Solve -(-36)/left(1+3xright)^2=1/3

$$- \frac{ (-36) }{ { \left(1+3x \right) }^{ 2 } } = \frac{ 1 }{ 3 }$$

Solve for x

$$x=-2\sqrt{3}-\frac{1}{3}\approx -3.797434948$$

$$x=2\sqrt{3}-\frac{1}{3}\approx 3.130768282$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 3\left(3x+1\right)^{2}, the least common multiple of \left(1+3x\right)^{2},3.

$$-3\left(-36\right)=\left(3x+1\right)^{2}$$

Multiply -3 and -36 to get 108.

$$108=\left(3x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$108=9x^{2}+6x+1$$

Swap sides so that all variable terms are on the left hand side.

$$9x^{2}+6x+1=108$$

Subtract 108 from both sides.

$$9x^{2}+6x+1-108=0$$

Subtract 108 from 1 to get -107.

$$9x^{2}+6x-107=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 6 for b, and -107 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\left(-107\right)}}{2\times 9}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 9\left(-107\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-6±\sqrt{36-36\left(-107\right)}}{2\times 9}$$

Multiply -36 times -107.

$$x=\frac{-6±\sqrt{36+3852}}{2\times 9}$$

Add 36 to 3852.

$$x=\frac{-6±\sqrt{3888}}{2\times 9}$$

Take the square root of 3888.

$$x=\frac{-6±36\sqrt{3}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-6±36\sqrt{3}}{18}$$

Now solve the equation x=\frac{-6±36\sqrt{3}}{18} when ± is plus. Add -6 to 36\sqrt{3}.

$$x=\frac{36\sqrt{3}-6}{18}$$

Divide -6+36\sqrt{3} by 18.

$$x=2\sqrt{3}-\frac{1}{3}$$

Now solve the equation x=\frac{-6±36\sqrt{3}}{18} when ± is minus. Subtract 36\sqrt{3} from -6.

$$x=\frac{-36\sqrt{3}-6}{18}$$

Divide -6-36\sqrt{3} by 18.

$$x=-2\sqrt{3}-\frac{1}{3}$$

The equation is now solved.

$$x=2\sqrt{3}-\frac{1}{3} x=-2\sqrt{3}-\frac{1}{3}$$

Steps for Completing the Square

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 3\left(3x+1\right)^{2}, the least common multiple of \left(1+3x\right)^{2},3.

$$-3\left(-36\right)=\left(3x+1\right)^{2}$$

Multiply -3 and -36 to get 108.

$$108=\left(3x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$108=9x^{2}+6x+1$$

Swap sides so that all variable terms are on the left hand side.

$$9x^{2}+6x+1=108$$

Subtract 1 from both sides.

$$9x^{2}+6x=108-1$$

Subtract 1 from 108 to get 107.

$$9x^{2}+6x=107$$

Divide both sides by 9.

$$\frac{9x^{2}+6x}{9}=\frac{107}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{6}{9}x=\frac{107}{9}$$

Reduce the fraction \frac{6}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{2}{3}x=\frac{107}{9}$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{107}{9}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{107+1}{9}$$

Add \frac{107}{9} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=12$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=12$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{12}$$

Simplify.

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

Subtract \frac{1}{3} from both sides of the equation.

$$x=2\sqrt{3}-\frac{1}{3} x=-2\sqrt{3}-\frac{1}{3}$$