Solve x-2/2=4x/x-2

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

Solve for x

$$x=4\sqrt{2}+6\approx 11.656854249$$

$$x=6-4\sqrt{2}\approx 0.343145751$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

Multiply 2 and 4 to get 8.

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

Subtract 8x from both sides.

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

Combine -4x and -8x to get -12x.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16}}{2}$$

Add 144 to -16.

$$x=\frac{-\left(-12\right)±\sqrt{128}}{2}$$

Take the square root of 128.

$$x=\frac{-\left(-12\right)±8\sqrt{2}}{2}$$

The opposite of -12 is 12.

$$x=\frac{12±8\sqrt{2}}{2}$$

Now solve the equation x=\frac{12±8\sqrt{2}}{2} when ± is plus. Add 12 to 8\sqrt{2}.

$$x=\frac{8\sqrt{2}+12}{2}$$

Divide 12+8\sqrt{2} by 2.

$$x=4\sqrt{2}+6$$

Now solve the equation x=\frac{12±8\sqrt{2}}{2} when ± is minus. Subtract 8\sqrt{2} from 12.

$$x=\frac{12-8\sqrt{2}}{2}$$

Divide 12-8\sqrt{2} by 2.

$$x=6-4\sqrt{2}$$

The equation is now solved.

$$x=4\sqrt{2}+6 x=6-4\sqrt{2}$$

Steps for Completing the Square

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

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

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

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

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

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

Multiply 2 and 4 to get 8.

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

Subtract 8x from both sides.

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

Combine -4x and -8x to get -12x.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-4+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-4+36$$

Add -4 to 36.

$$x^{2}-12x+36=32$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=32$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{32}$$

Simplify.

$$x-6=4\sqrt{2} x-6=-4\sqrt{2}$$

Add 6 to both sides of the equation.

$$x=4\sqrt{2}+6 x=6-4\sqrt{2}$$