Solve for y y^-2-y^-1-20=0

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

Step 1

Simplify each term.

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

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

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

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

Step 2

Find the LCD of the terms in the equation.

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

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

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

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.

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

Not prime

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.

$1$

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.

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

$y^{1} = y$

$y$ occurs $1$ time.

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.

$y \cdot y$

Multiply $y$ by $y$ .

$y^{2}$

Step 3

Multiply each term in $\frac{1}{y^{2}} - \frac{1}{y} - 20 = 0$ by $y^{2}$ to eliminate the fractions.

Multiply each term in $\frac{1}{y^{2}} - \frac{1}{y} - 20 = 0$ by $y^{2}$ .

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

Simplify the left side.

Simplify each term.

Cancel the common factor of $y^{2}$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $y$ .

Move the leading negative in $- \frac{1}{y}$ into the numerator.

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

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

$1 + \frac{- 1}{y} (y \cdot y) - 20 y^{2} = 0 y^{2}$

Cancel the common factor.

$1 + \frac{- 1}{y} (y \cdot y) - 20 y^{2} = 0 y^{2}$

Rewrite the expression.

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

Simplify the right side.

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

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

Step 4

Solve the equation.

Factor the left side of the equation.

Factor $- 1$ out of $1 - y - 20 y^{2}$ .

Reorder the expression.

Move $1$ .

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

Reorder $- y$ and $- 20 y^{2}$ .

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

Factor $- 1$ out of $- 20 y^{2}$ .

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

Factor $- 1$ out of $- y$ .

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

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

$- (20 y^{2}) - (y) - 1 \cdot - 1 = 0$

Factor $- 1$ out of $- (20 y^{2}) - (y)$ .

$- (20 y^{2} + y) - 1 \cdot - 1 = 0$

Factor $- 1$ out of $- (20 y^{2} + y) - 1 (- 1)$ .

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

Factor.

Factor by grouping.

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 $a \cdot c = 20 \cdot - 1 = - 20$ and whose sum is $b = 1$ .

Multiply by $1$ .

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

Rewrite $1$ as $- 4$ plus $5$

$- (20 y^{2} + (- 4 + 5) y - 1) = 0$

Apply the distributive property.

$- (20 y^{2} - 4 y + 5 y - 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((20 y^{2} - 4 y) + 5 y - 1) = 0$

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

$- (4 y (5 y - 1) + 1 (5 y - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 y - 1$ .

$- ((5 y - 1) (4 y + 1)) = 0$

Remove unnecessary parentheses.

$- (5 y - 1) (4 y + 1) = 0$

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

$5 y - 1 = 0$

$4 y + 1 = 0$

Set $5 y - 1$ equal to $0$ and solve for $y$ .

Set $5 y - 1$ equal to $0$ .

$5 y - 1 = 0$

Solve $5 y - 1 = 0$ for $y$ .

Add $1$ to both sides of the equation.

$5 y = 1$

Divide each term in $5 y = 1$ by $5$ and simplify.

Divide each term in $5 y = 1$ by $5$ .

$\frac{5 y}{5} = \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y}{5} = \frac{1}{5}$

Divide $y$ by $1$ .

$y = \frac{1}{5}$

Set $4 y + 1$ equal to $0$ and solve for $y$ .

Set $4 y + 1$ equal to $0$ .

$4 y + 1 = 0$

Solve $4 y + 1 = 0$ for $y$ .

Subtract $1$ from both sides of the equation.

$4 y = - 1$

Divide each term in $4 y = - 1$ by $4$ and simplify.

Divide each term in $4 y = - 1$ by $4$ .

$\frac{4 y}{4} = \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{- 1}{4}$

Divide $y$ by $1$ .

$y = \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{1}{4}$

The final solution is all the values that make $- (5 y - 1) (4 y + 1) = 0$ true.

$y = \frac{1}{5}, - \frac{1}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$y = \frac{1}{5}, - \frac{1}{4}$

Decimal Form:

$y = 0.2, - 0.25$

Solve for y y^-2-y^-1-20=0

Related Questions