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Namu, you lost one of the solutions in the middle of your working out! If 4a^2 = b^2, then b = 2a or b=-2a. Btw, the question specifies that a and b are integers (which is why you need to look at the other solution.)

You could solve these questions algebraically, but I think a geometric proof would probably be more enlightening. We should think of z and w as 2D vectors lying in the Argand plane. If (z+w) / (z-w) is pure imaginary, then arg(z+w) – arg(z-w) = +/- pi/2. Let |z| = |w| = r. (I think we...

Thanks, everyone. I do (did?) Science Education, with a major in mathematics. I'm not actually going very far at all, as I am probably starting a masters (part time) in applied maths next year. (So much more maths to learn, so little time...)

I don't care if they are late - I have just checked my results on MyUNSW, and they've changed my graduation status to : Awarded! Now I get to graduate! So happy....

About the 2nd Q, the way I was thinking is this: You can see all but one face of each tetrahedra (the one that is facing down) so if you can only see 1 green one, then clearly 2 of the tetrahedra must land with green facing down. Of the remaining tetrahedron, it must have green facing up, but...

Firstly, general maths is a pathetic subject, so if you can't cope with/don't want to cope with 2 unit, why not just drop maths altogether and do something else? Secondly, the perfectionism thing is going to die at uni anyway (because if you go to one that is even half way decent you are going...

About a week ago, I got this e-mail from uni (UNSW) about changes to the structure of the BSc/BEd course. Since I've just finished, it is all a bit irrelevant to me, but anyway... Apparently the course had to be restructured due to increased requirements from NSWIT. In sum, they are...

The Faculty of Arts and Social Sciences is in the middle of a restructure - they've cut the number of schools in the faculty from 12 to 5. This is probably why you are having trouble finding any info about them. I don't know how much the restructure will affect the undergraduate program. My...

I'll show you how to do the first one. I think you'll see how to do the second one yourself then. We are given f(x-1) = x^2 -1 (1) Let y=x-1. Then x=y+1, and subsituting into (1), we get f(y) = (y+1)^2 -1 =y^2 +2y +1 -1 = y^2 +2y But y is just a "dummy variable," so we can really call it...

You can't prove it as it is not necessarily true. Besides, you don't need to prove it: Let angle ACB = x, as suggested by Iyounamu Then angle ACB = angle CBA = x (base angles in isosceles triangle ACB) Also, angle AEB = angle ACB = x (angles subtending the same segment) Furthermore, angle...

And there precisely is your tacit assumption. You haven't proved that these triangles ARE congruent. We know that AB=AC and that OA is common to both triangles, but we need to know either the included angles, or the length of the other sides before we can say that they are congruent.

Try doing what I just suggested - draw the diagram with E in between A and B. Then you will see that the point of intersetion between BC and AE does not even lie within the circle.

No, that is not the case. If he proved that AE is a diameter then he is making some tacit assumptions that are not in the question. Basically you can locate E anywhere on the circle you like, so long as you also move D so that AD and BE are parallel. You can in fact draw the diagram with E...

There is no reason why AE should be a diameter. You have to join BC. Then use properties of cyclic quads (among other things) to show that angles CDA and DAE are supplementary.

I have met several people at uni doing double degrees in maths and law, and I have often wondered the same thing. I think that probably there aren't many areas of law where mathematics is directly applicable, but maybe the critical thinking and problem solving skills that you aquire while doing...

Don't be put off by some of the comments in this thread - if you really love mathematics, and you have some talent for the subject, then that's what you should study. I don't know why you would want to do a BA in it though (unless there are some other arts subjects you are interested in...

How about the function y = x^2 if x is rational, 0 if x is not rational. I think that would be differentiable at x=0 and nowhere else. Analysis is full of pathological counter examples.

OK, so kx(x+4)=(x-1)(x+2) Then (k-1)x^2 + (4k-1)x +2=0 so the discriminant is (4k-1)^2 - 8(k-1)= 16k^2 -16k +9. If we complete the square, we get 16k^2 -16k +9 = (4k - 2)^2 +5 > 0 for all k.

Work out the discriminant, which is a quadratic in k. Then show that this quadratic is positive definite.

Sydney girls doesn't offer general maths, so it's 2u, 3u, or no maths at all.