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Algebra

Hot goIITian

 Joined: 15 Jul 2011 Post: 135
28 Jul 2011 22:45:00 IST
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if 1 lies btwn d roots of x^2+px+q=0,den p+q is always >-1,....hw??????/
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Algebra

if 1 lies btwn d roots of x^2+px+q=0,den p+q is always >-1,....hw??????/

Hot goIITian

Joined: 13 May 2011
Posts: 138
29 Jul 2011 15:34:21 IST
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question is wrong.

take p=-4, q=2

1 lies between them but their sum is less than -1

Hot goIITian

Joined: 15 Jul 2011
Posts: 135
30 Jul 2011 14:49:34 IST
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bt dese values of p,q satisfy d eqn?????/

New kid on the Block

Joined: 1 Apr 2011
Posts: 22
1 Aug 2011 13:05:35 IST
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p+q is always less  than -1 and not greater than -1

Consider the graph of the function f(x) = x^2+px+q

Since the coeffiecient of x^2 is greater than zero the parabola would have the shape of a "bowl" and not that of a "mountain"

Now notice that when a bowl shaped parabola cuts the X-Axis at two distinct points, it will take negative values for all values of x in between the roots.

[If you are having a hardtime visualizing it,just grab a pen and paper and see it for yourself!]

So f(1) < 0 or 1+p+q < 0 and hence the answer

You can also solve this problem by brute force.Not as elegant as the above one but yeah! it will get you the answer

Express the roots in terms of coefficients using the familiar formula for the roots of a quadratic equation

and then simplify the inequality

Alpha < 1 < Beta

Hot goIITian

Joined: 15 Jul 2011
Posts: 135
1 Aug 2011 18:08:58 IST
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yes dude.i also applied d same method...bt d ans is >-1

New kid on the Block

Joined: 1 Apr 2011
Posts: 22
2 Aug 2011 09:16:18 IST
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Well ... have you considered the possibility of the answer being wrong,given that it has been proved wrong?  :)

New kid on the Block

Joined: 18 Aug 2011
Posts: 16
23 Aug 2011 23:31:17 IST
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a. f(1) is smaller than zero so u get da following condition .....

Forum Expert
Joined: 19 Feb 2009
Posts: 1958
24 Aug 2011 09:35:05 IST
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First of all lets write the co-efficients ..!

a = 1 , b = p  and c = q

SInce a > 0  the equation will open upward ...!

Hence for every x between root r1 and r2 ...the graph will have a negative value ..!
Given 1 lies between the root
So f(1) < 0

1 + p + q < 0

p+q < -1

And it will always be less than -1 ..never greater than -1 ..so please check out your question once again !

New kid on the Block

Joined: 24 Aug 2011
Posts: 1
24 Aug 2011 11:21:59 IST
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hey i think the question is wrong.. the condition for k lies between the roots is af(k)<0 and the ans is coming p+q<-1

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