4U Cambridge solutions (2 Viewers)

[Grey Council]

Here goes. As i said, i've just started 4u maths, and the complex numbers topic seems crazy to me. heehee, maybe i'm just shit.

Anyway, here is the question. Its question 5, from exercise 2.1 of the cambridge textbook. As you can see, i'm stuck fairly early on.

let m = theta (that wierd greek sign)
let n = theta bar (thera with that bar thing on top)

am^2 + bm + c = 0, where a, b, c, are real and x is a complex number. Show that an^2 + bn + c = 0

Deduce that if m is a non-real root of ax^2 + bx + c = 0, where a, b and c are real, then n is the other root of this quadratic equation.

I dont know how to approach this. That why i was after the worked solutions. I mean, what Dr Buchanan said is true, but how am i supposed to go about doing a question i dont know how to approach? Usually i'd ask a teacher, but i dont have a teacher over the holidays.

 

[Wohzazz]

I think i done this question be4

an^2+bn+c = (am^2+bm+c) bar = (0) bar =0

Use the fact that for complex quadratic equations, roots occur in complex conjugates and m and n are complex conjugate that satisfy previous equation to deduce n is the other root

 

[korry]

so, is there absolutely nowhere where we can get solutions for cambridge?

 

[Grey Council]

oh, Wohzazz?

an^2+bn+c = (am^2+bm+c) bar = (0) bar =0

How'd you get from the first statement to the second one? Isnt it asking us to PROVE that this is true? Or is it a rule that if Theta is a root of a quadratic equation, (theta)bar is the other root? Hmm, if its a rule, and i can just state it, then it makes it a hell of a lot easier. But the thing is, this question is asking us to show/deduce this. :S

 

[Rorix]

There is a rule that for polynomials with real coefficients, if a+bi is a root then a-bi is also a root.

You can prove it using the properties of conjugates (for the following, consider [x] as meaning the conjugate of x). We'll also assume a quadratic, but you can prove the general case using the same method.

ax^2 + bx + c = 0
[ax^2 + bx + c] = [0]
[a + b] = [a] + and [0]=0
so [ax^2] + [bx] + [c] = 0
[ab] = a when a is real - the coefficents are real so
a[x^2] + b[x] + [c] = 0
[x^2] = [x]^2, [c] = c
So a[x]^2 + b[x] + c = 0
i.e. the conjugate of a root is also a root of the quadratic

 

[Grey Council]

OOO! Brilliant. You've hit the spot, Rorix. lol, this is prolly an easy question, and i got stuck. gee, i'm gonna die in 4u maths. :|

 

[Wohzazz]

Originally posted by GuardiaN
oh, Wohzazz?


How'd you get from the first statement to the second one? Isnt it asking us to PROVE that this is true? Or is it a rule that if Theta is a root of a quadratic equation, (theta)bar is the other root? Hmm, if its a rule, and i can just state it, then it makes it a hell of a lot easier. But the thing is, this question is asking us to show/deduce this. :S

sorry, i've jumped too many steps, its what Rorix did, use all the conjugate results for multiplication and addition to get it

 

[wildtiger]

hi all - i have a problem with a question in cambridge's CONICS chapter on page 92, would anyone be able to help me out? Thanks!

I'm stuck on question 7b)

7. P(a cos X, b sin X) and Q (a cos Y, b sin Y) lie on the ellipse x^2/a^2 + y^2/b^2 =1.

a) If PQ subtends a right angle at (0,0), show that tan A tan B = - a^2/b^2

b) If PQ subjects a right angle at (a, 0) show that tan (A/2) tan (B/2) = - b^2/a^2

 

[Grey Council]

I advise you to start a new thread. Don't follow on here, the topic of this thread gets some people annoying. (kinda).

 

[J0n]

Maybe all these people are posting questions to show how much they need the "4U Cambridge solutions" - or they think that the solutions are in this thread, find out they're not and just post there anyway.

 

[wildtiger]

haha ok how do i move the thread thing? sorry im heaps n00b at this

 

[J0n]

I don't think you can move it yourself - you'll have to ask a mod to do it for you.

 

[Wohzazz]

Are you saying people who get good marks for maths but aren't natural math geniuses shouldn't? So rather than trying to find the answer if your proficient at 4U, those of us who can't find the answers ourselves, accept failure and declare ourselves as incompetent to undertake the course??!?

 

[abdooooo!!!]

i do... cambridge solutions all of them... but i need to trade it with fitz's solution...

or you can just buy them from sami

 

[coroneosrules]

I have Fitz Chapter 35 solutions if anyone wants them.

 

[:: ck ::]

me me me!

 

[wildtiger]

i got them off a friend last week... but now that i got them, i havent even unzipped them yet

 

[abdooooo!!!]

not this again... TRASH THEM!!!

edit: you deleting p0st is really annoying me... why are you so scared of others ie teachers or sami reading it?

 

[Wohzazz]

who's this sami you're mentioning abdhooooo!! How did you get Fitz answers?

 

[freaking_out]

Originally posted by abdooooo!!!
....edit: you deleting p0st is really annoying me... why are you so scared of others ie teachers or sami reading it?

why don't u just quoting him.