Solve for k x^2+kx+9=0

$x^{2} + k x + 9 = 0$

Step 1

Move all terms not containing $k$ to the right side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$k x + 9 = - x^{2}$

Subtract $9$ from both sides of the equation.

$k x = - x^{2} - 9$

Step 2

Divide each term in $k x = - x^{2} - 9$ by $x$ and simplify.

Divide each term in $k x = - x^{2} - 9$ by $x$ .

$\frac{k x}{x} = \frac{- x^{2}}{x} + \frac{- 9}{x}$

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{k x}{x} = \frac{- x^{2}}{x} + \frac{- 9}{x}$

Divide $k$ by $1$ .

$k = \frac{- x^{2}}{x} + \frac{- 9}{x}$

Simplify the right side.

Simplify each term.

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

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

$k = \frac{x (- x)}{x} + \frac{- 9}{x}$

Cancel the common factors.

Raise $x$ to the power of $1$ .

$k = \frac{x (- x)}{x^{1}} + \frac{- 9}{x}$

Factor $x$ out of $x^{1}$ .

$k = \frac{x (- x)}{x \cdot 1} + \frac{- 9}{x}$

Cancel the common factor.

$k = \frac{x (- x)}{x \cdot 1} + \frac{- 9}{x}$

Rewrite the expression.

$k = \frac{- x}{1} + \frac{- 9}{x}$

Divide $- x$ by $1$ .

$k = - x + \frac{- 9}{x}$

Move the negative in front of the fraction.

$k = - x - \frac{9}{x}$