ftjee aits - 15/2/09.

if the equation x4+ax3+bx2+cx+d=0 has four real positive roots then,

a) ac >= 16d

b) b2 >= 36d

c) ac >= 16b

d) b2 >= 36c

14 Answers

39
Dr.House ·

come on , if u want to know how to do in examination, let the roots be 1,2,3,4.

and gven equation is (x-1)(x-2)(x-3)(x-4)

now find a,b,c,d. thats it done.{ in fact this is how i did in exam}

1
skygirl ·

nahi yaar :(

multilple choice mein risk nahi lena chaiye... :(

24
eureka123 ·

ise kehte hain dhake se question karna.........[3]
but friends yehi cheez exam hall mein kaam aati hai........jis bhi method se ans aaye..bas aana chahiye..........[4]

1
skygirl ·

waise... solution diya hai ...

koi solution samjha do naa ..... please...

1
skygirl ·

eureka... am in full support wid you [3]

and am sure anyone who disagrees ... will be purely lying .. sry 'anyone'...

24
eureka123 ·

tried Rolle Thorm????

341
Hari Shankar ·

When they say positive roots inequalities should occur to you.

-a = Σr, -c = d Σ1/r

Hence from AM-GM ac = (-a)(-c) = d Σr Σ1/r ≥ 16d

1
skygirl ·

kk ...

24
eureka123 ·

this is called perfection...........[1]

39
Dr.House ·

bhai ab how can a usless student be compared with a genious teacher?????????????

24
eureka123 ·

u r also correct mathie...[3]

21
tapanmast Vora ·

Wer did the factor of '16' come frm in prophet sir's soln???
can sum1 xpand....

did it cum frm [x1 + x2 + x3 + x4]/4 = AM or somthin.... pl xplain

341
Hari Shankar ·

AM-GM is one of the ways of proving that

(x1+x2+x3+x4)(1/x1+1/x2+1/x3+1/x4) ≥ 16

This follows from (x1+x2+x3+x4) ≥ 4 (x1x2x3x4)1/4 and (1/x1+1/x2+1/x3+1/x4) ≥ 4/(x1x2x3x4)1/4

and multiplying the two.

Cauchy Schwarz, AM-HM are two other ways

11
rkrish ·

x4+ax3+bx2+cx+d=0

a<0,b>0,c<0,d>0

Roots : x1,x2,x3,x4

a= - (x1+x2+x3+x4)
b= (x1x2+x2x3+x3x4+x4x1+x1x3+x2x4)
c= - (x1x2x3+x2x3x4+x3x4x1+x4x1x2)
d= (x1x2x3x4)

If x1=x2=x3=x4=k(let)

a= - 4k
b= 6k2
c= - 4k3
d= k4

ac= 16k4= 16d
b2= 36k4= 36d

So (C) & (D) are incorrect.

If x1=x2=k1(let) & x3=x4=k2(let)

a= -2(k1+k2)
b= (k12+k22+4k1k2)
c= -2k1k2(k1+k2)
d= k12k22

ac= 4k1k2(k1+k2)2 > 16d since (k1+k2)2 > 4k1k2
b2= [ k14+k24+8k1k2(k12+k22) ]+18k12k22 > 36d
since [ k14+k24+ 8k1k2(k12+k22) ] > 18d

Hence (A) & (B) are correct !!

A bit longer but correct approach !!

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