Solve 2/x+3+1/12=1/2x-1

$$\frac { 2 } { x + 3 } + \frac { 1 } { 12 } = \frac { 1 } { 2 x - 1 }$$

Solve for x

$$x = \frac{\sqrt{2185} - 41}{4} \approx 1.435995893$$

$$x=\frac{-\sqrt{2185}-41}{4}\approx -21.935995893$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 12\left(2x-1\right)\left(x+3\right), the least common multiple of x+3,12,2x-1.

$$\left(24x-12\right)\times 2+12\left(2x-1\right)\left(x+3\right)\times \frac{1}{12}=12x+36$$

Use the distributive property to multiply 24x-12 by 2.

$$48x-24+12\left(2x-1\right)\left(x+3\right)\times \frac{1}{12}=12x+36$$

Multiply 12 and \frac{1}{12} to get 1.

$$48x-24+\left(2x-1\right)\left(x+3\right)=12x+36$$

Use the distributive property to multiply 2x-1 by x+3 and combine like terms.

$$48x-24+2x^{2}+5x-3=12x+36$$

Combine 48x and 5x to get 53x.

$$53x-24+2x^{2}-3=12x+36$$

Subtract 3 from -24 to get -27.

$$53x-27+2x^{2}=12x+36$$

Subtract 12x from both sides.

$$53x-27+2x^{2}-12x=36$$

Combine 53x and -12x to get 41x.

$$41x-27+2x^{2}=36$$

Subtract 36 from both sides.

$$41x-27+2x^{2}-36=0$$

Subtract 36 from -27 to get -63.

$$41x-63+2x^{2}=0$$

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.

$$2x^{2}+41x-63=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 41 for b, and -63 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-41±\sqrt{41^{2}-4\times 2\left(-63\right)}}{2\times 2}$$

Square 41.

$$x=\frac{-41±\sqrt{1681-4\times 2\left(-63\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-41±\sqrt{1681-8\left(-63\right)}}{2\times 2}$$

Multiply -8 times -63.

$$x=\frac{-41±\sqrt{1681+504}}{2\times 2}$$

Add 1681 to 504.

$$x=\frac{-41±\sqrt{2185}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-41±\sqrt{2185}}{4}$$

Now solve the equation x=\frac{-41±\sqrt{2185}}{4} when ± is plus. Add -41 to \sqrt{2185}.

$$x=\frac{\sqrt{2185}-41}{4}$$

Now solve the equation x=\frac{-41±\sqrt{2185}}{4} when ± is minus. Subtract \sqrt{2185} from -41.

$$x=\frac{-\sqrt{2185}-41}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{2185}-41}{4} x=\frac{-\sqrt{2185}-41}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 12\left(2x-1\right)\left(x+3\right), the least common multiple of x+3,12,2x-1.

$$\left(24x-12\right)\times 2+12\left(2x-1\right)\left(x+3\right)\times \frac{1}{12}=12x+36$$

Use the distributive property to multiply 24x-12 by 2.

$$48x-24+12\left(2x-1\right)\left(x+3\right)\times \frac{1}{12}=12x+36$$

Multiply 12 and \frac{1}{12} to get 1.

$$48x-24+\left(2x-1\right)\left(x+3\right)=12x+36$$

Use the distributive property to multiply 2x-1 by x+3 and combine like terms.

$$48x-24+2x^{2}+5x-3=12x+36$$

Combine 48x and 5x to get 53x.

$$53x-24+2x^{2}-3=12x+36$$

Subtract 3 from -24 to get -27.

$$53x-27+2x^{2}=12x+36$$

Subtract 12x from both sides.

$$53x-27+2x^{2}-12x=36$$

Combine 53x and -12x to get 41x.

$$41x-27+2x^{2}=36$$

Add 27 to both sides.

$$41x+2x^{2}=36+27$$

Add 36 and 27 to get 63.

$$41x+2x^{2}=63$$

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.

$$2x^{2}+41x=63$$

Divide both sides by 2.

$$\frac{2x^{2}+41x}{2}=\frac{63}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{41}{2}x=\frac{63}{2}$$

Divide \frac{41}{2}, the coefficient of the x term, by 2 to get \frac{41}{4}. Then add the square of \frac{41}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{41}{2}x+\left(\frac{41}{4}\right)^{2}=\frac{63}{2}+\left(\frac{41}{4}\right)^{2}$$

Square \frac{41}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{41}{2}x+\frac{1681}{16}=\frac{63}{2}+\frac{1681}{16}$$

Add \frac{63}{2} to \frac{1681}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{41}{2}x+\frac{1681}{16}=\frac{2185}{16}$$

Factor x^{2}+\frac{41}{2}x+\frac{1681}{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}.

$$\left(x+\frac{41}{4}\right)^{2}=\frac{2185}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{41}{4}\right)^{2}}=\sqrt{\frac{2185}{16}}$$

Simplify.

$$x+\frac{41}{4}=\frac{\sqrt{2185}}{4} x+\frac{41}{4}=-\frac{\sqrt{2185}}{4}$$

Subtract \frac{41}{4} from both sides of the equation.

$$x=\frac{\sqrt{2185}-41}{4} x=\frac{-\sqrt{2185}-41}{4}$$