# How do you solve a^2-sqrt(3)a+1 = 0 ?

May 12, 2015

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

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

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

So we have:

$0 = {a}^{2} - \sqrt{3} a + 1 = {a}^{2} - \sqrt{3} a + \frac{3}{4} + \frac{1}{4}$

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

Subtracting 1/4 from both sides, we get:

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

This has no real number solutions since the square of any real number is non-negative.

If you want complex solutions,

$a - \frac{\sqrt{3}}{2} = \pm \sqrt{- \frac{1}{4}} = \pm \frac{i}{2}$

Adding $\sqrt{\frac{3}{2}}$ to both sides, we get

$a = \frac{\sqrt{3}}{2} \pm \frac{i}{2}$.

May 12, 2015

I would start applying the formula to solve quadratic equations (in fact, this is a quadratic equation in "a"):

a=(-b+-sqrt(b^2-4ac))/(2a) => a=(sqrt3+-sqrt((sqrt3)^2-4·1·1))/(2·1) => a=(sqrt3+-sqrt(3-4))/2 => a=(sqrt3+-sqrt(-1))/2

As you can see, the equation has no real solution, since it has a square root of a negative number ($\sqrt{- 1}$).

• So, if you are working with real numbers, the answer is that there is no $a \in \mathbb{R}$ which makes ${a}^{2} - \sqrt{3} a + 1 = 0$.

• But if you are working with complex numbers, then there are two solutions:
${a}_{1} = \frac{\sqrt{3} + i}{2}$ and ${a}_{2} = \frac{\sqrt{3} - i}{2}$.