Solve x^2=(x+3)^2+(x-1)^2

$$x ^ { 2 } = ( x + 3 ) ^ { 2 } + ( x - 1 ) ^ { 2 }$$

Solve for x (complex solution)

$$x=-\sqrt{6}i-2\approx -2-2.449489743i$$

$$x=-2+\sqrt{6}i\approx -2+2.449489743i$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

Combine 6x and -2x to get 4x.

$$x^{2}=2x^{2}+4x+9+1$$

Add 9 and 1 to get 10.

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

Subtract 2x^{2} from both sides.

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

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

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

Subtract 4x from both sides.

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

Subtract 10 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\left(-10\right)}}{2\left(-1\right)}$$

Multiply 4 times -10.

$$x=\frac{-\left(-4\right)±\sqrt{16-40}}{2\left(-1\right)}$$

Add 16 to -40.

$$x=\frac{-\left(-4\right)±\sqrt{-24}}{2\left(-1\right)}$$

Take the square root of -24.

$$x=\frac{-\left(-4\right)±2\sqrt{6}i}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{6}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{6}i}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{6}i}{-2} when ± is plus. Add 4 to 2i\sqrt{6}.

$$x=\frac{4+2\sqrt{6}i}{-2}$$

Divide 4+2i\sqrt{6} by -2.

$$x=-\sqrt{6}i-2$$

Now solve the equation x=\frac{4±2\sqrt{6}i}{-2} when ± is minus. Subtract 2i\sqrt{6} from 4.

$$x=\frac{-2\sqrt{6}i+4}{-2}$$

Divide 4-2i\sqrt{6} by -2.

$$x=-2+\sqrt{6}i$$

The equation is now solved.

$$x=-\sqrt{6}i-2 x=-2+\sqrt{6}i$$

Steps for Completing the Square

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

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

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

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

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

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

Combine 6x and -2x to get 4x.

$$x^{2}=2x^{2}+4x+9+1$$

Add 9 and 1 to get 10.

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

Subtract 2x^{2} from both sides.

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

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

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

Subtract 4x from both sides.

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

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=\frac{10}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{4}{-1}\right)x=\frac{10}{-1}$$

Divide -4 by -1.

$$x^{2}+4x=\frac{10}{-1}$$

Divide 10 by -1.

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

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.

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

Square 2.

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

Add -10 to 4.

$$x^{2}+4x+4=-6$$

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{-6}$$

Simplify.

$$x+2=\sqrt{6}i x+2=-\sqrt{6}i$$

Subtract 2 from both sides of the equation.

$$x=-2+\sqrt{6}i x=-\sqrt{6}i-2$$