Question

How do you factor ${x^2} + \dfrac{1}{2}x + \dfrac{1}{4}$?

Answer 1

Hint: First compare the given quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Formula used: Chain Rule:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, first we will compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} + \dfrac{1}{2}x + \dfrac{1}{4}$ with $a{x^2} + bx + c$, we get
$a = 1$, $b = \dfrac{1}{2}$ and $c = \dfrac{1}{4}$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$ \Rightarrow D = {\left( {\dfrac{1}{2}} \right)^2} - 4\left( 1 \right)\left( {\dfrac{1}{4}} \right)$
After simplifying the result, we get
$ \Rightarrow D = \dfrac{{ - 3}}{4}$
Which means the given equation has imaginary roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$ \Rightarrow x = \dfrac{{ - \dfrac{1}{2} \pm \dfrac{{\sqrt 3 }}{2}i}}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$ \Rightarrow x = \dfrac{{ - 1}}{4} \pm \dfrac{{\sqrt 3 }}{4}i$
So, $x = \dfrac{{ - 1}}{4} + \dfrac{{\sqrt 3 }}{4}i$ and $x = \dfrac{{ - 1}}{4} - \dfrac{{\sqrt 3 }}{4}i$ are roots of equation ${x^2} + \dfrac{1}{2}x + \dfrac{1}{4}$.

Therefore, the trinomial ${x^2} + \dfrac{1}{2}x + \dfrac{1}{4}$ can be factored as $\left( {x + \dfrac{1}{4} - \dfrac{{\sqrt 3 }}{4}i} \right)\left( {x + \dfrac{1}{4} + \dfrac{{\sqrt 3 }}{4}i} \right)$..

Note: Factorisation: It is simply the resolution of an integer or polynomial into factors such that when multiplied together they will result in original or initial the integer or polynomial. In the factorisation method, we reduce any algebraic or quadratic equation into its simpler form, where the equations are represented as the product of factors instead of expanding the brackets. The factors of any equation can be an integer, a variable or an algebraic expression itself.
In above question, it should be noted that we get $x = \dfrac{{ - 1}}{4} + \dfrac{{\sqrt 3 }}{4}i$ and $x = \dfrac{{ - 1}}{4} - \dfrac{{\sqrt 3 }}{4}i$ as the roots of equation ${x^2} + \dfrac{1}{2}x + \dfrac{1}{4}$. No other roots will satisfy the condition. If we take wrong factors, then we will not get a trinomial on their product. So, carefully find the roots.