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

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{7}i+1}{8}\approx 0.125-0.330718914i$$

$$x=\frac{1+\sqrt{7}i}{8}\approx 0.125+0.330718914i$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply 2x-1 by 2.

$$4x-2-\left(2x+1\right)\times 2x=\left(2x-1\right)\left(2x+1\right)$$

Use the distributive property to multiply 2x+1 by 2.

$$4x-2-\left(4x+2\right)x=\left(2x-1\right)\left(2x+1\right)$$

Use the distributive property to multiply 4x+2 by x.

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

To find the opposite of 4x^{2}+2x, find the opposite of each term.

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

Combine 4x and -2x to get 2x.

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

Consider \left(2x-1\right)\left(2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Expand \left(2x\right)^{2}.

$$2x-2-4x^{2}=2^{2}x^{2}-1$$

Calculate 2 to the power of 2 and get 4.

$$2x-2-4x^{2}=4x^{2}-1$$

Subtract 4x^{2} from both sides.

$$2x-2-4x^{2}-4x^{2}=-1$$

Combine -4x^{2} and -4x^{2} to get -8x^{2}.

$$2x-2-8x^{2}=-1$$

Add 1 to both sides.

$$2x-2-8x^{2}+1=0$$

Add -2 and 1 to get -1.

$$2x-1-8x^{2}=0$$

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.

$$-8x^{2}+2x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 2 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-8\right)\left(-1\right)}}{2\left(-8\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-8\right)\left(-1\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-2±\sqrt{4+32\left(-1\right)}}{2\left(-8\right)}$$

Multiply 32 times -1.

$$x=\frac{-2±\sqrt{4-32}}{2\left(-8\right)}$$

Add 4 to -32.

$$x=\frac{-2±\sqrt{-28}}{2\left(-8\right)}$$

Take the square root of -28.

$$x=\frac{-2±2\sqrt{7}i}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-2±2\sqrt{7}i}{-16}$$

Now solve the equation x=\frac{-2±2\sqrt{7}i}{-16} when ± is plus. Add -2 to 2i\sqrt{7}.

$$x=\frac{-2+2\sqrt{7}i}{-16}$$

Divide -2+2i\sqrt{7} by -16.

$$x=\frac{-\sqrt{7}i+1}{8}$$

Now solve the equation x=\frac{-2±2\sqrt{7}i}{-16} when ± is minus. Subtract 2i\sqrt{7} from -2.

$$x=\frac{-2\sqrt{7}i-2}{-16}$$

Divide -2-2i\sqrt{7} by -16.

$$x=\frac{1+\sqrt{7}i}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{7}i+1}{8} x=\frac{1+\sqrt{7}i}{8}$$

Steps for Completing the Square

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

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

Use the distributive property to multiply 2x-1 by 2.

$$4x-2-\left(2x+1\right)\times 2x=\left(2x-1\right)\left(2x+1\right)$$

Use the distributive property to multiply 2x+1 by 2.

$$4x-2-\left(4x+2\right)x=\left(2x-1\right)\left(2x+1\right)$$

Use the distributive property to multiply 4x+2 by x.

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

To find the opposite of 4x^{2}+2x, find the opposite of each term.

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

Combine 4x and -2x to get 2x.

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

Consider \left(2x-1\right)\left(2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Expand \left(2x\right)^{2}.

$$2x-2-4x^{2}=2^{2}x^{2}-1$$

Calculate 2 to the power of 2 and get 4.

$$2x-2-4x^{2}=4x^{2}-1$$

Subtract 4x^{2} from both sides.

$$2x-2-4x^{2}-4x^{2}=-1$$

Combine -4x^{2} and -4x^{2} to get -8x^{2}.

$$2x-2-8x^{2}=-1$$

Add 2 to both sides.

$$2x-8x^{2}=-1+2$$

Add -1 and 2 to get 1.

$$2x-8x^{2}=1$$

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.

$$-8x^{2}+2x=1$$

Divide both sides by -8.

$$\frac{-8x^{2}+2x}{-8}=\frac{1}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{2}{-8}x=\frac{1}{-8}$$

Reduce the fraction \frac{2}{-8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{4}x=\frac{1}{-8}$$

Divide 1 by -8.

$$x^{2}-\frac{1}{4}x=-\frac{1}{8}$$

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.

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

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

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=-\frac{1}{8}+\frac{1}{64}$$

Add -\frac{1}{8} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=-\frac{7}{64}$$

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}.

$$\left(x-\frac{1}{8}\right)^{2}=-\frac{7}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{-\frac{7}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{\sqrt{7}i}{8} x-\frac{1}{8}=-\frac{\sqrt{7}i}{8}$$

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

$$x=\frac{1+\sqrt{7}i}{8} x=\frac{-\sqrt{7}i+1}{8}$$