Question

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

Answer 1

`(a-sqrt(3)/2)^2 = (a-sqrt(3)/2)(a-sqrt(3)/2)`

`= a^2-(sqrt(3)/2+sqrt(3)/2)a+(sqrt(3)/2)(sqrt(3)/2)`

`= a^2-sqrt(3)a+3/4`

So we have:

`0 = a^2-sqrt(3)a+1 = a^2-sqrt(3)a+3/4+1/4`

`=(a-sqrt(3)/2)^2+1/4`

Subtracting 1/4 from both sides, we get:

`(a-sqrt(3)/2)^2 = -1/4`

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

If you want complex solutions,

`a-sqrt(3)/2 = +-sqrt(-1/4) = +-i/2`

Adding `sqrt(3/2)` to both sides, we get

`a = sqrt(3)/2 +- i/2`.

Answer 2

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 RR` which makes `a^2-sqrt3a+1 = 0`.

• But if you are working with complex numbers, then there are two solutions:
  `a_1=(sqrt3+i)/2` and `a_2=(sqrt3-i)/2`.