I need help with a maths question...can you explain your avatar? =pYip said:Since no one has pointed it out yet user Bank$, your approach, though mathematically correct, is also inefficient. Your answer is also partly incorrect. It is not required to expand the expressions then factorize the cubic. It is much more efficient to go like this:
(x-1)(x-3)+[(x-1)(x-3)]^2<0
(x-1)(x-3)[1+(x-1)(x-3)]<0
(x-1)(x-3)(x^2-4x+4)<0
(x-1)(x-3)(x-2)^2<0
1<x<2, 2<x<3
Note that x cannot equal 2 as the inequality is <, not <=.
Thanks for the help Yip.Yip said:Passion, which is a function of time
Where ? If u are talking about x not equal to 2 i added that in right after my post under edit.Yip said:Your answer is also partly incorrect.
I do agree with the inefficientcy of my methord i guess ill have to go back a revise on yr 11 lol.Yip said:(x-1)(x-3)+[(x-1)(x-3)]^2<0
(x-1)(x-3)[1+(x-1)(x-3)]<0
(x-1)(x-3)(x^2-4x+4)<0
(x-1)(x-3)(x-2)^2<0
1< x <2, 2< x <3