Solve 4x^2+kx+9=0

$$4 x ^ { 2 } + k x + 9 = 0$$

Solve for k

$$k=-4x-\frac{9}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 4x^{2} from both sides. Anything subtracted from zero gives its negation.

$$kx+9=-4x^{2}$$

Subtract 9 from both sides.

$$kx=-4x^{2}-9$$

The equation is in standard form.

$$xk=-4x^{2}-9$$

Divide both sides by x.

$$\frac{xk}{x}=\frac{-4x^{2}-9}{x}$$

Dividing by x undoes the multiplication by x.

$$k=\frac{-4x^{2}-9}{x}$$

Divide -4x^{2}-9 by x.

$$k=-4x-\frac{9}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{k^{2}-144}-k}{8}$$

$$x=\frac{-\sqrt{k^{2}-144}-k}{8}$$

Solve for x

$$x=\frac{\sqrt{k^{2}-144}-k}{8}$$

$$x=\frac{-\sqrt{k^{2}-144}-k}{8}\text{, }|k|\geq 12$$