Solve 4left(3x+1right)^2-1=0

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

Solve for x

$$x=-\frac{1}{6}\approx -0.166666667$$

$$x=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

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

$$4\left(9x^{2}+6x+1\right)-1=0$$

Use the distributive property to multiply 4 by 9x^{2}+6x+1.

$$36x^{2}+24x+4-1=0$$

Subtract 1 from 4 to get 3.

$$36x^{2}+24x+3=0$$

Divide both sides by 3.

$$12x^{2}+8x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=8 ab=12\times 1=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=2 b=6$$

Rewrite 12x^{2}+8x+1 as \left(12x^{2}+2x\right)+\left(6x+1\right).

$$\left(12x^{2}+2x\right)+\left(6x+1\right)$$

Factor out 2x in 12x^{2}+2x.

$$2x\left(6x+1\right)+6x+1$$

Factor out common term 6x+1 by using distributive property.

$$\left(6x+1\right)\left(2x+1\right)$$

To find equation solutions, solve 6x+1=0 and 2x+1=0.

$$x=-\frac{1}{6} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

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

$$4\left(9x^{2}+6x+1\right)-1=0$$

Use the distributive property to multiply 4 by 9x^{2}+6x+1.

$$36x^{2}+24x+4-1=0$$

Subtract 1 from 4 to get 3.

$$36x^{2}+24x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, 24 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\times 36\times 3}}{2\times 36}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\times 36\times 3}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-24±\sqrt{576-144\times 3}}{2\times 36}$$

Multiply -144 times 3.

$$x=\frac{-24±\sqrt{576-432}}{2\times 36}$$

Add 576 to -432.

$$x=\frac{-24±\sqrt{144}}{2\times 36}$$

Take the square root of 144.

$$x=\frac{-24±12}{2\times 36}$$

Multiply 2 times 36.

$$x=\frac{-24±12}{72}$$

Now solve the equation x=\frac{-24±12}{72} when ± is plus. Add -24 to 12.

$$x=-\frac{12}{72}$$

Reduce the fraction \frac{-12}{72} to lowest terms by extracting and canceling out 12.

$$x=-\frac{1}{6}$$

Now solve the equation x=\frac{-24±12}{72} when ± is minus. Subtract 12 from -24.

$$x=-\frac{36}{72}$$

Reduce the fraction \frac{-36}{72} to lowest terms by extracting and canceling out 36.

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

The equation is now solved.

$$x=-\frac{1}{6} x=-\frac{1}{2}$$

Steps for Completing the Square

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

$$4\left(9x^{2}+6x+1\right)-1=0$$

Use the distributive property to multiply 4 by 9x^{2}+6x+1.

$$36x^{2}+24x+4-1=0$$

Subtract 1 from 4 to get 3.

$$36x^{2}+24x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$36x^{2}+24x=-3$$

Divide both sides by 36.

$$\frac{36x^{2}+24x}{36}=-\frac{3}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$x^{2}+\frac{24}{36}x=-\frac{3}{36}$$

Reduce the fraction \frac{24}{36} to lowest terms by extracting and canceling out 12.

$$x^{2}+\frac{2}{3}x=-\frac{3}{36}$$

Reduce the fraction \frac{-3}{36} to lowest terms by extracting and canceling out 3.

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

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.

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

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

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

Add -\frac{1}{12} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{1}{36}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+\frac{1}{3}=\frac{1}{6} x+\frac{1}{3}=-\frac{1}{6}$$

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

$$x=-\frac{1}{6} x=-\frac{1}{2}$$