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

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

Step 1

Subtract $6$ from both sides of the equation.

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

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.

$1, x, 1$

The LCM of one and any expression is the expression.

$x$

Step 3

Multiply each term in $2 x = - \frac{5}{x} - 6$ by $x$ to eliminate the fractions.

Multiply each term in $2 x = - \frac{5}{x} - 6$ by $x$ .

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

Simplify the left side.

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Simplify the right side.

Cancel the common factor of $x$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Solve the equation.

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

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

Add $5$ to both sides of the equation.

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

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.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 2 \cdot 5$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 8 \cdot 5}}{2 \cdot 2}$

Multiply $- 8$ by $5$ .

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

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.

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

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

Step 5

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

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