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Algebra

New kid on the Block

 Joined: 10 Jun 2009 Post: 20
1 Mar 2012 10:17:23 IST
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Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Algebra

1. If the equations and have a common root and a,b,c belong to N, find the minimum value of a+b+c.

2. If the equation does not have two distinct real roots and a + c>b, the prove that f(x) >= 0, for all x belonging to R.

New kid on the Block

Joined: 10 Jun 2009
Posts: 20
1 Mar 2012 10:19:45 IST
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1. equations are

2. x2 + 3x +5=0 and ax2 +bx +c=0

New kid on the Block

Joined: 10 Jun 2009
Posts: 20
1 Mar 2012 10:20:12 IST
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I wonder why equations are not shown in my first post.

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
1 Mar 2012 15:25:24 IST
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x^2 + 3x+ 5 =0

clearly the roots are complex conjugates of each other.

so if ax^2 + bx + c = 0 and x^2 + 3x + 5 have a common root then their other roots are also equal.

so, a/1 = b/3 = c/5 = (a+b+c)/(1+3+5) = (a+b+c)/9.

a,b,c are natual numbers.

the minimum value is 5+3+1 = 9.

for the second question,

a + c - b > 0.

or, f(-1) > 0.

this might help.

i wanted to ask if a,b,c  are natural numers here like the first question?

sorry, but i am too dizzy to think anything...

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Joined: 21 Mar 2012
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21 Mar 2012 13:24:44 IST
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Forum Expert
Joined: 23 Jan 2007
Posts: 7958
5 Apr 2012 15:36:37 IST
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2. Since the equation doesn't have two distinct real roots : discriminant , D<=0

So the graph will either be above x-axis or touch it or it will be below x-axis or touch it

So, f(x) is either >=0 or <=0 for all x

f(-1) = a-b+c > 0, since for x=-1 it is positive therefore f(x)>=0

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