Quadratic Equation and its Solutions

25th Oct 2018
MOHD NASEEM HASHMI

We know that any equation which has the highest exponent of its variable as 2 is known as a quadratic equation. The standard quadratic equation is written as \(ax^2+bx+c=0\)

It has three terms. The coefficient of $x^2$ is a, the coefficient of x is b and the constant term is c.
We can get the solution of a quadratic equation using many methods such as: 

(i) Completing square method

(ii) Using the quadratic formula
(iii) Factoring Method

    etc. 

Let's go through the completing square method. We can derive Quadratic formula using this method.
Consider the standard equation $ax^2+bx+c=0$. We will solve it for x in the following steps:

We have
\(ax^2+bx+c=0\)                              (Given) 

$ax^2+bx=-c$                                  (Subtract c from both sides)

$4a^2x^2+4abx=-4ac$                     (Multiply both sides by 4a)

$4a^2x^2+4abx+b^2=b^2-4ac$        (Add $b^2$ both sides)

$(2ax)^2+4abx+b^2=b^2-4ac$

We can see that left side is a perfect square so we can write it as 

$(2ax+b)^2=b^2-4ac$

It gives us two values 

$2ax+b=\pm \sqrt{b^2-4ac}$  

$2ax=-b \pm \sqrt{b^2-4ac}$                   (subtract b from both sides)

$ x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} $            (Divide both sides by 2a) 

This the solution that we get using completing the square method from the standard quadratic equation.
It's also known as the quadratic formula. 
Let's solve a quadratic equation using quadratic formula. 
Consider a quadratic equation $2x^2+3x-6=0$ 
To solve this equation using the quadratic formula, we will find the values of a, b, and c by comparing the given quadratic equation with the standard quadratic equation, then we will substitute the values of a, b and c in the quadratic formula and simplify to get the solution. 
We have 
 $2x^2+3x-6=0$ 
and standard quadratic equation 
$ax^2+bx+c=0$
We see that the coefficient of $x^2$ is 2, therefore, a=2
the coefficient of x is 3, therefore, b=3
and the constant term is -6, therefore, c=-6
Substituting these values in quadratic formula, we get

$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

$ x=\dfrac{-3 \pm \sqrt{3^2-4\times2\times(-6)}}{2\times 2} $

$ x=\dfrac{-3 \pm \sqrt{9+48}}{4} $

$ x=\dfrac{-3 \pm \sqrt{57}}{4} $

Thus, we have two solutions for the given quadratic equation, these two values are also known as the roots of the quadratic equation.
$ x_1=\dfrac{-3+ \sqrt{57}}{4} $   and   $ x_2=\dfrac{-3 - \sqrt{57}}{4} $

Similarly, we can get the solutions to any quadratic equation using the quadratic formula.

We will discuss the quadratic formula and the nature of the roots in the upcoming sections. 

Nature of the roots of a quadratic equation:

We know that roots of a quadratic equation is calculated using the quadratic formula

And quadratic formula contains a radical in it i.e. $\sqrt{b^2-4ac}$

We also know that square root can be real for only positive numbers only. The square root of a negative number is an imaginary number. Hence the quantity under the radical sign must be equal to or greater than 0. It means the quantity $b^2-4ac$ must be greater than or equal to zero for a quadratic equation to have real roots.
Let's say $b^2-4ac=D$. Here D stands for discriminant.

Then, roots are $ x=\dfrac{-b \pm \sqrt{D}}{2a} $ 

(i) If D>0, we get two distinct real roots

$ x_1=\dfrac{-b- \sqrt{D}}{2a} $   and   $ x_2=\dfrac{-b + \sqrt{D}}{2a} $

(ii)   If D=0, we get only one root since
       
     $x_1=x_2=\dfrac{-b}{2a}$

It is because the radical part becomes zero when D=0

(iii) If D<0, then the roots are imaginary


$ x_1=\dfrac{-b - i\sqrt{D}}{2a} $   and $ x_2=\dfrac{-b + i\sqrt{D}}{2a} $   

In this case roots are not real.
Now lets take some example.
Find number of roots and explain whether they real or  imaginary.
$1.\, \,   2x^2+3x+1=0$

$2.\, \, 3x^2+5x+6=0 $

$3.\, \, 5x^2-2x-7=0$