Solve Using the Quadratic Formula 15/x-(11x+5)/(x^2)=-1

$\frac{15}{x} - \frac{11 x + 5}{x^{2}} = - 1$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $\frac{15}{x} - \frac{11 x + 5}{x^{2}}$ .

Simplify each term.

Split the fraction $\frac{11 x + 5}{x^{2}}$ into two fractions.

$\frac{15}{x} - (\frac{11 x}{x^{2}} + \frac{5}{x^{2}}) = - 1$

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

Factor $x$ out of $11 x$ .

$\frac{15}{x} - (\frac{x \cdot 11}{x^{2}} + \frac{5}{x^{2}}) = - 1$

Cancel the common factors.

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

$\frac{15}{x} - (\frac{x \cdot 11}{x \cdot x} + \frac{5}{x^{2}}) = - 1$

Cancel the common factor.

$\frac{15}{x} - (\frac{x \cdot 11}{x \cdot x} + \frac{5}{x^{2}}) = - 1$

Rewrite the expression.

$\frac{15}{x} - (\frac{11}{x} + \frac{5}{x^{2}}) = - 1$

Apply the distributive property.

$\frac{15}{x} - \frac{11}{x} - \frac{5}{x^{2}} = - 1$

Combine fractions.

Combine the numerators over the common denominator.

$\frac{15 - 11}{x} + \frac{- 5}{x^{2}} = - 1$

Simplify the expression.

Subtract $11$ from $15$ .

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

Move the negative in front of the fraction.

$\frac{4}{x} - \frac{5}{x^{2}} = - 1$

Add $1$ to both sides of the equation.

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

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.

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

Since $x, x^{2}, 1, 1$ 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 $x^{1}, x^{2}$ .

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

$x^{1} = x$

$x$ occurs $1$ time.

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.

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

$x \cdot x$

Multiply $x$ by $x$ .

$x^{2}$

Step 3

Multiply each term in $\frac{4}{x} - \frac{5}{x^{2}} + 1 = 0$ by $x^{2}$ to eliminate the fractions.

Multiply each term in $\frac{4}{x} - \frac{5}{x^{2}} + 1 = 0$ by $x^{2}$ .

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

Simplify the left side.

Simplify each term.

Cancel the common factor of $x$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

$4 x - 5 + 1 x^{2} = 0 x^{2}$

Multiply $x^{2}$ by $1$ .

$4 x - 5 + x^{2} = 0 x^{2}$

Simplify the right side.

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

$4 x - 5 + x^{2} = 0$

Step 4

Solve the equation.

Factor $4 x - 5 + x^{2}$ 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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$(x - 1) (x + 5) = 0$

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

$x - 1 = 0$

$x + 5 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

$x = 1, - 5$

Solve Using the Quadratic Formula 15/x-(11x+5)/(x^2)=-1

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