IIT JEE , AIEEE Forum - Mathematics , Algebra

LAMPARD

f(x) is a cubic polynomial x³+ax²+bx+c such that f(x)=0 has 3 distinct integral roots and f(g(x))=0 does not have real roots where g(x)=x²+2x-5,then minimum value of a+b+c is=?
The solution given is as follows-
Let roots be p,q,r such that p<q<r.Since g(x) can take values from [-6,inf.],p<= -7,
q<= -8 and r<= -9.
I didnt understand how they got this from the info that g(x) can take the specified values.
Can anyone plz explain??You can also post any other method to solve the sum.

computer001

pl correct ur f(x) and g(x)
u missed out the x^2..u have typed as x i think

LAMPARD

Oh yes...corrected it now.

akhil_o

x^2+2x-5=g(x)
minima=> 2x+2=0 or x=-1
so M=1-2-5=-6
it is unbounded for maxima

so -6<=g(x)< infinity

now putting
f(g(x))=f(-6) for minima

=-216+36a-6b+c
since it has no real roots, it lies totally above y-axis for all g(x)

LAMPARD

Akhil,i know that g(x)>=-6 but how p,q and r are predicted from that??

akhil_o

is the answer -1?

since g(x) also takes 1 in the range

f(1)=a+b+c+1
but f(1)>0
so a+b+c+1>0,
a+b+c>-1

LAMPARD

The value of a+b+c is asked.If you want the solution,i will post it but i want to know how p,q and r are got because the remaining solution relies on it.

LAMPARD

Someone please try it!!!

LAMPARD

C'mon,what happened to all the math scholars???

elastiboysai

Lampard
g(x) will have zeroes if it is >-6
see i can factorizef(g(x))
as [g(x)-p][g(x)-q][g(x)-r]
if p,q,r are less than -6 then g(x) cant have real roots
but gvn f(x) has 3 dist. integral roots,..
so the gvn assumptions rt
edited

elastiboysai

k isee it now
f(x) has to have 3 distinct integral roots
makes sense.
so p<=7,q<=8,r<=9
and also
p<q<r

the fact that f(x) has integral roots need not imply g(x) has real roots
this was what was preluding me

LAMPARD

Oh yes...i have got it now...thanx elastiboy...

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