Solve 4x^2-3kx+1=0

$$4 x ^ { 2 } - 3 k x + 1 = 0$$

Solve for k

$$k=\frac{4x}{3}+\frac{1}{3x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$-3kx+1=-4x^{2}$$

Subtract 1 from both sides.

$$-3kx=-4x^{2}-1$$

The equation is in standard form.

$$\left(-3x\right)k=-4x^{2}-1$$

Divide both sides by -3x.

$$\frac{\left(-3x\right)k}{-3x}=\frac{-4x^{2}-1}{-3x}$$

Dividing by -3x undoes the multiplication by -3x.

$$k=\frac{-4x^{2}-1}{-3x}$$

Divide -4x^{2}-1 by -3x.

$$k=\frac{4x}{3}+\frac{1}{3x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{9k^{2}-16}+3k}{8}$$

$$x=\frac{-\sqrt{9k^{2}-16}+3k}{8}$$

Solve for x

$$x=\frac{\sqrt{9k^{2}-16}+3k}{8}$$

$$x=\frac{-\sqrt{9k^{2}-16}+3k}{8}\text{, }|k|\geq \frac{4}{3}$$