Solve (3x+1)^2/9-(6x+1)^2/36+(2x+1)^2/4=0

$$\frac { ( 3 x + 1 ) ^ { 2 } } { 9 } - \frac { ( 6 x + 1 ) ^ { 2 } } { 36 } + \frac { ( 2 x + 1 ) ^ { 2 } } { 4 } = 0$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=-1$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 36, the least common multiple of 9,36,4.

$$4\left(3x+1\right)^{2}-\left(6x+1\right)^{2}+9\left(2x+1\right)^{2}=0$$

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

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

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

$$36x^{2}+24x+4-\left(6x+1\right)^{2}+9\left(2x+1\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6x+1\right)^{2}.

$$36x^{2}+24x+4-\left(36x^{2}+12x+1\right)+9\left(2x+1\right)^{2}=0$$

To find the opposite of 36x^{2}+12x+1, find the opposite of each term.

$$36x^{2}+24x+4-36x^{2}-12x-1+9\left(2x+1\right)^{2}=0$$

Combine 36x^{2} and -36x^{2} to get 0.

$$24x+4-12x-1+9\left(2x+1\right)^{2}=0$$

Combine 24x and -12x to get 12x.

$$12x+4-1+9\left(2x+1\right)^{2}=0$$

Subtract 1 from 4 to get 3.

$$12x+3+9\left(2x+1\right)^{2}=0$$

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

$$12x+3+9\left(4x^{2}+4x+1\right)=0$$

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

$$12x+3+36x^{2}+36x+9=0$$

Combine 12x and 36x to get 48x.

$$48x+3+36x^{2}+9=0$$

Add 3 and 9 to get 12.

$$48x+12+36x^{2}=0$$

Divide both sides by 12.

$$4x+1+3x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3x^{2}+4x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\times 1=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=3$$

Rewrite 3x^{2}+4x+1 as \left(3x^{2}+x\right)+\left(3x+1\right).

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

Factor out x in 3x^{2}+x.

$$x\left(3x+1\right)+3x+1$$

Factor out common term 3x+1 by using distributive property.

$$\left(3x+1\right)\left(x+1\right)$$

To find equation solutions, solve 3x+1=0 and x+1=0.

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 36, the least common multiple of 9,36,4.

$$4\left(3x+1\right)^{2}-\left(6x+1\right)^{2}+9\left(2x+1\right)^{2}=0$$

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

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

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

$$36x^{2}+24x+4-\left(6x+1\right)^{2}+9\left(2x+1\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6x+1\right)^{2}.

$$36x^{2}+24x+4-\left(36x^{2}+12x+1\right)+9\left(2x+1\right)^{2}=0$$

To find the opposite of 36x^{2}+12x+1, find the opposite of each term.

$$36x^{2}+24x+4-36x^{2}-12x-1+9\left(2x+1\right)^{2}=0$$

Combine 36x^{2} and -36x^{2} to get 0.

$$24x+4-12x-1+9\left(2x+1\right)^{2}=0$$

Combine 24x and -12x to get 12x.

$$12x+4-1+9\left(2x+1\right)^{2}=0$$

Subtract 1 from 4 to get 3.

$$12x+3+9\left(2x+1\right)^{2}=0$$

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

$$12x+3+9\left(4x^{2}+4x+1\right)=0$$

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

$$12x+3+36x^{2}+36x+9=0$$

Combine 12x and 36x to get 48x.

$$48x+3+36x^{2}+9=0$$

Add 3 and 9 to get 12.

$$48x+12+36x^{2}=0$$

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.

$$36x^{2}+48x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, 48 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-48±\sqrt{48^{2}-4\times 36\times 12}}{2\times 36}$$

Square 48.

$$x=\frac{-48±\sqrt{2304-4\times 36\times 12}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-48±\sqrt{2304-144\times 12}}{2\times 36}$$

Multiply -144 times 12.

$$x=\frac{-48±\sqrt{2304-1728}}{2\times 36}$$

Add 2304 to -1728.

$$x=\frac{-48±\sqrt{576}}{2\times 36}$$

Take the square root of 576.

$$x=\frac{-48±24}{2\times 36}$$

Multiply 2 times 36.

$$x=\frac{-48±24}{72}$$

Now solve the equation x=\frac{-48±24}{72} when ± is plus. Add -48 to 24.

$$x=-\frac{24}{72}$$

Reduce the fraction \frac{-24}{72} to lowest terms by extracting and canceling out 24.

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

Now solve the equation x=\frac{-48±24}{72} when ± is minus. Subtract 24 from -48.

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

Divide -72 by 72.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 36, the least common multiple of 9,36,4.

$$4\left(3x+1\right)^{2}-\left(6x+1\right)^{2}+9\left(2x+1\right)^{2}=0$$

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

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

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

$$36x^{2}+24x+4-\left(6x+1\right)^{2}+9\left(2x+1\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6x+1\right)^{2}.

$$36x^{2}+24x+4-\left(36x^{2}+12x+1\right)+9\left(2x+1\right)^{2}=0$$

To find the opposite of 36x^{2}+12x+1, find the opposite of each term.

$$36x^{2}+24x+4-36x^{2}-12x-1+9\left(2x+1\right)^{2}=0$$

Combine 36x^{2} and -36x^{2} to get 0.

$$24x+4-12x-1+9\left(2x+1\right)^{2}=0$$

Combine 24x and -12x to get 12x.

$$12x+4-1+9\left(2x+1\right)^{2}=0$$

Subtract 1 from 4 to get 3.

$$12x+3+9\left(2x+1\right)^{2}=0$$

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

$$12x+3+9\left(4x^{2}+4x+1\right)=0$$

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

$$12x+3+36x^{2}+36x+9=0$$

Combine 12x and 36x to get 48x.

$$48x+3+36x^{2}+9=0$$

Add 3 and 9 to get 12.

$$48x+12+36x^{2}=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$48x+36x^{2}=-12$$

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.

$$36x^{2}+48x=-12$$

Divide both sides by 36.

$$\frac{36x^{2}+48x}{36}=-\frac{12}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$x^{2}+\frac{48}{36}x=-\frac{12}{36}$$

Reduce the fraction \frac{48}{36} to lowest terms by extracting and canceling out 12.

$$x^{2}+\frac{4}{3}x=-\frac{12}{36}$$

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

$$x^{2}+\frac{4}{3}x=-\frac{1}{3}$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Add -\frac{1}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{2}{3} from both sides of the equation.

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