Solve by Factoring x^-2-9x^-1+20=0

$x^{- 2} - 9 x^{- 1} + 20 = 0$

Step 1

Factor $x^{- 2} - 9 x^{- 1} + 20$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $20$ and whose sum is $- 9$ .

$- 5, - 4$

Write the factored form using these integers.

$(x^{- 1} - 5) (x^{- 1} - 4) = 0$

Step 2

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

$(\frac{1}{x} - 5) (x^{- 1} - 4) = 0$

Step 3

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

$(\frac{1}{x} - 5) (\frac{1}{x} - 4) = 0$

Step 4

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

$(\frac{1}{x} - 5 \cdot \frac{x}{x}) (\frac{1}{x} - 4) = 0$

Step 5

Combine $- 5$ and $\frac{x}{x}$ .

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

Step 6

Combine the numerators over the common denominator.

$\frac{1 - 5 x}{x} \cdot (\frac{1}{x} - 4) = 0$

Step 7

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

$\frac{1 - 5 x}{x} \cdot (\frac{1}{x} - 4 \cdot \frac{x}{x}) = 0$

Step 8

Combine $- 4$ and $\frac{x}{x}$ .

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

Step 9

Combine the numerators over the common denominator.

$\frac{1 - 5 x}{x} \cdot \frac{1 - 4 x}{x} = 0$

Step 10

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

$\frac{1 - 5 x}{x} = 0$

$\frac{1 - 4 x}{x} = 0$

Step 11

Set $\frac{1 - 5 x}{x}$ equal to $0$ and solve for $x$ .

Set $\frac{1 - 5 x}{x}$ equal to $0$ .

$\frac{1 - 5 x}{x} = 0$

Solve $\frac{1 - 5 x}{x} = 0$ for $x$ .

Set the numerator equal to zero.

$1 - 5 x = 0$

Solve the equation for $x$ .

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

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

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Simplify the right side.

Dividing two negative values results in a positive value.

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

Step 12

Set $\frac{1 - 4 x}{x}$ equal to $0$ and solve for $x$ .

Set $\frac{1 - 4 x}{x}$ equal to $0$ .

$\frac{1 - 4 x}{x} = 0$

Solve $\frac{1 - 4 x}{x} = 0$ for $x$ .

Set the numerator equal to zero.

$1 - 4 x = 0$

Solve the equation for $x$ .

Subtract $1$ from both sides of the equation.

$- 4 x = - 1$

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

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

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

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{1}{4}$

Step 13

The final solution is all the values that make $\frac{1 - 5 x}{x} \cdot \frac{1 - 4 x}{x} = 0$ true.

$x = \frac{1}{5}, \frac{1}{4}$

Solve by Factoring x^-2-9x^-1+20=0

Related Questions

Support LCMGCF.com by disabling your adblocker.