help wit a few questions from cambridge (1 Viewer)

[prsce24]

This is in exercise 2.2 question 15.

15.a) Obtain in the form a + ib the roots of the equation x^2 + 2x + 3 = 0. Find the modulus and argument of each root and represent the roots on an Argand diagram by the points A and B.

b) Let H and K be the points representing the roots of x^2 + 2px + q =0, where p and q are real and p^2 < q. Find the algebraic relation satisfied by p and q in each of the following cases: i) Angle HOK is a right angle. ii) A,B,H, and K are equidistant from O.

answer a) too easy b) i)2p^2 - q=0 ii) q=3

for me in b) i), i got square root(2p^2-q), but i dont know how to get rid of square and end up with equation 2p^2-q=0.

 

[Rorix]

man this question is hard
it could be because i dont see parts i or ii

 

[prsce24]

Rorix what do u mean u cant see

anyways what was that buchanan guy's site again? my link just died

 

[prsce24]

hmm wtf, i wrote the full question , but it doesnt load up.
hmm could you guys who have the book please solve it for me thanx

 

[Rorix]

*points at time differences between my post and the edit*

 

[McLake]

x^2 + 2px + q = 0

so x = [-2p +/- sqrt (4p^2 - 4*1*q)]/[2*1]
x = -2p +/- 2sqrt(p^2 - q)/2
x = -p +/- sqrt(p^2 - q)
so x1 = sqrt(p^2 - q) - p
and x2 = - sqrt(p^2 - q) - p

now m1*m2 = -1
so (sqrt(p^2 - q) - p) * (- sqrt(p^2 - q) - p) = -1
so -(p^2 - q) - p^2 = -1
so 2p^2 - q =1

or something ...

 

[steverulz55]

(i)x = -p +/- rt(p^2 - q)
= -p +/- rt(q-p^2)i (since p^2 - q <0)

now, arg (H) - arg (K) = pi/2
but arg (H) = -arg(K) (since H is conj of K)
=> 2arg (H) = pi/2
=> arg (H) = pi/4
=> rt (q-p^2) / -p = 1
=> q-p^2 = p^2
=> 2p^2 - q = 0 as required

(ii) H. K = q
=> |H||K| = |q|
but |H|= |K| = rt3 (A, B, H, K equidistant from O)
therefore |q| = |3|
=> |q| = 3
hmmm....lemme see if q is strictly > 0

ahh yes, 2p^2 - q = 0, 2p^2 = q
therefore q >= 0 (p and q are real)
hence q = 3