How to approach this problem ?

AdkAdi

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I have tried selecting random numbers and then proving LHS=RHS but I suppose, that is hota legit method!
 

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Let's see if I understand what you are saying. Assume 4BC>1. Then 1 - (more than 1) is positive, making the root reals??
 
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Let's see if I understand what you are saying. Assume 4BC>1. Then 1 - (more than 1) is positive, making the root reals??
Sorry I messed it up ..I typed the wrong statement..I do know about the nature of roots and the properties of determinants
 
Sorry I messed it up ..I typed the wrong statement..I do know about the nature of roots and the properties of determinants
Good, so please state the nature of the roots. Hint- sometimes you get exactly one real root (when?), sometimes you get two two real roots (when) and other times you get two imaginary roots (when?).

How do determinants come into this problem?
 
See when the discriminant (b²-4ac=D)
D=0 real equal roots
D>0 Real distinct roots
D<0 Imaginary roots
 
You really should say how many roots.
Saying D>0 implies Real distinct roots does not say that there are exactly two distinct roots.
 
Sorry, but I did not think that you meant discriminant when you wrote determinants
 
You really should say how many roots.
Saying D>0 implies Real distinct roots does not say that there are exactly two distinct roots.
If a quadratic is given the. It is implied that there will be 2 roots
 
By the Fundamental Theorem of Algebra if a polynomial has n as it's highest power, then it will have n roots, where a root can be repeated.
You are absolutely correct but most of us have a tendency to write 1 root if the root is repeated.

-Dan
 
The first thing I would have done in this problem is NAME THE VARIABLES, and note that they are distinct

[math]0 < p < q < r < s \text { and all are real numbers.}[/math]
If all three have real roots then

[math]1 \ge 4BC,\ 1 \ge 4AC, \text { and } 1 \ge 4AB.[/math]
What then can we say about q and r, given that they are both positive and q < r?
 
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