1r there any condition set for b and c
Find all integers 'a' such that
(x - a)(x - 12) + 2
can b factored as (x + b)(x + c)
where 'b' and 'c' are integers.
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7 Answers
62(x - a)(x - 12) + 2 = (x+b)(x+c) for all x
x2-12x-ax+12a+2 = x2+(b+c)x+bc
12a+2-bc = (12+b+c+a)x
for all x
thus
12a+2-bc=0
and 12+b+c+a = 0
a= -12-b-c
a,b,c are integers.
clearly, bc=-12(12+b+c)+2
now can you find all integer solutions for b and c?
62bc+12b=-12(12+c)+2
b(c+12) = -12(12+c)+2
thus (b+12)(c+12) = 2
now we have only few cases b=-11, b=-10, b=-13 and b=-14
why?
1i know what nishant sir has done is absolutely right..but cant we do it this way also...
we have (x-a)(x-12)+2
=x2 -x(12+a) +(12a+2)
now we need this to be factorised as (x-a)(x-b) means that we want the above function to have integral roots...
or D=K2 where K is an integer...
we thus have, (12+a)2-4(12a+2)=K2
or basically (a-12)2=K2+8
or (a-12)2-K2=8
(a-12-K)(a-12+k)=8
we have cases like
1,8
-1,-8
-2,-4
2,4
for a product of 8, where the 2 factors are integers..
we discard the 2 cases as they donot give us integral values of a and K
using the 3rd and the 4th cases we can solve to get values of a and K
please tell me if this rite...if not tell where i am messing it up...
cheers!!