Solve 2(2/1+2x)+x-4/(1+2x)(1+x)+4=0

$$2 ( \frac { 2 } { 1 + 2 x } ) + \frac { x - 4 } { ( 1 + 2 x ) ( 1 + x ) } + 4 = 0$$

Solve for x

$$x=\frac{\sqrt{161}-17}{16}\approx -0.269463904$$

$$x=\frac{-\sqrt{161}-17}{16}\approx -1.855536096$$

Steps Using the Quadratic Formula

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

$$2\left(x+1\right)\times 2+x-4+\left(x+1\right)\left(2x+1\right)\times 4=0$$

Multiply 2 and 2 to get 4.

$$4\left(x+1\right)+x-4+\left(x+1\right)\left(2x+1\right)\times 4=0$$

Use the distributive property to multiply 4 by x+1.

$$4x+4+x-4+\left(x+1\right)\left(2x+1\right)\times 4=0$$

Combine 4x and x to get 5x.

$$5x+4-4+\left(x+1\right)\left(2x+1\right)\times 4=0$$

Subtract 4 from 4 to get 0.

$$5x+\left(x+1\right)\left(2x+1\right)\times 4=0$$

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

$$5x+\left(2x^{2}+3x+1\right)\times 4=0$$

Use the distributive property to multiply 2x^{2}+3x+1 by 4.

$$5x+8x^{2}+12x+4=0$$

Combine 5x and 12x to get 17x.

$$17x+8x^{2}+4=0$$

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.

$$8x^{2}+17x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 17 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4\times 8\times 4}}{2\times 8}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 8\times 4}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-17±\sqrt{289-32\times 4}}{2\times 8}$$

Multiply -32 times 4.

$$x=\frac{-17±\sqrt{289-128}}{2\times 8}$$

Add 289 to -128.

$$x=\frac{-17±\sqrt{161}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-17±\sqrt{161}}{16}$$

Now solve the equation x=\frac{-17±\sqrt{161}}{16} when ± is plus. Add -17 to \sqrt{161}.

$$x=\frac{\sqrt{161}-17}{16}$$

Now solve the equation x=\frac{-17±\sqrt{161}}{16} when ± is minus. Subtract \sqrt{161} from -17.

$$x=\frac{-\sqrt{161}-17}{16}$$

The equation is now solved.

$$x=\frac{\sqrt{161}-17}{16} x=\frac{-\sqrt{161}-17}{16}$$

Steps for Completing the Square

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

$$2\left(x+1\right)\times 2+x-4+\left(x+1\right)\left(2x+1\right)\times 4=0$$

Multiply 2 and 2 to get 4.

$$4\left(x+1\right)+x-4+\left(x+1\right)\left(2x+1\right)\times 4=0$$

Use the distributive property to multiply 4 by x+1.

$$4x+4+x-4+\left(x+1\right)\left(2x+1\right)\times 4=0$$

Combine 4x and x to get 5x.

$$5x+4-4+\left(x+1\right)\left(2x+1\right)\times 4=0$$

Subtract 4 from 4 to get 0.

$$5x+\left(x+1\right)\left(2x+1\right)\times 4=0$$

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

$$5x+\left(2x^{2}+3x+1\right)\times 4=0$$

Use the distributive property to multiply 2x^{2}+3x+1 by 4.

$$5x+8x^{2}+12x+4=0$$

Combine 5x and 12x to get 17x.

$$17x+8x^{2}+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$17x+8x^{2}=-4$$

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.

$$8x^{2}+17x=-4$$

Divide both sides by 8.

$$\frac{8x^{2}+17x}{8}=-\frac{4}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{17}{8}x=-\frac{4}{8}$$

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{17}{8}x=-\frac{1}{2}$$

Divide \frac{17}{8}, the coefficient of the x term, by 2 to get \frac{17}{16}. Then add the square of \frac{17}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{17}{8}x+\left(\frac{17}{16}\right)^{2}=-\frac{1}{2}+\left(\frac{17}{16}\right)^{2}$$

Square \frac{17}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{17}{8}x+\frac{289}{256}=-\frac{1}{2}+\frac{289}{256}$$

Add -\frac{1}{2} to \frac{289}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{17}{8}x+\frac{289}{256}=\frac{161}{256}$$

Factor x^{2}+\frac{17}{8}x+\frac{289}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{17}{16}\right)^{2}=\frac{161}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{16}\right)^{2}}=\sqrt{\frac{161}{256}}$$

Simplify.

$$x+\frac{17}{16}=\frac{\sqrt{161}}{16} x+\frac{17}{16}=-\frac{\sqrt{161}}{16}$$

Subtract \frac{17}{16} from both sides of the equation.

$$x=\frac{\sqrt{161}-17}{16} x=\frac{-\sqrt{161}-17}{16}$$