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Quadratic Equations: Made Easier for JEE

Quadratic Equation lesson 2

In the previous lesson we learned on how to solve an equation. We found three methods to solve the equation. If not visit http://scipsycho.blogspot.in/2016/08/quadratic-equations.html

Now, we will go into details about quadratic equation, the importance of parameter a, b and c and other things that are important for JEE aspirants. Let's begin.

As we saw earlier, a quadratic equation can be represented by 

ax2 + bx +c =0. 
Let's start with 'a'.

1.) Coefficient of x

Case 1. : when a < 0

let's inspect the graphical form of the equation.

If x -> ∞, only the x2  term will decide the outcome of the expression as it's magnitude will overwhelm the other factors.

Now, as x -> ∞ the value of expression will tend to infinity if a>0.
However, a is less than zero therefore, the expression will tend to 
-∞.

when x -> -∞, x-> ∞, and thus even then the expression tends to 
-∞. ( Since a is negative therefore ax2 < 0)

As we all know the graph of a quadratic expression is a parabola,
therefore, the parabola will be downward facing when a < 0
Let us take an example.

y= -x+ 5x -6  will look something like this

graph of y= -x^2 + 5x -6 facing downward

















Case 2. : a>0 

When a > 0 as we would already guess would make the parabola upward facing let's see how.

when x-> ∞ or when x-> -∞, then x->; ∞ and thus the expression tends to ∞. ( Since a is positive therefore ax2 > 0).

Let's take the example of the above equation with just one difference i.e. a>0.
Therefore,

y= x- 5x -6  will look something like this
y= x2- 5x -6 is facing upward























Conclusion: Thus the sign of a or the coefficient of  xdecides whether the equation will be upward facing or downward facing.

As we all know a ≠ 0,
therefore,we have covered all the cases. Let's talk about 'b'.

2.) Coefficient of x 

Let us assume a quadratic expression f(x) where it denotes a function in x having roots as Î± and Î² respectively.

Therefore, the equation becomes, f(x) = (x - Î±)(x - Î²).

Now, solving gives, f(x) =  x - (α + Î²)x + Î±Î².

Comparing it with ax+ bx + c.
gives,

b/a = -(α + Î²)

and 
 c/a= Î±Î²

Thus, we can say that the roots of the quadratic equation show a very amazing relation as,

Sum of roots = -b/a                 .....................(1)

Product of roots = c/a              .....................(2)


We will be using line ...(2) later.
Let's focus on line ....(1) for now.

So, we know that the sum of roots is -b/a. Well it pretty much says about the use of the coefficient of b. 

Also, following very vital points come in light:

a.)
     The average of the roots of a quadratic equation is -b/2a.


This is a very useful deduction.At this point it might seem obvious but these are those things that are mostly asked in IIT JEE of course indirectly.

b.) if the roots are irrational than they will be of the form  m + √n .Now if the coefficient of the equation are rational therefore, the sum of the roots which is -b/a. Since, only sum of two irrational numbers can be rational therefore, the other root must be ,

m - √n.

This is also valid for complex roots. 
You must be thinking that why the rational part is also equal.
Well, since all the parameter are rational therefore 'c' is also rational which is used to find the product of the roots.

c.) For the graph of the form y = ax2 + bx +c the parabola is symmetric about the midpoint of the roots i.e. (α + Î²)/2.
Thus, it is equal to the average of the roos i.e. -b/a.

Thus the line, x=-b/2a cuts the parabola y= ax2 + bx +c in symmetry.
For eg:
y= x+ 5x -6 is cut by x=-5/2*1 i.e. x=-5/2 in symmetry.

x2 + 5x -6 is symmetrically intersected by x=-5/2


















d.) The value of 'b' is used to find out the discriminant of the quadratic equation. As we all know this discriminant tells us about the nature of the roots, 'b' term becomes very important.

3.) The constant 'c'

Well, you might think that there is not much use of this constant but there are many important points that we can deduce from it.

Case a) Let us take c=0

If c=0, then the following points come to light,

i.) The roots of the quadratic equation are real and distinct.
Well, it's a very important fact. As we all know that if c=0,
the Discriminant is always positive and thus real and distinct roots of the equation exits.

ii.) The quadratic equation passes through the origin i.e. '0' is one of the roots of the equation.

For eg: the graph of the equation  2x2 + 5x=0, is something like this,
graph of 2x2 + 5x is passing through origin


Case b) Now, c>0,

i.) The graph of the equation intersects the positive y-axis.
Well, it can easily be understood if we put x=0 in the general equation y = ax2 + bx +c. which gives, y= c.
This won't seem much of a point but it will become fairly important once we start question which ask the range of values of any parameter.
We will be seeing such question lately.

Case c) Similarly, for c <0 the equation intersects the negative
y-axis.

Now, let us introduce some specific types of questions and various ways to solve it.

1.) The roots of the equation ax2 + bx +c=0 are m and n. Find out the equation whose roots are m+1 and n+1.

Now, we are going to discuss an important method to do so. However, keep in mind that this method works only when the expression is identical for the both the roots i.e. if m is changed to f(m) then n 
should also change to f(n).

We call this method as transformation of roots.

Let the roots m and n are now changed to (a1m + a2)/(b1m + b2) and (a1n + a2)/(b1n + b2) respectively.
Then write,

y = (a1x + a2)/(b1x + b2
Now, find x as a function of y.
Once, you have done that replace the values of  x in ax2 + bx +c with that of y.

After that you have done that, you will be left with a quadratic equation in variable y. All you have to do is to replace y by x.

For eg:

Let the equation be x2 + x + 1. As you know the roots of the equation are complex and thus finding an equation with roots m+1 and n+1 where n and m are the roots of this equation is very difficult. So, let's follow the above method.

y = x+1 
This implies, x = y-1

Putting it in the equation gives,

(y-1)2 + (y-1) + 1,
simplifying it gives,

y2 -y + 1
Now, replacing y by x gives,

x2 -x + 1.

This is the required equation.

The other type of question that I am going to discuss is utterly difficult to explain here. So, I will be posting a video about this on my newly made youtube channel SciPsycho very soon. 
Don't forget to like and subscribe.

Thank you.





  


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