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

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

Solve for k

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

$$x\neq -2\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply k-4 by x^{2}.

$$kx^{2}-4x^{2}+2\left(k-4\right)x+4=0$$

Use the distributive property to multiply 2 by k-4.

$$kx^{2}-4x^{2}+\left(2k-8\right)x+4=0$$

Use the distributive property to multiply 2k-8 by x.

$$kx^{2}-4x^{2}+2kx-8x+4=0$$

Add 4x^{2} to both sides. Anything plus zero gives itself.

$$kx^{2}+2kx-8x+4=4x^{2}$$

Add 8x to both sides.

$$kx^{2}+2kx+4=4x^{2}+8x$$

Subtract 4 from both sides.

$$kx^{2}+2kx=4x^{2}+8x-4$$

Combine all terms containing k.

$$\left(x^{2}+2x\right)k=4x^{2}+8x-4$$

Divide both sides by x^{2}+2x.

$$\frac{\left(x^{2}+2x\right)k}{x^{2}+2x}=\frac{4x^{2}+8x-4}{x^{2}+2x}$$

Dividing by x^{2}+2x undoes the multiplication by x^{2}+2x.

$$k=\frac{4x^{2}+8x-4}{x^{2}+2x}$$

Divide 4x^{2}+8x-4 by x^{2}+2x.

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

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(k-8\right)\left(k-4\right)}-k+4}{k-4}$$

$$x=\frac{-\sqrt{\left(k-8\right)\left(k-4\right)}-k+4}{k-4}\text{, }k\neq 4$$

Solve for x

$$x=\frac{\sqrt{\left(k-8\right)\left(k-4\right)}-k+4}{k-4}$$

$$x=\frac{-\sqrt{\left(k-8\right)\left(k-4\right)}-k+4}{k-4}\text{, }k\geq 8\text{ or }k<4$$