Solve x-2/2x=2/2-x+4/x^2-2x

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

Solve for x

$$x=-2$$

Steps Using Difference of Squares

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 2x\left(x-2\right), the least common multiple of 2x,2-x,x^{2}-2x.

$$\left(x-2\right)\left(x-2\right)=-2x\times 2+2\times 4$$

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

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

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

$$x^{2}-4x+4=-2x\times 2+2\times 4$$

Multiply -2 and 2 to get -4.

$$x^{2}-4x+4=-4x+2\times 4$$

Multiply 2 and 4 to get 8.

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

Add 4x to both sides.

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

Combine -4x and 4x to get 0.

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

Subtract 8 from both sides.

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

Subtract 8 from 4 to get -4.

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

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

To find equation solutions, solve x-2=0 and x+2=0.

$$x=2 x=-2$$

Variable x cannot be equal to 2.

$$x=-2$$

Steps by Finding Square Root

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 2x\left(x-2\right), the least common multiple of 2x,2-x,x^{2}-2x.

$$\left(x-2\right)\left(x-2\right)=-2x\times 2+2\times 4$$

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

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

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

$$x^{2}-4x+4=-2x\times 2+2\times 4$$

Multiply -2 and 2 to get -4.

$$x^{2}-4x+4=-4x+2\times 4$$

Multiply 2 and 4 to get 8.

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

Add 4x to both sides.

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

Combine -4x and 4x to get 0.

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

Subtract 4 from both sides.

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

Subtract 4 from 8 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Variable x cannot be equal to 2.

$$x=-2$$

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 2x\left(x-2\right), the least common multiple of 2x,2-x,x^{2}-2x.

$$\left(x-2\right)\left(x-2\right)=-2x\times 2+2\times 4$$

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

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

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

$$x^{2}-4x+4=-2x\times 2+2\times 4$$

Multiply -2 and 2 to get -4.

$$x^{2}-4x+4=-4x+2\times 4$$

Multiply 2 and 4 to get 8.

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

Add 4x to both sides.

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

Combine -4x and 4x to get 0.

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

Subtract 8 from both sides.

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

Subtract 8 from 4 to get -4.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-4\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{0±4}{2}$$

Now solve the equation x=\frac{0±4}{2} when ± is plus. Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{0±4}{2} when ± is minus. Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=2 x=-2$$

Variable x cannot be equal to 2.

$$x=-2$$