(x - 9753)(x -7531) + 234321
this can be factored by multiplying out and solving as quadratic however let is take x –a = t where a = (9753+7531)/2 = 8642
now 9753-8642 = 8642-7531 = 1111
so (x - 9753)(x -7531) + 234321
= (x-8642-1111)(x-8642 + 1111) + 234321
= (t-1111)(t+1111) + 234321 taking t = x- 8642
= t^2 – 1111^2 + 234321
= t^2- 1234321 + 234321
= t^2- 1000000
= t^2- 1000^2
= (t+1000)(t-1000)
= (x-8642+1000)(x-8642-1000)
= (x-7642)(x-9642)
This trick helps in keeping expression simple
some short and selected math problems of different levels in random order I try to keep the ans simple
Tuesday, September 6, 2011
2011/068) Find all ordered pairs of integers (x,y) which satisfy x^3 + y^3 - 3x^2 + 6y^2 + 3x + 12y + 6 = 0
Rewrite this using x terms and y terms in groups
(x^3 - 3x^2 + 3x) + (y^3 + 6y^2 + 12y) = -6
==> (x^3 - 3x^2 + 3x - 1) + (y^3 + 6y^2 + 12y + 8) = -6 - 1 + 8
==> (x - 1)^3 + (y + 2)^3 = 1
as difference of 2 cubes cannot be 1 so one of them is zero and another is 1 and there is no other choice
x - 1 = 1 and y + 2 = 0 ==> (x, y) = (2, -2), or
x - 1 = 0 and y + 2 = 1 ==> (x, y) = (1, -1).
so solution sets are (2,-2) and (1,-1)
(x^3 - 3x^2 + 3x) + (y^3 + 6y^2 + 12y) = -6
==> (x^3 - 3x^2 + 3x - 1) + (y^3 + 6y^2 + 12y + 8) = -6 - 1 + 8
==> (x - 1)^3 + (y + 2)^3 = 1
as difference of 2 cubes cannot be 1 so one of them is zero and another is 1 and there is no other choice
x - 1 = 1 and y + 2 = 0 ==> (x, y) = (2, -2), or
x - 1 = 0 and y + 2 = 1 ==> (x, y) = (1, -1).
so solution sets are (2,-2) and (1,-1)
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