Factorize ${x^2} + \dfrac{1}{2}x + \dfrac{1}{4}$?

Hint: In this question, we have to factor the quadratic equation, we can factor the equation using the quadratic formula and using the concepts of determinants, and the formula for the quadratic equation is given by\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], where "\[a\]", "\[b\]", and “\[c\]" are real numbers, and the discriminant formula is given by$D = {b^2} - 4ac$, now the quadratic formula becomes \[\begin{gathered}
  x = \dfrac{{ - b \pm \sqrt D }}{{2a}} \\
  a = 1,b = \dfrac{1}{2},c = \dfrac{1}{4} \\
  \left( {x - \left( {\dfrac{{ - 1 + \sqrt 3 i}}{4}} \right)} \right)\left( {x - \left( {\dfrac{{ - 1 - \sqrt 3 i}}{4}} \right)} \right) \\
   \Rightarrow \left( {\left( {x + \dfrac{1}{4}} \right) - \left( {\dfrac{{\sqrt 3 i}}{4}} \right)} \right)\left( {\left( {x + \dfrac{1}{4}} \right) + \left( {\dfrac{{\sqrt 3 i}}{4}} \right)} \right)\left( {a - b} \right) \\
  {x^2} + \dfrac{1}{2}x + \left( {\dfrac{1}{{16}}} \right) - \left( {\dfrac{{3\left( { - 1} \right)}}{{16}}} \right) \\
    \\
\end{gathered} \] $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$,and if the value of discriminant is less than 0, I,e., negative then there is no possible solution for the given equation.

Complete step-by-step solution:
Given quadratic equation is ${x^2} + \dfrac{1}{2}x + \dfrac{1}{4}$,
Now using the discriminant formula which is given by $D = {b^2} - 4ac$, we will first find the discriminant,
So, here $a = 1$,$b = \dfrac{1}{2}$ and $c = \dfrac{1}{4}$,
Now substituting the values in the formula we get,
$D = {\left( {\dfrac{1}{2}} \right)^2} - 4\left( 1 \right)\left( {\dfrac{1}{4}} \right)$,
Now simplifying we get,
\[ \Rightarrow D = \dfrac{1}{4} - 1\],
So, further simplifying we get,
\[ \Rightarrow \]\[D = \dfrac{{ - 3}}{4} < 0\],
Now, we know that root of the negative number will be a complex number which is represented as 0 \[i\] known as iota and is equal to \[\sqrt { - 1} \],
\[ \Rightarrow D = \dfrac{3}{4}{i^2}\],
Now using the value in the formula we get,
\[ \Rightarrow \]\[x = \dfrac{{ - \dfrac{1}{2} \pm \sqrt {\left( {\dfrac{3}{4}{i^2}} \right)} }}{2}\],
Now simplifying we get,
\[ \Rightarrow x = \dfrac{{\dfrac{{ - 1}}{2} \pm \dfrac{{\sqrt 3 }}{2}i}}{2}\],
Now further simplifying we get,
\[ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt 3 i}}{4}\],
Now we get the two values for \[x\] and they are,\[x = \dfrac{{ - 1 + \sqrt 3 i}}{4}\]and\[x = \dfrac{{ - 1 - \sqrt 3 i}}{4}\],
Now factoring the equation using the two values we get,
\[\left( {x - \left( {\dfrac{{ - 1 + \sqrt 3 i}}{4}} \right)} \right)\left( {x - \left( {\dfrac{{ - 1 - \sqrt 3 i}}{4}} \right)} \right)\],
We can check if we multiply the terms we will get the quadratic equation given, i.e.,
Now simplifying the expression we get,
\[ \Rightarrow \left( {\left( {x + \dfrac{1}{4}} \right) - \left( {\dfrac{{\sqrt 3 i}}{4}} \right)} \right)\left( {\left( {x + \dfrac{1}{4}} \right) + \left( {\dfrac{{\sqrt 3 i}}{4}} \right)} \right)\],
So using the identity\[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\], we get,
\[ \Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} - {\left( {\dfrac{{\sqrt 3 i}}{4}} \right)^2}\],
Now simplifying we get using identity \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] we get,
\[ \Rightarrow {x^2} + 2\left( x \right)\left( {\dfrac{1}{4}} \right) + {\left( {\dfrac{1}{4}} \right)^2} - \left( {\dfrac{{3{i^2}}}{{16}}} \right)\],
Now further simplifying we get,
\[ \Rightarrow {x^2} + \dfrac{1}{2}x + \left( {\dfrac{1}{{16}}} \right) - \left( {\dfrac{{3\left( { - 1} \right)}}{{16}}} \right)\],
Now further simplifying we get,
\[ \Rightarrow \]\[{x^2} + \dfrac{1}{2}x + \dfrac{1}{{16}} + \dfrac{3}{{16}}\],
Now simplifying by adding we get,
\[ \Rightarrow \]\[{x^2} + \dfrac{1}{2}x + \dfrac{4}{{16}}\],
So the equation becomes \[{x^2} + \dfrac{1}{2}x + \dfrac{1}{4}\] which is the given equation.

\[\therefore \]Then by factoring the equation \[{x^2} + \dfrac{1}{2}x + \dfrac{1}{4}\] we get \[\left( {x - \left( {\dfrac{{ - 1 + \sqrt 3 i}}{4}} \right)} \right)\left( {x - \left( {\dfrac{{ - 1 - \sqrt 3 i}}{4}} \right)} \right)\].

Note: The discriminant is the part under the square root in the quadratic formula,$D = {b^2} - 4ac$, and we should remember that If it’s value is more than 0, the equation has two real solutions. If its value is less than 0, there are no solutions. If it's equal to 0, there is one solution.