Solve n^2-sqrt{3}n+1=0

$$n ^ { 2 } - \sqrt { 3 } n + 1 = 0$$

Solve for n

$$n=\frac{\sqrt{3}+i}{2}\approx 0.866025404+0.5i$$

$$n=\frac{\sqrt{3}-i}{2}\approx 0.866025404-0.5i$$

Steps Using the Quadratic Formula

Reorder the terms.

$$n^{2}-\sqrt{3}n+1=0$$

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.

$$n^{2}+\left(-\sqrt{3}\right)n+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\sqrt{3} for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-\sqrt{3}\right)±\sqrt{\left(-\sqrt{3}\right)^{2}-4}}{2}$$

Square -\sqrt{3}.

$$n=\frac{-\left(-\sqrt{3}\right)±\sqrt{3-4}}{2}$$

Add 3 to -4.

$$n=\frac{-\left(-\sqrt{3}\right)±\sqrt{-1}}{2}$$

Take the square root of -1.

$$n=\frac{-\left(-\sqrt{3}\right)±i}{2}$$

The opposite of -\sqrt{3} is \sqrt{3}.

$$n=\frac{\sqrt{3}±i}{2}$$

Now solve the equation n=\frac{\sqrt{3}±i}{2} when ± is plus. Add \sqrt{3} to i.

$$n=\frac{\sqrt{3}+i}{2}$$

Divide \sqrt{3}+i by 2.

$$n=\frac{\sqrt{3}}{2}+\frac{1}{2}i$$

Now solve the equation n=\frac{\sqrt{3}±i}{2} when ± is minus. Subtract i from \sqrt{3}.

$$n=\frac{\sqrt{3}-i}{2}$$

Divide \sqrt{3}-i by 2.

$$n=\frac{\sqrt{3}}{2}-\frac{1}{2}i$$

The equation is now solved.

$$n=\frac{\sqrt{3}}{2}+\frac{1}{2}i n=\frac{\sqrt{3}}{2}-\frac{1}{2}i$$

Steps for Completing the Square

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$n^{2}-\sqrt{3}n=-1$$

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.

$$n^{2}+\left(-\sqrt{3}\right)n=-1$$

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

$$n^{2}+\left(-\sqrt{3}\right)n+\left(-\frac{\sqrt{3}}{2}\right)^{2}=-1+\left(-\frac{\sqrt{3}}{2}\right)^{2}$$

Square -\frac{\sqrt{3}}{2}.

$$n^{2}+\left(-\sqrt{3}\right)n+\frac{3}{4}=-1+\frac{3}{4}$$

Add -1 to \frac{3}{4}.

$$n^{2}+\left(-\sqrt{3}\right)n+\frac{3}{4}=-\frac{1}{4}$$

Factor n^{2}+\left(-\sqrt{3}\right)n+\frac{3}{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}.

$$\left(n-\frac{\sqrt{3}}{2}\right)^{2}=-\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{\sqrt{3}}{2}\right)^{2}}=\sqrt{-\frac{1}{4}}$$

Simplify.

$$n-\frac{\sqrt{3}}{2}=\frac{1}{2}i n-\frac{\sqrt{3}}{2}=-\frac{1}{2}i$$

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

$$n=\frac{\sqrt{3}}{2}+\frac{1}{2}i n=\frac{\sqrt{3}}{2}-\frac{1}{2}i$$