How to solve Kx(x-2)+6=0

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve Kx(x-2)+6=0 | Equatix

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Question

$$Kx(x-2)+6=0$$

Solve for K

$K=-\frac{6}{x\left(x-2\right)}$
$x\neq 2\text{ and }x\neq 0$

Steps for Solving Linear Equation

Use the distributive property to multiply $Kx$ by $x-2$.

$$Kx^{2}-2Kx+6=0$$

Subtract $6$ from both sides. Anything subtracted from zero gives its negation.

$$Kx^{2}-2Kx=-6$$

Combine all terms containing $K$.

$$\left(x^{2}-2x\right)K=-6$$

Divide both sides by $x^{2}-2x$.

$$\frac{\left(x^{2}-2x\right)K}{x^{2}-2x}=-\frac{6}{x^{2}-2x}$$

Dividing by $x^{2}-2x$ undoes the multiplication by $x^{2}-2x$.

$$K=-\frac{6}{x^{2}-2x}$$

Divide $-6$ by $x^{2}-2x$.

$$K=-\frac{6}{x\left(x-2\right)}$$

Show Solution

Solve for x (complex solution)

$x=\frac{\sqrt{K\left(K-6\right)}+K}{K}$
$x=\frac{-\sqrt{K\left(K-6\right)}+K}{K}\text{, }K\neq 0$

Solve for x

$x=\frac{\sqrt{K\left(K-6\right)}+K}{K}$
$x=\frac{-\sqrt{K\left(K-6\right)}+K}{K}\text{, }K<0\text{ or }K\geq 6$