Photo Credit: Victorrochajr (Flickr.com)
I am quite sure that most of you, if not all of you would be aware of the basics of quadratic equation but just to provide a brief refresher:
For a quadratic equation given represented by ax^2 + bx + c = 0
Roots are given by (-b + sqrt(b^2 – 4ac))/2a & (-b – sqrt(b^2 – 4ac))/2a
Sum of the roots is -b/a
Product of the roots is c/a
You can expect a problem or two based upon the idea of roots and solving equations in CAT and other exams. If you are lucky, it would be something on the simpler side like this:
Example 1: Amar, Akbar and Antony solve a given quadratic equation. Amar commits a mistake in the constant term and finds the roots as 8 and 2. Akbar commits a mistake in the coefficient of x and finds the roots as -9 and -1. If Antony solves the equations without making any mistakes, what roots does he find?
Amar’s equation:
(x – 8)(x – 2) = 0
x^2 – 10x + 16 = 0
Akbar’s equation:
(x + 9)(x + 1) = 0
x^2 + 10x + 9 = 0
Antony will form the correct equation by taking the coefficient of x from Amar’s equation and the constant term from Akbar’s equation.
So, Antony’s equation will be:
x^2 – 10x + 9 = 0
(x – 1)(x – 9) = 0
Roots are going to 9 and 1.
But sometimes, questions can be a little more complex and deal with polynomials of higher degree. It is important to note that the pattern of sum of the roots, product of the roots, etc. is followed in polynomials of higher degree as well.
Given below are couple of examples of Cubic and Biquadratic equations which are basically equations of degree 3 and 4 respectively.
Cubic equation
ax^3 + bx^2 + cx + d = 0
Sum of the roots = – b/a
Sum of the product of the roots taken two at a time = c/a
Product of the roots = -d/a
Biquadratic or a Quartic equation
ax^4 + bx^3 + cx^2 + dx + e = 0
Sum of the roots = – b/a
Sum of the product of the roots taken two at a time = c/a
Sum of the product of the roots taken three at a time = -d/a
Product of the roots = e/a
I guess that with the help of the above two equations you would be able to identify the fact that for a polynomial:
Sum of the roots is given by -(Coefficient of second highest power of x)/ (Coefficient of highest power of x)
Product of the roots is given by (+ or -) Constant/ (Coefficient of highest power of x)
The + / – sign alternates and the other coefficients help us determine the sum of product of roots taken 2 at a time, 3 at a time, etc.
Let us look at couple of examples to understand how this can be used.
Example 2: If a, b and c are the roots of the equation x^3 – 3x^2 + x + 1 = 0 find the value of
a) 1/a + 1/b + 1/c
b) 1/ab + 1/bc + 1/ca
c) (a + 1)(b + 1)(c + 1)
Solution:
a) 1/a + 1/b + 1/c = (ab + bc + ca)/abc = 1/-1 = -1
b) 1/ab + 1/bc + 1/ca = (a + b + c)/abc = -(-3)/(-1) = -3
c) If a, b & c are the roots of the equation, then:
f(x) = x^3 – 3x^2 + x + 1 = (x – a)(x – b)(x – c) = 0
f(-1) = – 1 – 3 – 1 + 1 = (- 1 – a)( – 1 – b)( – 1 – c)
f(-1) = – 4 = – (a + 1)(b + 1)(c + 1)
So, (a + 1)(b + 1)(c + 1) = 4
As you can see, by using the ideas listed above in the post, the questions have become a lot easier and there is no need to actually find out the value of the roots.
Example 3: Given that three roots of f(x) = x^4 + ax^2 + bx + c are 2, – 3, and 5. What is the value of a + b + c?
We have to find out a + b + c
f(1) is 1 + a + b + c
So, we need to find out f(1) – 1
Let the 4th root be r
Coefficient of x^3 = -(Sum of the roots)
0 = – (r + 2 -3 + 5)
r = – 4
So, f(x) = (x – 2) (x + 3) (x + 4)(x – 5)
f(1) = (-1)*4*5*(-4) = 80
a + b + c = f(1) – 1 = 80 – 1 = 79
I have a feeling that these ideas would be more helpful in XAT than CAT as on most occasions, quantitative aptitude is tougher in XAT. May be I have a sample bias but I feel that XAT paper setters favor topics like algebra and functions than their CAT counterparts.
Let me end with this:
And then Satan said, Put the alphabet in math….
Ravi Handa, an alumnus of IIT Kharagpur, has been teaching for CAT and various other competitive exams for around a decade. He currently runs an online CAT coaching and CAT Preparation course on his website http://www.handakafunda.com