Solve x/2+2/x-x/12=x/4-frac{5}{4}

$$\frac { x } { 2 } + \frac { 2 } { x } - \frac { x } { 12 } = \frac { x } { 4 } - \frac { 5 } { 4 }$$

Solve for x

$$x=\frac{\sqrt{33}-15}{4}\approx -2.313859338$$

$$x=\frac{-\sqrt{33}-15}{4}\approx -5.186140662$$

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 12x, the least common multiple of 2,x,12,4.

$$6xx+12\times 2-xx=3xx+12x\left(-\frac{5}{4}\right)$$

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

$$6xx+12\times 2-x^{2}=3xx+12x\left(-\frac{5}{4}\right)$$

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

$$6x^{2}+12\times 2-x^{2}=3xx+12x\left(-\frac{5}{4}\right)$$

Multiply 12 and 2 to get 24.

$$6x^{2}+24-x^{2}=3xx+12x\left(-\frac{5}{4}\right)$$

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

$$6x^{2}+24-x^{2}=3x^{2}+12x\left(-\frac{5}{4}\right)$$

Express 12\left(-\frac{5}{4}\right) as a single fraction.

$$6x^{2}+24-x^{2}=3x^{2}+\frac{12\left(-5\right)}{4}x$$

Multiply 12 and -5 to get -60.

$$6x^{2}+24-x^{2}=3x^{2}+\frac{-60}{4}x$$

Divide -60 by 4 to get -15.

$$6x^{2}+24-x^{2}=3x^{2}-15x$$

Subtract 3x^{2} from both sides.

$$6x^{2}+24-x^{2}-3x^{2}=-15x$$

Combine 6x^{2} and -3x^{2} to get 3x^{2}.

$$3x^{2}+24-x^{2}=-15x$$

Add 15x to both sides.

$$3x^{2}+24-x^{2}+15x=0$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}+24+15x=0$$

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.

$$2x^{2}+15x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 15 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-15±\sqrt{15^{2}-4\times 2\times 24}}{2\times 2}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 2\times 24}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-15±\sqrt{225-8\times 24}}{2\times 2}$$

Multiply -8 times 24.

$$x=\frac{-15±\sqrt{225-192}}{2\times 2}$$

Add 225 to -192.

$$x=\frac{-15±\sqrt{33}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-15±\sqrt{33}}{4}$$

Now solve the equation x=\frac{-15±\sqrt{33}}{4} when ± is plus. Add -15 to \sqrt{33}.

$$x=\frac{\sqrt{33}-15}{4}$$

Now solve the equation x=\frac{-15±\sqrt{33}}{4} when ± is minus. Subtract \sqrt{33} from -15.

$$x=\frac{-\sqrt{33}-15}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{33}-15}{4} x=\frac{-\sqrt{33}-15}{4}$$

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 12x, the least common multiple of 2,x,12,4.

$$6xx+12\times 2-xx=3xx+12x\left(-\frac{5}{4}\right)$$

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

$$6xx+12\times 2-x^{2}=3xx+12x\left(-\frac{5}{4}\right)$$

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

$$6x^{2}+12\times 2-x^{2}=3xx+12x\left(-\frac{5}{4}\right)$$

Multiply 12 and 2 to get 24.

$$6x^{2}+24-x^{2}=3xx+12x\left(-\frac{5}{4}\right)$$

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

$$6x^{2}+24-x^{2}=3x^{2}+12x\left(-\frac{5}{4}\right)$$

Express 12\left(-\frac{5}{4}\right) as a single fraction.

$$6x^{2}+24-x^{2}=3x^{2}+\frac{12\left(-5\right)}{4}x$$

Multiply 12 and -5 to get -60.

$$6x^{2}+24-x^{2}=3x^{2}+\frac{-60}{4}x$$

Divide -60 by 4 to get -15.

$$6x^{2}+24-x^{2}=3x^{2}-15x$$

Subtract 3x^{2} from both sides.

$$6x^{2}+24-x^{2}-3x^{2}=-15x$$

Combine 6x^{2} and -3x^{2} to get 3x^{2}.

$$3x^{2}+24-x^{2}=-15x$$

Add 15x to both sides.

$$3x^{2}+24-x^{2}+15x=0$$

Subtract 24 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-x^{2}+15x=-24$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}+15x=-24$$

Divide both sides by 2.

$$\frac{2x^{2}+15x}{2}=-\frac{24}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{15}{2}x=-\frac{24}{2}$$

Divide -24 by 2.

$$x^{2}+\frac{15}{2}x=-12$$

Divide \frac{15}{2}, the coefficient of the x term, by 2 to get \frac{15}{4}. Then add the square of \frac{15}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{15}{2}x+\left(\frac{15}{4}\right)^{2}=-12+\left(\frac{15}{4}\right)^{2}$$

Square \frac{15}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{15}{2}x+\frac{225}{16}=-12+\frac{225}{16}$$

Add -12 to \frac{225}{16}.

$$x^{2}+\frac{15}{2}x+\frac{225}{16}=\frac{33}{16}$$

Factor x^{2}+\frac{15}{2}x+\frac{225}{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}.

$$\left(x+\frac{15}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{15}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$x+\frac{15}{4}=\frac{\sqrt{33}}{4} x+\frac{15}{4}=-\frac{\sqrt{33}}{4}$$

Subtract \frac{15}{4} from both sides of the equation.

$$x=\frac{\sqrt{33}-15}{4} x=\frac{-\sqrt{33}-15}{4}$$