Solve left(x+1right)^2+left(x+2right)^2=x+12

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

Solve for x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$x=1$$

Steps Using Factoring By Grouping

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

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

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

$$x^{2}+2x+1+x^{2}+4x+4=x+12$$

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

$$2x^{2}+2x+1+4x+4=x+12$$

Combine 2x and 4x to get 6x.

$$2x^{2}+6x+1+4=x+12$$

Add 1 and 4 to get 5.

$$2x^{2}+6x+5=x+12$$

Subtract x from both sides.

$$2x^{2}+6x+5-x=12$$

Combine 6x and -x to get 5x.

$$2x^{2}+5x+5=12$$

Subtract 12 from both sides.

$$2x^{2}+5x+5-12=0$$

Subtract 12 from 5 to get -7.

$$2x^{2}+5x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=5 ab=2\left(-7\right)=-14$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -14.

$$-1,14 -2,7$$

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 5.

$$a=-2 b=7$$

Rewrite 2x^{2}+5x-7 as \left(2x^{2}-2x\right)+\left(7x-7\right).

$$\left(2x^{2}-2x\right)+\left(7x-7\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(x-1\right)+7\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(2x+7\right)$$

To find equation solutions, solve x-1=0 and 2x+7=0.

$$x=1 x=-\frac{7}{2}$$

Steps Using the Quadratic Formula

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

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

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

$$x^{2}+2x+1+x^{2}+4x+4=x+12$$

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

$$2x^{2}+2x+1+4x+4=x+12$$

Combine 2x and 4x to get 6x.

$$2x^{2}+6x+1+4=x+12$$

Add 1 and 4 to get 5.

$$2x^{2}+6x+5=x+12$$

Subtract x from both sides.

$$2x^{2}+6x+5-x=12$$

Combine 6x and -x to get 5x.

$$2x^{2}+5x+5=12$$

Subtract 12 from both sides.

$$2x^{2}+5x+5-12=0$$

Subtract 12 from 5 to get -7.

$$2x^{2}+5x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\left(-7\right)}}{2\times 2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 2\left(-7\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-5±\sqrt{25-8\left(-7\right)}}{2\times 2}$$

Multiply -8 times -7.

$$x=\frac{-5±\sqrt{25+56}}{2\times 2}$$

Add 25 to 56.

$$x=\frac{-5±\sqrt{81}}{2\times 2}$$

Take the square root of 81.

$$x=\frac{-5±9}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-5±9}{4}$$

Now solve the equation x=\frac{-5±9}{4} when ± is plus. Add -5 to 9.

$$x=\frac{4}{4}$$

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{-5±9}{4} when ± is minus. Subtract 9 from -5.

$$x=-\frac{14}{4}$$

Reduce the fraction \frac{-14}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{2}$$

The equation is now solved.

$$x=1 x=-\frac{7}{2}$$

Steps for Completing the Square

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

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

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

$$x^{2}+2x+1+x^{2}+4x+4=x+12$$

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

$$2x^{2}+2x+1+4x+4=x+12$$

Combine 2x and 4x to get 6x.

$$2x^{2}+6x+1+4=x+12$$

Add 1 and 4 to get 5.

$$2x^{2}+6x+5=x+12$$

Subtract x from both sides.

$$2x^{2}+6x+5-x=12$$

Combine 6x and -x to get 5x.

$$2x^{2}+5x+5=12$$

Subtract 5 from both sides.

$$2x^{2}+5x=12-5$$

Subtract 5 from 12 to get 7.

$$2x^{2}+5x=7$$

Divide both sides by 2.

$$\frac{2x^{2}+5x}{2}=\frac{7}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{5}{2}x=\frac{7}{2}$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=\frac{7}{2}+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{7}{2}+\frac{25}{16}$$

Add \frac{7}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{81}{16}$$

Factor x^{2}+\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{9}{4} x+\frac{5}{4}=-\frac{9}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$x=1 x=-\frac{7}{2}$$