Solve 1-1/9x+1-1/(9x+1)^2=0

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

Solve for x

$$x=\frac{-\sqrt{5}-1}{18}\approx -0.179781554$$

$$x=\frac{\sqrt{5}-1}{18}\approx 0.068670443$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -\frac{1}{9} since division by zero is not defined. Multiply both sides of the equation by \left(9x+1\right)^{2}, the least common multiple of 9x+1,\left(9x+1\right)^{2}.

$$\left(9x+1\right)^{2}-\left(9x+1\right)-1=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(9x+1\right)^{2}.

$$81x^{2}+18x+1-\left(9x+1\right)-1=0$$

To find the opposite of 9x+1, find the opposite of each term.

$$81x^{2}+18x+1-9x-1-1=0$$

Combine 18x and -9x to get 9x.

$$81x^{2}+9x+1-1-1=0$$

Subtract 1 from 1 to get 0.

$$81x^{2}+9x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 81 for a, 9 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 81\left(-1\right)}}{2\times 81}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 81\left(-1\right)}}{2\times 81}$$

Multiply -4 times 81.

$$x=\frac{-9±\sqrt{81-324\left(-1\right)}}{2\times 81}$$

Multiply -324 times -1.

$$x=\frac{-9±\sqrt{81+324}}{2\times 81}$$

Add 81 to 324.

$$x=\frac{-9±\sqrt{405}}{2\times 81}$$

Take the square root of 405.

$$x=\frac{-9±9\sqrt{5}}{2\times 81}$$

Multiply 2 times 81.

$$x=\frac{-9±9\sqrt{5}}{162}$$

Now solve the equation x=\frac{-9±9\sqrt{5}}{162} when ± is plus. Add -9 to 9\sqrt{5}.

$$x=\frac{9\sqrt{5}-9}{162}$$

Divide -9+9\sqrt{5} by 162.

$$x=\frac{\sqrt{5}-1}{18}$$

Now solve the equation x=\frac{-9±9\sqrt{5}}{162} when ± is minus. Subtract 9\sqrt{5} from -9.

$$x=\frac{-9\sqrt{5}-9}{162}$$

Divide -9-9\sqrt{5} by 162.

$$x=\frac{-\sqrt{5}-1}{18}$$

The equation is now solved.

$$x=\frac{\sqrt{5}-1}{18} x=\frac{-\sqrt{5}-1}{18}$$

Steps for Completing the Square

Variable x cannot be equal to -\frac{1}{9} since division by zero is not defined. Multiply both sides of the equation by \left(9x+1\right)^{2}, the least common multiple of 9x+1,\left(9x+1\right)^{2}.

$$\left(9x+1\right)^{2}-\left(9x+1\right)-1=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(9x+1\right)^{2}.

$$81x^{2}+18x+1-\left(9x+1\right)-1=0$$

To find the opposite of 9x+1, find the opposite of each term.

$$81x^{2}+18x+1-9x-1-1=0$$

Combine 18x and -9x to get 9x.

$$81x^{2}+9x+1-1-1=0$$

Subtract 1 from 1 to get 0.

$$81x^{2}+9x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$81x^{2}+9x=1$$

Divide both sides by 81.

$$\frac{81x^{2}+9x}{81}=\frac{1}{81}$$

Dividing by 81 undoes the multiplication by 81.

$$x^{2}+\frac{9}{81}x=\frac{1}{81}$$

Reduce the fraction \frac{9}{81} to lowest terms by extracting and canceling out 9.

$$x^{2}+\frac{1}{9}x=\frac{1}{81}$$

Divide \frac{1}{9}, the coefficient of the x term, by 2 to get \frac{1}{18}. Then add the square of \frac{1}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{9}x+\left(\frac{1}{18}\right)^{2}=\frac{1}{81}+\left(\frac{1}{18}\right)^{2}$$

Square \frac{1}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{9}x+\frac{1}{324}=\frac{1}{81}+\frac{1}{324}$$

Add \frac{1}{81} to \frac{1}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{9}x+\frac{1}{324}=\frac{5}{324}$$

Factor x^{2}+\frac{1}{9}x+\frac{1}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{18}\right)^{2}=\frac{5}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{18}\right)^{2}}=\sqrt{\frac{5}{324}}$$

Simplify.

$$x+\frac{1}{18}=\frac{\sqrt{5}}{18} x+\frac{1}{18}=-\frac{\sqrt{5}}{18}$$

Subtract \frac{1}{18} from both sides of the equation.

$$x=\frac{\sqrt{5}-1}{18} x=\frac{-\sqrt{5}-1}{18}$$