# Q11: IIT Level Solved II

Now we come to the harder problems which occur in the Engineering Entrance exams.
In this question, three different Q functions with one root mentioned in each. We have to find the relationship between these three roots.

A quadratic expression becomes zero when x = the roots, alpha & beta. So here we have put x = the roots given for each Q and arrive at three equations 1,2 & 3. The trick here is to put alpha & beta for x in the third equation. Taking the third equation as f(x) we find f(alpha) and  f(beta) and combining with the other three equations that we have, we come to the conclusion that
f(alpha) <0 while f(beta)>0.
​Since it is given that 0<alpha<beta, we can draw the graph with gamma lying between alpha & beta and that is the relation we can conclude regarding the relationship of these three roots.

This diagram simply shows that y value is negative at x=q and the y values hence the quadratic function itself is greater than 0 when x=p or x=r. Also, when x=alpha or beta, the y values are zero as that is where the quadratic curve cuts the x axis at two points.

The discriminant was carefully computed using elementary but complex algebra as shown and we find D is positive and thus there are Real Roots.

Note the vertical placement of the three algebraic entities in the equation given. Writing vertically has significant virtues especially large algebraic equations.

This diagram shows the complex roots of a quadratic if the Discriminant, D < 0 i.e. D is a negative quantity. In such cases the square root of  – 1 is taken as i and the results are explicit and we get two roots as usual. We are not going to deal with Complex Numbers yet but only gave you a taste of this fascinating subject called Complex Numbers.

Since a b c are mentioned as positive, the axis given by
x=-b/2a will be negative since both a and b are positive. So, the positive part of the roots, be it real or complex, is -b/2a.

So, having an understanding of how a, b & c affects the position of the graph helps you to crack tough questitons

In the diagram above the question was to find the nature of the roots. And we clearly see that both the roots are negative is what we can certainly conclude as c>0. Alternately we can have no real roots if D<0 i.e. when the graph is entirely above x axis as c>0
 In these diagrams what we learn is that between the roots, the y values are shown in red, they are downwards and are negative in value i.e. when we say Q<0 it means that the y values of the Q function are <0 for x values lying between the two roots. And vice versa. when the x values are outside the roots the y values are all upwards shown in green and are >0.
 Here we have a quadratic inside another quadratic. When you compute D you end up with another quadratic which gives the values of D. Basically we need to know whether D is positive or negative. Now since D is also a quadratic, we will follow the standard practice of looking at the roots of this D quadratic and when ‘a’ will lie between the roots, D will be negative which will lead to negative roots for the original quadratic function. And D will be positive for ‘a’ taking values beyond the roots on two sides of it.

 The Mod or Modulus function is a two in one function.Mod X is always positive. However, if X is positive or X>0 means Mod X = Xwhile if X is negative i.e. X<0, in that case​Mod X = – X since X being negative, – X is positive.  Whenever an equation has a Mod function inside it you have to take two Cases separately and work out the equations. Finally you have to combine both the Cases and come to your final conclusion.