Quadratic Equations are first taught to us in 6th or 7th class and most of us are able to score good marks in it because we are able to solve 90% of the questions by just using that formula. And that formula is:

The above formula gives us the roots of the quadratic equation ax^2 + bx + c = 0

For this post, I am assuming that you are aware of the basics of quadratic equations and know how to use the above mentioned formula. In case you are not, spending five minutes on the wiki page of Quadratic Equations won’t hurt. Wikipedia can be daunting at times, so come back here as soon as you start feeling woozy.

In this particular post, I am going to discuss ideas related to maxima and minima with respect to equations and quadratic equations in particular.

The first thing that you need to understand is that a quadratic equation will either have a maxima or a minima but it cannot have both. The reason for that is, a quadratic equation is shaped like a parabola which is either open upwards or downwards.

As you can see here, the parabola is open upwards and it will only have a minima

As you can see here, the parabola is open downwards and it will only have a maxima

The nature of the graph of the quadratic equation is decided by the co-efficient of x^2.

• If the co-efficient of x^2 is greater than 0, the parabola will be open upwards and hence it will have a minima.
• If the co-efficient of x^2 is less than 0, the parabola will be open downwards and hence it will have a maxima.

Once you have decided whether it will be a maxima or a minima the next task is to figure out two things:

a) The point at which the maxima / minima occurs

b) The maximum / minimum value of the quadratic equation

Let us try to figure these values out with the help of an example of both types i.e. when co-efficent of x^2 is greater than 0 and when it is less than 0

Case 1: a > 0

Now, we know that a perfect square is always non-negative

=> The lowest value that (x-2)^2 can take is 0

=>The minimum value that the entire quadratic can take is 3

=>The minima occurs at x = 2 when x-2 is 0

Case 2: a

Now, we know that a perfect square is always non-negative

=> The lowest value that (x+3)^2 can take is 0

=> The maximum value that the entire quadratic can take is 16

=> The maxima occurs at x = -3 when x+3 is 0

As you might have realised, the above idea of ‘completing a square’ can be used to find out the maximum / minimum value and also the point at which the maxima/minima occurs in any quadratic equation. Based upon the logic given above, there is also a set of formulas that you can use for a quadratic equation represented by ax^2 + bx + c = 0

Maxima / Minima occurs at -b/2a

Maximum / Minimum value is (4ac – b^2) / 4a

To make things a little more complicated, you might get a question in which the quadratic equation is in the denominator. It is sometimes also referred to as the rational function.

Example 1: Constant / Quadratic Rational Function

Find out the maximum or minimum value of 3 / (2x^2 – 5x + 7)

We know that the denominator is a quadratic equation with a > 0

=> The denominator will have a minimum value at -b/2a = 5/4 and the value will be (4ac – b^2) / 4a = 31/8

=> The function will have a maximum value at 5/4 and the value will be 3/(31/8) = 24/31

Example 2: Linear / Quadratic Rational Function

For x to be real, the discrimnant of the above equation should be non-negative

Maximum value of the function is 1 and the minimum value of the function is -1/7

Example 3: Quadratic / Quadratic Rational Function

For x to be real, the discrimnant of the above equation should be non-negative

So, the least value of the function is -1/3

Let me add that it is not necessary that you will always be able to find out the maximum and / or the minimum value of a rational function.

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