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[pwoh]

I've done parts a, b, c but I have no idea about (d).

Let be a root of the quartic polynomial

(a) Show that

(b) Show that is also a root of

(c) With , show that becomes

(d) For certain values of A and B, P(x) has no real roots.
Let D be the region of the AB plane where P(x) has no real roots and .
The region D is shaded in the figure. Specify the bounding straight line segment and curved segment c. Determine the coordinates of T.

 

[Affinity]

if you solve for u you get

u = [- A +- sqrt(A^2 - 4(B-2))] / 2

besides the usual constraint that the discriminant is positive, you have an additional condition that would preclude some roots : |u| > 2 (think why)

so the curve is A^2 - 4(B-2) = 0

you can get he straight line from the other condition

From memory this is from the 1993 or 1994 hsc .

 

[khorne]

|u| >= 2, as x = 1, u =2, all other values of x result in u >2,...is that what it should be, or am I mistaken.

 

[Affinity]

something along those lines (sqrt(|x|) - 1/sqrt(|x|) )^2 >= 0 then expand tidy up and consider the case when x < 0

 

[pwoh]

If I use (sqrt(|x|) - 1/sqrt(|x|) )^2 and compare that with [- A +- sqrt(A^2 - 4(B-2))] / 2, I get something like B =< 2A -2.
But that doesn't seem right - then B =< -2, unless I've mixed up a sign somewhere.

Also, if x not real, then I'm not sure how that comparison works. Taking |x| would be taking the modulus?

 

[Affinity]

1. Don't care about complex u's -> looking at real x's hence u's

[- A +- sqrt(A^2 - 4(B-2))] / 2


now since A is positive, the negative sign with the square root results in a bigger absolute value

if there's a real root for x, then |u| must be >= 2

want |[- A - sqrt(A^2 - 4(B-2))] / 2| >= 2

equivalent to: A + sqrt(A^2 - 4(B-2)) >= 4

equivalent to: sqrt(A^2 - 4(B-2)) >= 4 - A

if A >= 4 then it holds/the constraint does not matter

otherwise the inequality is equivalent to

A^2 - 4(B-2) >= (4-A)^2
-4(B-2) >= -8A + 16
B<= 2A - 4
B <= 2(A-1)

hence if B > 2(A-1), A <4, (A>=0) there will not be a real solution

so the straight line is B = 2(A-1) (up to A < 4 !!!)

and the curved line is A^2 - 4(B-2) >= 0
or B = A^2/4 + 2

8(A-1) = A^2 + 8
A^2 - 8A + 16 = 0
A = 4
(which means the curve does indeed take over from the point where the line ceases to place a limitation)

so T is (4,6)

Make sure you understand how the constraints work

 

[mirakon]

Thanks from me as well as I was doing this question just the other day and also got confused for part d.

 

[pwoh]

Thanks!