Solve (4-x)^2+(x)^2=5

$$( 4 - x ) ^ { 2 } + ( x ) ^ { 2 } = 5$$

Solve for x (complex solution)

$$x=\frac{\sqrt{6}i}{2}+2\approx 2+1.224744871i$$

$$x=-\frac{\sqrt{6}i}{2}+2\approx 2-1.224744871i$$

Steps Using the Quadratic Formula

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

$$16-8x+x^{2}+x^{2}=5$$

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

$$16-8x+2x^{2}=5$$

Subtract 5 from both sides.

$$16-8x+2x^{2}-5=0$$

Subtract 5 from 16 to get 11.

$$11-8x+2x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-8x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\times 11}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 11}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\times 11}}{2\times 2}$$

Multiply -8 times 11.

$$x=\frac{-\left(-8\right)±\sqrt{64-88}}{2\times 2}$$

Add 64 to -88.

$$x=\frac{-\left(-8\right)±\sqrt{-24}}{2\times 2}$$

Take the square root of -24.

$$x=\frac{-\left(-8\right)±2\sqrt{6}i}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{6}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±2\sqrt{6}i}{4}$$

Now solve the equation x=\frac{8±2\sqrt{6}i}{4} when ± is plus. Add 8 to 2i\sqrt{6}.

$$x=\frac{8+2\sqrt{6}i}{4}$$

Divide 8+2i\sqrt{6} by 4.

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

Now solve the equation x=\frac{8±2\sqrt{6}i}{4} when ± is minus. Subtract 2i\sqrt{6} from 8.

$$x=\frac{-2\sqrt{6}i+8}{4}$$

Divide 8-2i\sqrt{6} by 4.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$16-8x+x^{2}+x^{2}=5$$

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

$$16-8x+2x^{2}=5$$

Subtract 16 from both sides.

$$-8x+2x^{2}=5-16$$

Subtract 16 from 5 to get -11.

$$-8x+2x^{2}=-11$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-8x=-11$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=-\frac{11}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=-\frac{11}{2}$$

Divide -8 by 2.

$$x^{2}-4x=-\frac{11}{2}$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=-\frac{11}{2}+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-\frac{11}{2}+4$$

Add -\frac{11}{2} to 4.

$$x^{2}-4x+4=-\frac{3}{2}$$

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{-\frac{3}{2}}$$

Simplify.

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

Add 2 to both sides of the equation.

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