Solve x=4+2xdiv(1+2x)

$$x=4+2x \div (1+2x)$$

Solve for x

$$x = \frac{\sqrt{113} + 9}{4} \approx 4.907536453$$

$$x=\frac{9-\sqrt{113}}{4}\approx -0.407536453$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{1+2x}{1+2x}.

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

Since \frac{4\left(1+2x\right)}{1+2x} and \frac{2x}{1+2x} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 4\left(1+2x\right)+2x.

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

Combine like terms in 4+8x+2x.

$$x=\frac{4+10x}{1+2x}$$

Subtract \frac{4+10x}{1+2x} from both sides.

$$x-\frac{4+10x}{1+2x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{1+2x}{1+2x}.

$$\frac{x\left(1+2x\right)}{1+2x}-\frac{4+10x}{1+2x}=0$$

Since \frac{x\left(1+2x\right)}{1+2x} and \frac{4+10x}{1+2x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(1+2x\right)-\left(4+10x\right)}{1+2x}=0$$

Do the multiplications in x\left(1+2x\right)-\left(4+10x\right).

$$\frac{x+2x^{2}-4-10x}{1+2x}=0$$

Combine like terms in x+2x^{2}-4-10x.

$$\frac{-9x+2x^{2}-4}{1+2x}=0$$

Variable x cannot be equal to -\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 2x+1.

$$-9x+2x^{2}-4=0$$

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.

$$2x^{2}-9x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\left(-4\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$x=\frac{-\left(-9\right)±\sqrt{81+32}}{2\times 2}$$

Add 81 to 32.

$$x=\frac{-\left(-9\right)±\sqrt{113}}{2\times 2}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{113}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{9±\sqrt{113}}{4}$$

Now solve the equation x=\frac{9±\sqrt{113}}{4} when ± is plus. Add 9 to \sqrt{113}.

$$x=\frac{\sqrt{113}+9}{4}$$

Now solve the equation x=\frac{9±\sqrt{113}}{4} when ± is minus. Subtract \sqrt{113} from 9.

$$x=\frac{9-\sqrt{113}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{113}+9}{4} x=\frac{9-\sqrt{113}}{4}$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{1+2x}{1+2x}.

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

Since \frac{4\left(1+2x\right)}{1+2x} and \frac{2x}{1+2x} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 4\left(1+2x\right)+2x.

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

Combine like terms in 4+8x+2x.

$$x=\frac{4+10x}{1+2x}$$

Subtract \frac{4+10x}{1+2x} from both sides.

$$x-\frac{4+10x}{1+2x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{1+2x}{1+2x}.

$$\frac{x\left(1+2x\right)}{1+2x}-\frac{4+10x}{1+2x}=0$$

Since \frac{x\left(1+2x\right)}{1+2x} and \frac{4+10x}{1+2x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(1+2x\right)-\left(4+10x\right)}{1+2x}=0$$

Do the multiplications in x\left(1+2x\right)-\left(4+10x\right).

$$\frac{x+2x^{2}-4-10x}{1+2x}=0$$

Combine like terms in x+2x^{2}-4-10x.

$$\frac{-9x+2x^{2}-4}{1+2x}=0$$

Variable x cannot be equal to -\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 2x+1.

$$-9x+2x^{2}-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$-9x+2x^{2}=4$$

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.

$$2x^{2}-9x=4$$

Divide both sides by 2.

$$\frac{2x^{2}-9x}{2}=\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{9}{2}x=\frac{4}{2}$$

Divide 4 by 2.

$$x^{2}-\frac{9}{2}x=2$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=2+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=2+\frac{81}{16}$$

Add 2 to \frac{81}{16}.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{113}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{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}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{113}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{113}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{\sqrt{113}}{4} x-\frac{9}{4}=-\frac{\sqrt{113}}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=\frac{\sqrt{113}+9}{4} x=\frac{9-\sqrt{113}}{4}$$