 - 5 y + 8 y^{2})}{4} = \frac{0}{4}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $4$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{4 (5 - 5 y + 8 y^{2})}{4} = \frac{0}{4}$

Step 4.2.2.1.2

Divide $5 - 5 y + 8 y^{2}$ by $1$ .

$5 - 5 y + 8 y^{2} = \frac{0}{4}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $0$ by $4$ .

$5 - 5 y + 8 y^{2} = 0$

Step 4.3

Use the quadratic formula to find the solutions.

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

Step 4.4

Substitute the values $a = 8$ , $b = - 5$ , and $c = 5$ into the quadratic formula and solve for $y$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (8 \cdot 5)}}{2 \cdot 8}$

Step 4.5

Simplify.

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Step 4.5.1

Simplify the numerator.

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Step 4.5.1.1

Raise $- 5$ to the power of $2$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 8 \cdot 5}}{2 \cdot 8}$

Step 4.5.1.2

Multiply $- 4 \cdot 8 \cdot 5$ .

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Step 4.5.1.2.1

Multiply $- 4$ by $8$ .

$y = \frac{5 \pm \sqrt{25 - 32 \cdot 5}}{2 \cdot 8}$

Step 4.5.1.2.2

Multiply $- 32$ by $5$ .

$y = \frac{5 \pm \sqrt{25 - 160}}{2 \cdot 8}$

Step 4.5.1.3

Subtract $160$ from $25$ .

$y = \frac{5 \pm \sqrt{- 135}}{2 \cdot 8}$

Step 4.5.1.4

Rewrite $- 135$ as $- 1 (135)$ .

$y = \frac{5 \pm \sqrt{- 1 \cdot 135}}{2 \cdot 8}$

Step 4.5.1.5

Rewrite $\sqrt{- 1 (135)}$ as $\sqrt{- 1} \cdot \sqrt{135}$ .

$y = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{135}}{2 \cdot 8}$

Step 4.5.1.6

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

$y = \frac{5 \pm i \cdot \sqrt{135}}{2 \cdot 8}$

Step 4.5.1.7

Rewrite $135$ as $3^{2} \cdot 15$ .

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Step 4.5.1.7.1

Factor $9$ out of $135$ .

$y = \frac{5 \pm i \cdot \sqrt{9 (15)}}{2 \cdot 8}$

Step 4.5.1.7.2

Rewrite $9$ as $3^{2}$ .

$y = \frac{5 \pm i \cdot \sqrt{3^{2} \cdot 15}}{2 \cdot 8}$

Step 4.5.1.8

Pull terms out from under the radical.

$y = \frac{5 \pm i \cdot (3 \sqrt{15})}{2 \cdot 8}$

Step 4.5.1.9

Move $3$ to the left of $i$ .

$y = \frac{5 \pm 3 i \sqrt{15}}{2 \cdot 8}$

Step 4.5.2

Multiply $2$ by $8$ .

$y = \frac{5 \pm 3 i \sqrt{15}}{16}$

Step 4.6

The final answer is the combination of both solutions.

$y = \frac{5 + 3 i \sqrt{15}}{16}, \frac{5 - 3 i \sqrt{15}}{16}$