My badMight need more of a lead-up for students.
In (ii) and (iii) you are missing another solution(i)By inspection x=-1 is a factor of P(x), by polynomial division we get the required result
(ii) For 3 real roots, we set the discriminant >=0 and we get B<=-A
(iii) The region is bounded by lines B>=-1, A>=-1, B<=-A
Yep, however B=> 3A and -A => B are only the first half of hte solutionsB>=3A?
Even for the average 4U student, that would be too difficult.
Not sure if I'm overdoing it with the guiding thing now
Yea something like thatso careless tonight sigh..
(ii) First case: B>=3A, B<=-A (both positive)
Second Case: B>=-A, B<=3A (both negative)
(iii)First Case: A>=-1, B>=-1, B>=3A, B<=-A
Second Case: A<=1, B<=1, B>=-A, B<=3A
Ah well, okEven for the average 4U student, that would be too difficult.
I'm probably going to sound stupid, but ok z=x+i.
Yep the line y=1 (When using x,y we are talking about reala numbers so it doesn't make sense to say y=i)I'm probably going to sound stupid, but ok z=x+i.
Locus of z=x+i, is that just all of the plane? 0.0 Liek all of x values right, so adding i doesn't really do much...
Man i need to revise my complex.
K, sub any x
Then will lie in y=i. So just line y=i? Horizontal line at i?
Ok, z^2=(x+i)^2Yep the line y=1 (When using x,y we are talking about reala numbers so it doesn't make sense to say y=i)
Think of it this way
z is some number on the complex plane, where the real part can be anything (x is real), but its imaginary part has to always be 1, this is of course y=1
Try the second one
SoOk, z^2=(x+i)^2
Using the same logic, it is the line squared so a parabola with vertex at y=1?