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value of abcd

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by Suyog » Tue Oct 09, 2007 2:49 pm
For R.H.S
if we expand the terms; the last term will be abcd

for L.H.S.
we can take it as
the ( a - 1) ^ 2 - 25 ; here the last term in the bracekts will be +1 always.
so 1 - 25 = -24

what are the options given.
Let me know if its wrong.
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by Gunjan99 » Thu Oct 11, 2007 2:24 am
how did you infered this.. here the last term in the bracekts will be +1 always?? in step:
the ( a - 1) ^ 2 - 25 ; here the last term in the bracekts will be +1 always
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by samirpandeyit62 » Thu Oct 11, 2007 3:19 am
Agree with Suyog OA should be -24

1st approach (fastest)

(x^2 - 5x - 1)^2 - 25

here when the term (x^2 - 5x - 1)^2 is expanded the result will be all multiples of x & only 1 as a constant term (without x coz it is a square)

when (x-a) * (x-b) * (x-c) * (x-d) is expanded then the constant term will be abcd

so some multiples of x + 1-25 = some multiples of x + abcd

so abcd =-24

2nd approach

(x^2 - 5x - 1)^2 - 25

= (x^2 - 5x - 1)^2 - 5^2

= (x^2 - 5x + 4)(x^2 - 5x - 6)

= (x-1)(x-4)(x-6)(x+1)

abcd =-24
Regards
Samir
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