Question

How do you solve ${x^2} + x + 1 = 0$ ?

Answer 1

Hint:The given problem requires us to solve a quadratic equation. There are various methods that can be employed to solve a quadratic equation like completing the square method, using quadratic formula and by splitting the middle term. Using the quadratic formula gives us the roots of the equation directly with ease. So, we will first compare the given quadratic equation with the standard form and then apply the quadratic formula to find the roots of the equation.

Complete step by step answer:
In the given question, we are required to solve the equation ${x^2} + x + 1 = 0$ with the help of quadratic formula. The quadratic formula can be employed for solving an equation only if we compare the given equation with standard form of quadratic equation.
Comparing with standard quadratic equation $a{x^2} + bx + c = 0$
Here,$a = 1$, $b = 1$ and$c = 1$.
Now, using the quadratic formula, we get the roots of the equation as:
$x = \dfrac{{( - b) \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the values of a, b, and c in the quadratic formula, we get,
$x = \dfrac{{ - \left( 1 \right) \pm \sqrt {{{(1)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}}$
Simplifying the calculations, we get,
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 - 4} }}{2}$
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2}$
Now, we know that $\sqrt { - 1} = i$. So, we get,
$ \Rightarrow x = \dfrac{{ - 1 \pm i\sqrt 3 }}{2}$
Separating the denominators for the real and imaginary part of the complex roots, we get,
$ \Rightarrow x = - \dfrac{1}{2} \pm i\dfrac{{\sqrt 3 }}{2}$
So, $x = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$ and $x = - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$ are the roots of the equation ${x^2} + x + 1 = 0$.

So, the roots of the equation ${x^2} + x + 1 = 0$ are: $x = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$ and $x = - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$.

Note:Quadratic equations are the polynomial equations with degree of the variable or unknown as 2. Quadratic equations may also be solved by a hit and trial method if the roots of the equation are easy to find. A quadratic equation may also have complex roots. But complex roots always occur in pairs. So, either the quadratic equation has two real roots or two imaginary roots.