Solve 1/2x+2+1/4x-4=3

$$\frac { 1 } { 2 x + 2 } + \frac { 1 } { 4 x - 4 } = 3$$

Solve for x

$$x=\frac{\sqrt{537}}{24}+\frac{1}{8}\approx 1.090552519$$

$$x=-\frac{\sqrt{537}}{24}+\frac{1}{8}\approx -0.840552519$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-1\right)\left(x+1\right), the least common multiple of 2x+2,4x-4.

$$2x-2+x+1=12\left(x-1\right)\left(x+1\right)$$

Combine 2x and x to get 3x.

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

Add -2 and 1 to get -1.

$$3x-1=12\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 12 by x-1.

$$3x-1=\left(12x-12\right)\left(x+1\right)$$

Use the distributive property to multiply 12x-12 by x+1 and combine like terms.

$$3x-1=12x^{2}-12$$

Subtract 12x^{2} from both sides.

$$3x-1-12x^{2}=-12$$

Add 12 to both sides.

$$3x-1-12x^{2}+12=0$$

Add -1 and 12 to get 11.

$$3x+11-12x^{2}=0$$

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.

$$-12x^{2}+3x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, 3 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-12\right)\times 11}}{2\left(-12\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-12\right)\times 11}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-3±\sqrt{9+48\times 11}}{2\left(-12\right)}$$

Multiply 48 times 11.

$$x=\frac{-3±\sqrt{9+528}}{2\left(-12\right)}$$

Add 9 to 528.

$$x=\frac{-3±\sqrt{537}}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-3±\sqrt{537}}{-24}$$

Now solve the equation x=\frac{-3±\sqrt{537}}{-24} when ± is plus. Add -3 to \sqrt{537}.

$$x=\frac{\sqrt{537}-3}{-24}$$

Divide -3+\sqrt{537} by -24.

$$x=-\frac{\sqrt{537}}{24}+\frac{1}{8}$$

Now solve the equation x=\frac{-3±\sqrt{537}}{-24} when ± is minus. Subtract \sqrt{537} from -3.

$$x=\frac{-\sqrt{537}-3}{-24}$$

Divide -3-\sqrt{537} by -24.

$$x=\frac{\sqrt{537}}{24}+\frac{1}{8}$$

The equation is now solved.

$$x=-\frac{\sqrt{537}}{24}+\frac{1}{8} x=\frac{\sqrt{537}}{24}+\frac{1}{8}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-1\right)\left(x+1\right), the least common multiple of 2x+2,4x-4.

$$2x-2+x+1=12\left(x-1\right)\left(x+1\right)$$

Combine 2x and x to get 3x.

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

Add -2 and 1 to get -1.

$$3x-1=12\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 12 by x-1.

$$3x-1=\left(12x-12\right)\left(x+1\right)$$

Use the distributive property to multiply 12x-12 by x+1 and combine like terms.

$$3x-1=12x^{2}-12$$

Subtract 12x^{2} from both sides.

$$3x-1-12x^{2}=-12$$

Add 1 to both sides.

$$3x-12x^{2}=-12+1$$

Add -12 and 1 to get -11.

$$3x-12x^{2}=-11$$

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.

$$-12x^{2}+3x=-11$$

Divide both sides by -12.

$$\frac{-12x^{2}+3x}{-12}=-\frac{11}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\frac{3}{-12}x=-\frac{11}{-12}$$

Reduce the fraction \frac{3}{-12} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{1}{4}x=-\frac{11}{-12}$$

Divide -11 by -12.

$$x^{2}-\frac{1}{4}x=\frac{11}{12}$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\frac{11}{12}+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{11}{12}+\frac{1}{64}$$

Add \frac{11}{12} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{179}{192}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{179}{192}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{179}{192}}$$

Simplify.

$$x-\frac{1}{8}=\frac{\sqrt{537}}{24} x-\frac{1}{8}=-\frac{\sqrt{537}}{24}$$

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{\sqrt{537}}{24}+\frac{1}{8} x=-\frac{\sqrt{537}}{24}+\frac{1}{8}$$