Solve (x+1)^2+(x+4)^2=4

$$( x + 1 ) ^ { 2 } + ( x + 4 ) ^ { 2 } = 4$$

Solve for x (complex solution)

$$x=-\frac{5}{2}+\frac{1}{2}i=-2.5+0.5i$$

$$x=-\frac{5}{2}-\frac{1}{2}i=-2.5-0.5i$$

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}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x^{2}+2x+1+x^{2}+8x+16=4$$

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

$$2x^{2}+2x+1+8x+16=4$$

Combine 2x and 8x to get 10x.

$$2x^{2}+10x+1+16=4$$

Add 1 and 16 to get 17.

$$2x^{2}+10x+17=4$$

Subtract 4 from both sides.

$$2x^{2}+10x+17-4=0$$

Subtract 4 from 17 to get 13.

$$2x^{2}+10x+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 10 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2\times 13}}{2\times 2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 2\times 13}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-10±\sqrt{100-8\times 13}}{2\times 2}$$

Multiply -8 times 13.

$$x=\frac{-10±\sqrt{100-104}}{2\times 2}$$

Add 100 to -104.

$$x=\frac{-10±\sqrt{-4}}{2\times 2}$$

Take the square root of -4.

$$x=\frac{-10±2i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-10±2i}{4}$$

Now solve the equation x=\frac{-10±2i}{4} when ± is plus. Add -10 to 2i.

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

Divide -10+2i by 4.

$$x=-\frac{5}{2}+\frac{1}{2}i$$

Now solve the equation x=\frac{-10±2i}{4} when ± is minus. Subtract 2i from -10.

$$x=\frac{-10-2i}{4}$$

Divide -10-2i by 4.

$$x=-\frac{5}{2}-\frac{1}{2}i$$

The equation is now solved.

$$x=-\frac{5}{2}+\frac{1}{2}i x=-\frac{5}{2}-\frac{1}{2}i$$

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}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x^{2}+2x+1+x^{2}+8x+16=4$$

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

$$2x^{2}+2x+1+8x+16=4$$

Combine 2x and 8x to get 10x.

$$2x^{2}+10x+1+16=4$$

Add 1 and 16 to get 17.

$$2x^{2}+10x+17=4$$

Subtract 17 from both sides.

$$2x^{2}+10x=4-17$$

Subtract 17 from 4 to get -13.

$$2x^{2}+10x=-13$$

Divide both sides by 2.

$$\frac{2x^{2}+10x}{2}=-\frac{13}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{10}{2}x=-\frac{13}{2}$$

Divide 10 by 2.

$$x^{2}+5x=-\frac{13}{2}$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-\frac{13}{2}+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=-\frac{13}{2}+\frac{25}{4}$$

Add -\frac{13}{2} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+5x+\frac{25}{4}=-\frac{1}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=-\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{-\frac{1}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{1}{2}i x+\frac{5}{2}=-\frac{1}{2}i$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=-\frac{5}{2}+\frac{1}{2}i x=-\frac{5}{2}-\frac{1}{2}i$$