Solve 9x^2+kx+16=0

$$9 x ^ { 2 } + k x + 16 = 0$$

Solve for k

$$k=-9x-\frac{16}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 9x^{2} from both sides. Anything subtracted from zero gives its negation.

$$kx+16=-9x^{2}$$

Subtract 16 from both sides.

$$kx=-9x^{2}-16$$

The equation is in standard form.

$$xk=-9x^{2}-16$$

Divide both sides by x.

$$\frac{xk}{x}=\frac{-9x^{2}-16}{x}$$

Dividing by x undoes the multiplication by x.

$$k=\frac{-9x^{2}-16}{x}$$

Divide -9x^{2}-16 by x.

$$k=-9x-\frac{16}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{k^{2}-576}-k}{18}$$

$$x=\frac{-\sqrt{k^{2}-576}-k}{18}$$

Solve for x

$$x=\frac{\sqrt{k^{2}-576}-k}{18}$$

$$x=\frac{-\sqrt{k^{2}-576}-k}{18}\text{, }|k|\geq 24$$