Solve 1+2/x^2-4x=1/2(x-4)

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

Solve for x

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,4 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-4\right), the least common multiple of x^{2}-4x,2\left(x-4\right).

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

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

$$2x^{2}-8x+2\times 2=x$$

Multiply 2 and 2 to get 4.

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

Subtract x from both sides.

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

Combine -8x and -x to get -9x.

$$2x^{2}-9x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-9 ab=2\times 4=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -9.

$$a=-8 b=-1$$

Rewrite 2x^{2}-9x+4 as \left(2x^{2}-8x\right)+\left(-x+4\right).

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

Factor out 2x in the first and -1 in the second group.

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

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(2x-1\right)$$

To find equation solutions, solve x-4=0 and 2x-1=0.

$$x=4 x=\frac{1}{2}$$

Variable x cannot be equal to 4.

$$x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,4 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-4\right), the least common multiple of x^{2}-4x,2\left(x-4\right).

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

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

$$2x^{2}-8x+2\times 2=x$$

Multiply 2 and 2 to get 4.

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

Subtract x from both sides.

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

Combine -8x and -x to get -9x.

$$2x^{2}-9x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\times 4}}{2\times 2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\times 4}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$x=\frac{-\left(-9\right)±\sqrt{81-32}}{2\times 2}$$

Add 81 to -32.

$$x=\frac{-\left(-9\right)±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$x=\frac{-\left(-9\right)±7}{2\times 2}$$

The opposite of -9 is 9.

$$x=\frac{9±7}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{9±7}{4}$$

Now solve the equation x=\frac{9±7}{4} when ± is plus. Add 9 to 7.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{9±7}{4} when ± is minus. Subtract 7 from 9.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{2}$$

The equation is now solved.

$$x=4 x=\frac{1}{2}$$

Variable x cannot be equal to 4.

$$x=\frac{1}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,4 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-4\right), the least common multiple of x^{2}-4x,2\left(x-4\right).

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

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

$$2x^{2}-8x+2\times 2=x$$

Multiply 2 and 2 to get 4.

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

Subtract x from both sides.

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

Combine -8x and -x to get -9x.

$$2x^{2}-9x+4=0$$

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

$$2x^{2}-9x=-4$$

Divide both sides by 2.

$$\frac{2x^{2}-9x}{2}=-\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{9}{2}x=-\frac{4}{2}$$

Divide -4 by 2.

$$x^{2}-\frac{9}{2}x=-2$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=-2+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=-2+\frac{81}{16}$$

Add -2 to \frac{81}{16}.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{49}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{7}{4} x-\frac{9}{4}=-\frac{7}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=4 x=\frac{1}{2}$$

Variable x cannot be equal to 4.

$$x=\frac{1}{2}$$