Solve x^2+(3-x)^2=sqrt{5}

$$x ^ { 2 } + ( 3 - x ) ^ { 2 } = \sqrt { 5 }$$

Solve for x (complex solution)

$$x=\frac{3+i\sqrt{9-2\sqrt{5}}}{2}\approx 1.5+1.063938913i$$

$$x=\frac{-i\sqrt{9-2\sqrt{5}}+3}{2}\approx 1.5-1.063938913i$$

Steps Using the Quadratic Formula

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

$$x^{2}+9-6x+x^{2}=\sqrt{5}$$

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

$$2x^{2}+9-6x=\sqrt{5}$$

Subtract \sqrt{5} from both sides.

$$2x^{2}+9-6x-\sqrt{5}=0$$

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.

$$2x^{2}-6x+9-\sqrt{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and 9-\sqrt{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\left(9-\sqrt{5}\right)}}{2\times 2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(9-\sqrt{5}\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\left(9-\sqrt{5}\right)}}{2\times 2}$$

Multiply -8 times 9-\sqrt{5}.

$$x=\frac{-\left(-6\right)±\sqrt{36+8\sqrt{5}-72}}{2\times 2}$$

Add 36 to -72+8\sqrt{5}.

$$x=\frac{-\left(-6\right)±\sqrt{8\sqrt{5}-36}}{2\times 2}$$

Take the square root of -36+8\sqrt{5}.

$$x=\frac{-\left(-6\right)±2i\sqrt{9-2\sqrt{5}}}{2\times 2}$$

The opposite of -6 is 6.

$$x=\frac{6±2i\sqrt{9-2\sqrt{5}}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±2i\sqrt{9-2\sqrt{5}}}{4}$$

Now solve the equation x=\frac{6±2i\sqrt{9-2\sqrt{5}}}{4} when ± is plus. Add 6 to 2i\sqrt{9-2\sqrt{5}}.

$$x=\frac{6+2i\sqrt{9-2\sqrt{5}}}{4}$$

Divide 6+2i\sqrt{9-2\sqrt{5}} by 4.

$$x=\frac{3+i\sqrt{9-2\sqrt{5}}}{2}$$

Now solve the equation x=\frac{6±2i\sqrt{9-2\sqrt{5}}}{4} when ± is minus. Subtract 2i\sqrt{9-2\sqrt{5}} from 6.

$$x=\frac{-2i\sqrt{9-2\sqrt{5}}+6}{4}$$

Divide 6-2i\sqrt{9-2\sqrt{5}} by 4.

$$x=\frac{-i\sqrt{9-2\sqrt{5}}+3}{2}$$

The equation is now solved.

$$x=\frac{3+i\sqrt{9-2\sqrt{5}}}{2} x=\frac{-i\sqrt{9-2\sqrt{5}}+3}{2}$$

Steps for Completing the Square

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

$$x^{2}+9-6x+x^{2}=\sqrt{5}$$

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

$$2x^{2}+9-6x=\sqrt{5}$$

Subtract 9 from both sides.

$$2x^{2}-6x=\sqrt{5}-9$$

Divide both sides by 2.

$$\frac{2x^{2}-6x}{2}=\frac{\sqrt{5}-9}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{6}{2}\right)x=\frac{\sqrt{5}-9}{2}$$

Divide -6 by 2.

$$x^{2}-3x=\frac{\sqrt{5}-9}{2}$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\frac{\sqrt{5}-9}{2}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{\sqrt{5}-9}{2}+\frac{9}{4}$$

Add \frac{\sqrt{5}-9}{2} to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{\sqrt{5}}{2}-\frac{9}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{\sqrt{5}}{2}-\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{\sqrt{5}}{2}-\frac{9}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{i\sqrt{9-2\sqrt{5}}}{2} x-\frac{3}{2}=-\frac{i\sqrt{9-2\sqrt{5}}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+i\sqrt{9-2\sqrt{5}}}{2} x=\frac{-i\sqrt{9-2\sqrt{5}}+3}{2}$$