Solve 1/x+x/2=x+4/2x

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

Solve for x

$$x=-1$$

$$x=2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of x,2,2x.

$$2+xx=x+4$$

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

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

Subtract x from both sides.

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

Subtract 4 from both sides.

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

Subtract 4 from 2 to get -2.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8}}{2}$$

Add 1 to 8.

$$x=\frac{-\left(-1\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$x=\frac{-\left(-1\right)±3}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±3}{2}$$

Now solve the equation x=\frac{1±3}{2} when ± is plus. Add 1 to 3.

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

Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{1±3}{2} when ± is minus. Subtract 3 from 1.

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=2 x=-1$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of x,2,2x.

$$2+xx=x+4$$

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

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

Subtract x from both sides.

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

Subtract 2 from both sides.

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

Subtract 2 from 4 to get 2.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{3}{2} x-\frac{1}{2}=-\frac{3}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=2 x=-1$$