Solve (6x+2)^2+(6x+2)^2=0

$$( 6 x + 2 ) ^ { 2 } + ( 6 x + 2 ) ^ { 2 } = 0$$

Solve for x

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

Steps Using Factoring By Grouping

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

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

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

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

Combine 36x^{2} and 36x^{2} to get 72x^{2}.

$$72x^{2}+24x+4+24x+4=0$$

Combine 24x and 24x to get 48x.

$$72x^{2}+48x+4+4=0$$

Add 4 and 4 to get 8.

$$72x^{2}+48x+8=0$$

Divide both sides by 8.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=6 ab=9\times 1=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

Rewrite 9x^{2}+6x+1 as \left(9x^{2}+3x\right)+\left(3x+1\right).

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

Factor out 3x in 9x^{2}+3x.

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

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

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

Rewrite as a binomial square.

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

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

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

Steps Using the Quadratic Formula

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

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

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

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

Combine 36x^{2} and 36x^{2} to get 72x^{2}.

$$72x^{2}+24x+4+24x+4=0$$

Combine 24x and 24x to get 48x.

$$72x^{2}+48x+4+4=0$$

Add 4 and 4 to get 8.

$$72x^{2}+48x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 72 for a, 48 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-48±\sqrt{48^{2}-4\times 72\times 8}}{2\times 72}$$

Square 48.

$$x=\frac{-48±\sqrt{2304-4\times 72\times 8}}{2\times 72}$$

Multiply -4 times 72.

$$x=\frac{-48±\sqrt{2304-288\times 8}}{2\times 72}$$

Multiply -288 times 8.

$$x=\frac{-48±\sqrt{2304-2304}}{2\times 72}$$

Add 2304 to -2304.

$$x=\frac{-48±\sqrt{0}}{2\times 72}$$

Take the square root of 0.

$$x=-\frac{48}{2\times 72}$$

Multiply 2 times 72.

$$x=-\frac{48}{144}$$

Reduce the fraction \frac{-48}{144} to lowest terms by extracting and canceling out 48.

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

Steps for Completing the Square

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

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

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

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

Combine 36x^{2} and 36x^{2} to get 72x^{2}.

$$72x^{2}+24x+4+24x+4=0$$

Combine 24x and 24x to get 48x.

$$72x^{2}+48x+4+4=0$$

Add 4 and 4 to get 8.

$$72x^{2}+48x+8=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$72x^{2}+48x=-8$$

Divide both sides by 72.

$$\frac{72x^{2}+48x}{72}=-\frac{8}{72}$$

Dividing by 72 undoes the multiplication by 72.

$$x^{2}+\frac{48}{72}x=-\frac{8}{72}$$

Reduce the fraction \frac{48}{72} to lowest terms by extracting and canceling out 24.

$$x^{2}+\frac{2}{3}x=-\frac{8}{72}$$

Reduce the fraction \frac{-8}{72} to lowest terms by extracting and canceling out 8.

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

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.

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

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

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

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

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=0$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+\frac{1}{3}=0 x+\frac{1}{3}=0$$

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

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

The equation is now solved. Solutions are the same.

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