Find the Roots (Zeros) 2x+6=-5/x

$2 x + 6 = - \frac{5}{x}$

Subtract $6$ from both sides of the equation.

$2 x = - \frac{5}{x} - 6$

Add $\frac{5}{x}$ to both sides of the equation.

$2 x + \frac{5}{x} = - 6$

Find the LCD of the terms in the equation.

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

$1, x, 1$

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

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

$x$

Multiply each term by $x$ and simplify.

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Multiply each term in $2 x + \frac{5}{x} = - 6$ by $x$ in order to remove all the denominators from the equation.

$2 x \cdot x + \frac{5}{x} \cdot x = - 6 \cdot x$

Simplify each term.

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Multiply $x$ by $x$ by adding the exponents.

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Move $x$ .

$2 (x \cdot x) + \frac{5}{x} \cdot x = - 6 \cdot x$

Multiply $x$ by $x$ .

$2 x^{2} + \frac{5}{x} \cdot x = - 6 \cdot x$

Cancel the common factor of $x$ .

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Cancel the common factor.

$2 x^{2} + \frac{5}{x} \cdot x = - 6 \cdot x$

Rewrite the expression.

$2 x^{2} + 5 = - 6 \cdot x$

$2 x^{2} + 5 = - 6 x$

Solve the equation.

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Add $6 x$ to both sides of the equation.

$2 x^{2} + 5 + 6 x = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = 6$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (2 \cdot 5)}}{2 \cdot 2}$

Simplify.

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Simplify the numerator.

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Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot (2 \cdot 5)}}{2 \cdot 2}$

Multiply $2$ by $5$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 10}}{2 \cdot 2}$

Multiply $- 4$ by $10$ .

$x = \frac{- 6 \pm \sqrt{36 - 40}}{2 \cdot 2}$

Subtract $40$ from $36$ .

$x = \frac{- 6 \pm \sqrt{- 4}}{2 \cdot 2}$

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{- 6 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 2}$

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

$x = \frac{- 6 \pm i \cdot \sqrt{4}}{2 \cdot 2}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 6 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 2}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 6 \pm i \cdot 2}{2 \cdot 2}$

Move $2$ to the left of $i$ .

$x = \frac{- 6 \pm 2 i}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 6 \pm 2 i}{4}$

Simplify $\frac{- 6 \pm 2 i}{4}$ .

$x = \frac{- 3 \pm i}{2}$

The final answer is the combination of both solutions.

$x = - \frac{3}{2} + \frac{i}{2}, - \frac{3}{2} - \frac{i}{2}$

$x = \frac{- 3 \pm i}{2}$

Find the Roots (Zeros) 2x+6=-5/x

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