Solve x/x-2-2=3(x-2)/x

$$\frac { x } { x - 2 } - 2 = \frac { 3 ( x - 2 ) } { x }$$

Solve for x

$$x=3$$

$$x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right), the least common multiple of x-2,x.

$$xx+x\left(x-2\right)\left(-2\right)=\left(x-2\right)\times 3\left(x-2\right)$$

Multiply x and x to get x^{2}.

$$x^{2}+x\left(x-2\right)\left(-2\right)=\left(x-2\right)\times 3\left(x-2\right)$$

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

$$x^{2}+\left(x^{2}-2x\right)\left(-2\right)=\left(x-2\right)\times 3\left(x-2\right)$$

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

$$x^{2}-2x^{2}+4x=\left(x-2\right)\times 3\left(x-2\right)$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+4x=\left(x-2\right)\times 3\left(x-2\right)$$

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

$$-x^{2}+4x=\left(x-2\right)^{2}\times 3$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$-x^{2}+4x=\left(x^{2}-4x+4\right)\times 3$$

Use the distributive property to multiply x^{2}-4x+4 by 3.

$$-x^{2}+4x=3x^{2}-12x+12$$

Subtract 3x^{2} from both sides.

$$-x^{2}+4x-3x^{2}=-12x+12$$

Combine -x^{2} and -3x^{2} to get -4x^{2}.

$$-4x^{2}+4x=-12x+12$$

Add 12x to both sides.

$$-4x^{2}+4x+12x=12$$

Combine 4x and 12x to get 16x.

$$-4x^{2}+16x=12$$

Subtract 12 from both sides.

$$-4x^{2}+16x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 16 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-4\right)\left(-12\right)}}{2\left(-4\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-4\right)\left(-12\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-16±\sqrt{256+16\left(-12\right)}}{2\left(-4\right)}$$

Multiply 16 times -12.

$$x=\frac{-16±\sqrt{256-192}}{2\left(-4\right)}$$

Add 256 to -192.

$$x=\frac{-16±\sqrt{64}}{2\left(-4\right)}$$

Take the square root of 64.

$$x=\frac{-16±8}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-16±8}{-8}$$

Now solve the equation x=\frac{-16±8}{-8} when ± is plus. Add -16 to 8.

$$x=-\frac{8}{-8}$$

Divide -8 by -8.

$$x=1$$

Now solve the equation x=\frac{-16±8}{-8} when ± is minus. Subtract 8 from -16.

$$x=-\frac{24}{-8}$$

Divide -24 by -8.

$$x=3$$

The equation is now solved.

$$x=1 x=3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right), the least common multiple of x-2,x.

$$xx+x\left(x-2\right)\left(-2\right)=\left(x-2\right)\times 3\left(x-2\right)$$

Multiply x and x to get x^{2}.

$$x^{2}+x\left(x-2\right)\left(-2\right)=\left(x-2\right)\times 3\left(x-2\right)$$

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

$$x^{2}+\left(x^{2}-2x\right)\left(-2\right)=\left(x-2\right)\times 3\left(x-2\right)$$

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

$$x^{2}-2x^{2}+4x=\left(x-2\right)\times 3\left(x-2\right)$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+4x=\left(x-2\right)\times 3\left(x-2\right)$$

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

$$-x^{2}+4x=\left(x-2\right)^{2}\times 3$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$-x^{2}+4x=\left(x^{2}-4x+4\right)\times 3$$

Use the distributive property to multiply x^{2}-4x+4 by 3.

$$-x^{2}+4x=3x^{2}-12x+12$$

Subtract 3x^{2} from both sides.

$$-x^{2}+4x-3x^{2}=-12x+12$$

Combine -x^{2} and -3x^{2} to get -4x^{2}.

$$-4x^{2}+4x=-12x+12$$

Add 12x to both sides.

$$-4x^{2}+4x+12x=12$$

Combine 4x and 12x to get 16x.

$$-4x^{2}+16x=12$$

Divide both sides by -4.

$$\frac{-4x^{2}+16x}{-4}=\frac{12}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{16}{-4}x=\frac{12}{-4}$$

Divide 16 by -4.

$$x^{2}-4x=\frac{12}{-4}$$

Divide 12 by -4.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add -3 to 4.

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

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$