Solve 3/x(2-1/2x)+x+3=64

$$\frac { 3 } { x } ( 2 - \frac { 1 } { 2 } x ) + x + 3 = 64$$

Solve for x

$$x = \frac{\sqrt{15529} + 125}{4} \approx 62.403852089$$

$$x=\frac{125-\sqrt{15529}}{4}\approx 0.096147911$$

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.

$$2\times 3\left(2-\frac{1}{2}x\right)+2xx+2x\times 3=128x$$

Multiply 2 and 3 to get 6.

$$6\left(2-\frac{1}{2}x\right)+2xx+2x\times 3=128x$$

Use the distributive property to multiply 6 by 2-\frac{1}{2}x.

$$12+6\left(-\frac{1}{2}\right)x+2xx+2x\times 3=128x$$

Express 6\left(-\frac{1}{2}\right) as a single fraction.

$$12+\frac{6\left(-1\right)}{2}x+2xx+2x\times 3=128x$$

Multiply 6 and -1 to get -6.

$$12+\frac{-6}{2}x+2xx+2x\times 3=128x$$

Divide -6 by 2 to get -3.

$$12-3x+2xx+2x\times 3=128x$$

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

$$12-3x+2x^{2}+2x\times 3=128x$$

Multiply 2 and 3 to get 6.

$$12-3x+2x^{2}+6x=128x$$

Combine -3x and 6x to get 3x.

$$12+3x+2x^{2}=128x$$

Subtract 128x from both sides.

$$12+3x+2x^{2}-128x=0$$

Combine 3x and -128x to get -125x.

$$12-125x+2x^{2}=0$$

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.

$$2x^{2}-125x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -125 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-125\right)±\sqrt{\left(-125\right)^{2}-4\times 2\times 12}}{2\times 2}$$

Square -125.

$$x=\frac{-\left(-125\right)±\sqrt{15625-4\times 2\times 12}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-125\right)±\sqrt{15625-8\times 12}}{2\times 2}$$

Multiply -8 times 12.

$$x=\frac{-\left(-125\right)±\sqrt{15625-96}}{2\times 2}$$

Add 15625 to -96.

$$x=\frac{-\left(-125\right)±\sqrt{15529}}{2\times 2}$$

The opposite of -125 is 125.

$$x=\frac{125±\sqrt{15529}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{125±\sqrt{15529}}{4}$$

Now solve the equation x=\frac{125±\sqrt{15529}}{4} when ± is plus. Add 125 to \sqrt{15529}.

$$x=\frac{\sqrt{15529}+125}{4}$$

Now solve the equation x=\frac{125±\sqrt{15529}}{4} when ± is minus. Subtract \sqrt{15529} from 125.

$$x=\frac{125-\sqrt{15529}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{15529}+125}{4} x=\frac{125-\sqrt{15529}}{4}$$

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.

$$2\times 3\left(2-\frac{1}{2}x\right)+2xx+2x\times 3=128x$$

Multiply 2 and 3 to get 6.

$$6\left(2-\frac{1}{2}x\right)+2xx+2x\times 3=128x$$

Use the distributive property to multiply 6 by 2-\frac{1}{2}x.

$$12+6\left(-\frac{1}{2}\right)x+2xx+2x\times 3=128x$$

Express 6\left(-\frac{1}{2}\right) as a single fraction.

$$12+\frac{6\left(-1\right)}{2}x+2xx+2x\times 3=128x$$

Multiply 6 and -1 to get -6.

$$12+\frac{-6}{2}x+2xx+2x\times 3=128x$$

Divide -6 by 2 to get -3.

$$12-3x+2xx+2x\times 3=128x$$

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

$$12-3x+2x^{2}+2x\times 3=128x$$

Multiply 2 and 3 to get 6.

$$12-3x+2x^{2}+6x=128x$$

Combine -3x and 6x to get 3x.

$$12+3x+2x^{2}=128x$$

Subtract 128x from both sides.

$$12+3x+2x^{2}-128x=0$$

Combine 3x and -128x to get -125x.

$$12-125x+2x^{2}=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$-125x+2x^{2}=-12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-125x=-12$$

Divide both sides by 2.

$$\frac{2x^{2}-125x}{2}=-\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{125}{2}x=-\frac{12}{2}$$

Divide -12 by 2.

$$x^{2}-\frac{125}{2}x=-6$$

Divide -\frac{125}{2}, the coefficient of the x term, by 2 to get -\frac{125}{4}. Then add the square of -\frac{125}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{125}{2}x+\left(-\frac{125}{4}\right)^{2}=-6+\left(-\frac{125}{4}\right)^{2}$$

Square -\frac{125}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{125}{2}x+\frac{15625}{16}=-6+\frac{15625}{16}$$

Add -6 to \frac{15625}{16}.

$$x^{2}-\frac{125}{2}x+\frac{15625}{16}=\frac{15529}{16}$$

Factor x^{2}-\frac{125}{2}x+\frac{15625}{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}.

$$\left(x-\frac{125}{4}\right)^{2}=\frac{15529}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{125}{4}\right)^{2}}=\sqrt{\frac{15529}{16}}$$

Simplify.

$$x-\frac{125}{4}=\frac{\sqrt{15529}}{4} x-\frac{125}{4}=-\frac{\sqrt{15529}}{4}$$

Add \frac{125}{4} to both sides of the equation.

$$x=\frac{\sqrt{15529}+125}{4} x=\frac{125-\sqrt{15529}}{4}$$