Solve x^2+(k-1)x+4=0

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

Solve for k

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

$$x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply k-1 by x.

$$x^{2}+kx-x+4=0$$

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

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

Add x to both sides.

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

Subtract 4 from both sides.

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

The equation is in standard form.

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

Divide both sides by x.

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

Dividing by x undoes the multiplication by x.

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

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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(k-5\right)\left(k+3\right)}-k+1}{2}$$

$$x=\frac{-\sqrt{\left(k-5\right)\left(k+3\right)}-k+1}{2}$$

Solve for x

$$x=\frac{\sqrt{\left(k-5\right)\left(k+3\right)}-k+1}{2}$$

$$x=\frac{-\sqrt{\left(k-5\right)\left(k+3\right)}-k+1}{2}\text{, }k\leq -3\text{ or }k\geq 5$$