this is a difficulty of my own

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sujoy

Junior Member
Joined
Apr 30, 2005
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110
Hi ,
everybody, how are you doing today ?
Here after a long time I come up with a difficulty of my own:
In a quadratic equation ax^2 +bx +c =0 , for both the roots to be positive, the rule is that : a & c should be of the same sign & b should be of the opposite sign.
My difficulty is that , how this small statement always gurantee that "b" is always greater than [b^2 - 4ac ]^1/2------------I mean in the root no 2 :
-b +[b^2 - 4ac]^ 1/2
----------------------------- .............................>( root no 1)
2a

-b - [b^2 - 4ac] ^ 1/2
------------------------------ .............................>(root no 2)
2a
........I understand that if "b" has a negetive value in the equation "-b" is positive
and vice-versa, what I mean is how it is that
|b| > - [b^2 - 4ac ]^1/2 ? :?
regards
Sujoy
 
Welcome back.
If a & c have the same sign then b²-4ac is subtracting something from b² so the sqrt is always less than |b|. But they should also say that b²-4ac must be > 0 for x to be positive, not complex.
 
Another way of putting it:
all quadratics representing (x - p)(x - q) where p and q > 0
will have 2 positive roots ... AHEM!
 
Thank you .
If a & c have the same sign then b²-4ac is subtracting something from b² so the sqrt is always less than |b|. But they should also say that b²-4ac must be > 0 for x to be positive, not complex.
Click to expand...
Could this be proved anyway algebrically
regards
Sujoy
 
What I meant is that I can understand what teacher Gene had written
It is :
x^2 > [x - constant]^2 & this is very logical i.e., we can follow this logically.
( after he explained)
Or we may have other methods also , is what I asked .
Probably I had confused somewhere .
Its OKAY now .
Thank you .
regards
Sujoy
 
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