IIT JEE , AIEEE Forum - Mathematics , Algebra

neeraj_agarwal_1990

If a,b are real roots of x²+px+1=0 and c,d are real roots of x²+qx+1=0 then
(a-c)(b-c)(a+d)(b+d) is divisible by:

(a) a+b+c+d

(b) a+b-c-d

(c) a-b+c-d

(d) a-b-c-d

harshit0909

hey neeraj........i will try 2 solve it,nyway,i tried it verbally,put p nd q as 2 nd -2,clearl d huge term we hv becomes 0,derefore our ans must not be o or infinity,only (b) nd (d) satisfy it,so either both r ans,or one of dem,rest u can try,i will do dat,i just wanted 2 introduce dis concept of challenging d ques framework,,,,,,,,,,bye

harshit

aditya_arora04

Hi neeraj. The question can be done as follows :

1. According to the equations given, we get that :

a+b= -p and ab = 1

c+d= -q and cd = 1

2. Now, (a-c)(b-c)(a+d)(b+d)

(a-c)(b+d)(b-c)(a+d)

(ab+ad-bc-cd)(ab+bd-ac-cd)

(ad-bc)(bd-ac) [ As ab-cd = 0 ]

abd²- a²cd - b²cd + abc²

abd²+ abc² - ( a²cd + b²cd )

ab( c² + d² ) - cd( a² + b² )

c² + d² - ( a² + b² ) [ As ab=cd=1 ]

( c+d )² - 2cd - { ( a+b)² - 2ab }

q² - 2 - { p² - 2 }

q² - p²

( q+p )( q-p )

( -a-b-c-d )( a+b-c-d )

- ( a+b+c+d )( a+b-c-d )

That is the expression is divisible by (a+b+c+d) and (a+b-c-d). So the answers are (a) and (b)

puneet

hiiiiiii neeraaj ...

Aditya has given a perfect solution .. it is absolutely correct ...

Good work aditya ... well done .. cheers to you ..

I hope it is clear to u .. neeraj .. the approach is fairly simple .. isn't it ??

I case of doubt .. get back ..

cheers

harshit0909

yeah dats a perfect solution,

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