O-Level Additional Math

[mum_sugoku]

Xiao Hu:

Hi mum_sugoku,
Thanks so much! Yes, on hindsight, now I know I should have proceeded by using the congruent triangle results from part a.

I really appreciate your help,
Xiao Hu.

You are welcome . BTW if I'm not wrong, when question asks you to \"hence\" solve the subsequent problem, you are expected to make use of previous result to work out the solution, else you may be penalised for not following instruction!!!

[Herbie]

How to solve 499^2

[koguma]

Herbie:

How to solve 499^2

Assume 499^2 = 499 * 499,
Since 499 = (500-1)
so solve (500-1)^2 using (x-y)^2 formulae in algebra topic

[Herbie]

koguma:

Herbie:

How to solve 499^2

Assume 499^2 = 499 * 499,
Since 499 = (500-1)
so solve (500-1)^2 using (x-y)^2 formulae in algebra topic

Thanks! Thanks

[CoffeeCat]

Belle2011:

Hello,

May I ask where can I get books/resources on math olympiad questions suitable for lower sec?
Thank-you.

cheers,
Belle.

There isn't much introductory guides on maths olympiad for lower sec unlike for primary levels...

A good one to impart the concepts will be
The art of problem solving volume 1 & 2
http://www.artofproblemsolving.com/Store/index.php


A local publication, the math olympiad series
http://sms.math.nus.edu.sg/Publication/Publication.aspx
will provide ample questions with solutions, the only barrier to learning is that it may assume the reader has certain prior knowledge (which the above recommendation might help to a certain extent).

[red rose]

Would appreciate help on part (b) of this question. I typed out the whole question for clarity.


The equation x^2/2-x-3/2=0 has roots a and b.

(a) Without solving for a and b, find the value(s) of
(i) a-b
(ii) 1/b^2-1/a^2

(b) If m=a-b, n=1/b^2 - 1/a^2, form all the possible quadratic equations with roots m and n (4 marks)

Thanks in advance!

[CoffeeCat]

red rose:

Would appreciate help on part (b) of this question. I typed out the whole question for clarity.


The equation x^2/2-x-3/2=0 has roots a and b.

(a) Without solving for a and b, find the value(s) of
(i) a-b
(ii) 1/b^2-1/a^2

(b) If m=a-b, n=1/b^2 - 1/a^2, form all the possible quadratic equations with roots m and n (4 marks)

Thanks in advance!

Hmm is (i) reallie a-b and not a+b?
The idea is coefficients of a quadratic equation can be expressed in terms of the roots.
So the strategy is to express every algebraic expressions in terms of a+b and ab (or their derivations, for e.g. part (ii) can use part (i) ).

(x-a)(x-b) = x^2 - (a+b)x + ab
From the eqn, x^2 - 2x - 3 = 0
a+b = 2
ab = -3

We can get a-b from sqrt of (a+b)^2 -4*ab
hence a-b = sqrt (4 + 12) = 4.
This qns is quite vague as it assumes a > b.


Similarly,
1/b^2 - 1/a^2 = (a^2 - b^2)/ (ab)^2
a^2 - b^2 = (a-b)(a+b) = 2*4 = 8
hence the ans is 8/(-3)^2 = 8/9

[JadeDry]

Hello,


I would appreciate assistance with the following question:

3 divided by (x-3) + (1 divided by (x+1)) divided by (2x divided by (x-3))

Is the answer:

(7(x squared) +9)/((2x)(x+1)(x-3))

or

2x/(x+1) ?

Thanks in advance.

[mum_sugoku]

JadeDry:

Hello,


I would appreciate assistance with the following question:

3 divided by (x-3) + (1 divided by (x+1)) divided by (2x divided by (x-3))

Is the answer:

(7(x squared) +9)/((2x)(x+1)(x-3))

or

2x/(x+1) ?

Thanks in advance.

I've got the first one

[ADoc]

red rose:

Would appreciate help on part (b) of this question. I typed out the whole question for clarity.


The equation x^2/2-x-3/2=0 has roots a and b.

(a) Without solving for a and b, find the value(s) of
(i) a-b
(ii) 1/b^2-1/a^2

(b) If m=a-b, n=1/b^2 - 1/a^2, form all the possible quadratic equations with roots m and n (4 marks)

Thanks in advance!

Hi. It's been a while since I last posted on KS.
Guess this may be too late for your question. Nevertheless...

(b) Simply substituting all the possible values from a (i) & (ii)
which are (i) = +/- 4 (since (a-b)^2 = 16, therefore a - b = +/-4);
(ii) = +/- 8/9

so you'll obtain 2^2 = 4 possibilities in total. I'll save some cyber space by not typing them out.

By the way, just some useful identities when solving questions involving product and sum of roots that are suitable for Sec level especially for some of the IP schools such as DHS in particular.

a^2 + b^2 = (a + b)^2 - 2ab

(a - b)^2 = (a + b)^2 - 4ab

a^4 - b^4 = (a^2 + b^2)(a + b)(a - b)

a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2

a^3 - b^3 = (a - b) [(a + b)^2 - ab]

a^3 + b^3 = (a + b) [(a + b)^2 - 3ab]

cheers
Eugene
Full-time tutor / Part-time Lecturer
Specialising in IP group tuitions for DHS & VS.