Solve the quadratic equation ${x^2} - \sqrt 3 x + 1 = 0$ for x.

Hint: The given problem requires us to solve a quadratic equation ${x^2} - \sqrt 3 x + 1 = 0$. We will use the quadratic formula to find the roots of the equation directly with ease. So, we will first compare the given equation with the standard form of the quadratic equation and put the values known quantities of in the formula.

Complete step by step solution: 
In the given question, we are required to solve the equation ${x^2} - \sqrt 3 x + 1 = 0$.
The quadratic formula can be employed for solving an equation only if we compare the given equation with the standard form of quadratic equation.
Comparing with standard quadratic equation $A{x^2} + Bx + C = 0$
Here,$A = 1$, $B = - \sqrt 3 $ and $C = 1$.
Now, we use the quadratic formula to find the roots of the given equation.
$x = \dfrac{{\left( { - B} \right) \pm \sqrt {{B^2} - 4AC} }}{{2A}}$
Now, we substitute the values of a, b and c to find the roots of the equation.
$x = \dfrac{{ - \left( { - \sqrt 3 } \right) \pm \sqrt {{{\left( { - \sqrt 3 } \right)}^2} - 4 \times \left( 1 \right) \times \left( 1 \right)} }}{{2 \times \left( 1 \right)}}$
Simplifying the calculations, we get,
$ \Rightarrow x = \dfrac{{\sqrt 3 \pm \sqrt {3 - 4} }}{2}$
Adding like terms under the square root, we get,
$ \Rightarrow x = \dfrac{{\sqrt 3 \pm \sqrt { - 1} }}{2}$
Now, we know that $\sqrt { - 1} = i$. So, we get,
$ \Rightarrow x = \dfrac{{\sqrt 3 \pm i}}{2}$
So, the roots of the equation ${x^2} - \sqrt 3 x + 1 = 0$ are: $x = \dfrac{{\sqrt 3 \pm i}}{2}$.

Note: Quadratic equations are the equations having the degree of the variable as two. These can also be solved using a hit and trial method if the roots of the equation are easy to find. If one root of a quadratic equation with rational coefficients is imaginary, then the other root will also be the conjugate pair of the first imaginary root. For example: If one root is $a + ib$, then the root will be $a - ib$.