Q7: sum, prod of roots

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Taking a break from the geometrical aspects of a quadratic, we turn to a study of a purely algebraic aspect which is again very elegant and also very useful in solving a variety of problems. 

As we have talked about earlier, alpha = A – B and
beta = A + B where A and B are as shown, A=-b/2a & B=sq rt D/2a. Thus, when we add,
alpha + beta = 2A = – b/a and 
alpha x beta = A^2 – B^2 = c/a


There is an exercise below. This is an essential practice in trying to fathom the values of these algebraic entities composed out of the two roots, alpha and beta. We will give you solution later
EXERCISE
Solved Examples

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Solved Examples and Problems

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As soon as you encounter a quadratic equation, you must be immediately able to find the Sum and Product of the roots, alpha & beta

We have shown a few solved examples and there a few practice exercise for you to attempt. 

Make a practice of finding the Sum & Product of the roots and also the Discriminant. 



Forming an Equation given the Sum & Product of the Roots
In mathematics we look at an entity (an expression) every which way, especially we look at the obverse side e.g. if it is a right angled triangle, the length of the three sides will abide by Pythagoras’ Theorem. To state the same thing obversely, we can state that if the three sides of any triangle follow the Pythagoras’ Theorem, the triangle is a right angled triangle.

Similarly here, if we are given a quadratic equation, we can find the sum & product of the two roots. 

Here we show its obverse theorem i.e. if we are given the sum and product of the two roots of a quadratic, we can write down the quadratic equation like this: 
x^2 – (Sum) x + Product =0

Thus knowing this, we can solve the problem shown in the diagram.