Solve left(3x-1+7right)^2+left(x+3-1right)^2=0

$${ \left(3x-1+7 \right) }^{ 2 } + { \left(x+3-1 \right) }^{ 2 } = 0$$

Solve for x

$$x=-2$$

Steps Using Factoring By Grouping

Add -1 and 7 to get 6.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+6\right)^{2}.

$$9x^{2}+36x+36+\left(x+3-1\right)^{2}=0$$

Subtract 1 from 3 to get 2.

$$9x^{2}+36x+36+\left(x+2\right)^{2}=0$$

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

$$9x^{2}+36x+36+x^{2}+4x+4=0$$

Combine 9x^{2} and x^{2} to get 10x^{2}.

$$10x^{2}+36x+36+4x+4=0$$

Combine 36x and 4x to get 40x.

$$10x^{2}+40x+36+4=0$$

Add 36 and 4 to get 40.

$$10x^{2}+40x+40=0$$

Divide both sides by 10.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

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

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 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

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

$$\left(x^{2}+2x\right)+\left(2x+4\right)$$

Factor out x in the first and 2 in the second group.

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

Factor out common term x+2 by using distributive property.

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

Rewrite as a binomial square.

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

To find equation solution, solve x+2=0.

$$x=-2$$

Steps Using the Quadratic Formula

Add -1 and 7 to get 6.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+6\right)^{2}.

$$9x^{2}+36x+36+\left(x+3-1\right)^{2}=0$$

Subtract 1 from 3 to get 2.

$$9x^{2}+36x+36+\left(x+2\right)^{2}=0$$

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

$$9x^{2}+36x+36+x^{2}+4x+4=0$$

Combine 9x^{2} and x^{2} to get 10x^{2}.

$$10x^{2}+36x+36+4x+4=0$$

Combine 36x and 4x to get 40x.

$$10x^{2}+40x+36+4=0$$

Add 36 and 4 to get 40.

$$10x^{2}+40x+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 40 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\times 10\times 40}}{2\times 10}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 10\times 40}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-40±\sqrt{1600-40\times 40}}{2\times 10}$$

Multiply -40 times 40.

$$x=\frac{-40±\sqrt{1600-1600}}{2\times 10}$$

Add 1600 to -1600.

$$x=\frac{-40±\sqrt{0}}{2\times 10}$$

Take the square root of 0.

$$x=-\frac{40}{2\times 10}$$

Multiply 2 times 10.

$$x=-\frac{40}{20}$$

Divide -40 by 20.

$$x=-2$$

Steps for Completing the Square

Add -1 and 7 to get 6.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+6\right)^{2}.

$$9x^{2}+36x+36+\left(x+3-1\right)^{2}=0$$

Subtract 1 from 3 to get 2.

$$9x^{2}+36x+36+\left(x+2\right)^{2}=0$$

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

$$9x^{2}+36x+36+x^{2}+4x+4=0$$

Combine 9x^{2} and x^{2} to get 10x^{2}.

$$10x^{2}+36x+36+4x+4=0$$

Combine 36x and 4x to get 40x.

$$10x^{2}+40x+36+4=0$$

Add 36 and 4 to get 40.

$$10x^{2}+40x+40=0$$

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

$$10x^{2}+40x=-40$$

Divide both sides by 10.

$$\frac{10x^{2}+40x}{10}=-\frac{40}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{40}{10}x=-\frac{40}{10}$$

Divide 40 by 10.

$$x^{2}+4x=-\frac{40}{10}$$

Divide -40 by 10.

$$x^{2}+4x=-4$$

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

$$x^{2}+4x+2^{2}=-4+2^{2}$$

Square 2.

$$x^{2}+4x+4=-4+4$$

Add -4 to 4.

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

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=0 x+2=0$$

Subtract 2 from both sides of the equation.

$$x=-2 x=-2$$

The equation is now solved. Solutions are the same.

$$x=-2$$