O-Level Additional Math

[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.

[ADoc]

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.

hihi! First answer too, assuming the correct order of operations from what you typed above.

cheers
eugene

[MathIzzzFun]

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

Hi

it is faster to use (x+1)(x-1) = x^2-1 --> x^2 = (x+1)(x-1)+1
499^2 = 500x 498 +1

cheers.

[bran36]

Hi, as the exams are near, I'm preparing for my papers. However, while doing my assessment book, there's this question I do not understand.

It's an inequality question. I don't understand why must the sign change. Perhaps anyone can enlighten me? Many thanks!!

My work:
http://i44.tinypic.com/2gxfqdv.jpg

The Answer:
http://i44.tinypic.com/35irdbp.jpg

[mum_sugoku]

bran36:

Hi, as the exams are near, I'm preparing for my papers. However, while doing my assessment book, there's this question I do not understand.

It's an inequality question. I don't understand why must the sign change. Perhaps anyone can enlighten me? Many thanks!!

My work:
http://i44.tinypic.com/2gxfqdv.jpg

The Answer:
http://i44.tinypic.com/35irdbp.jpg

(Actually it's quite hard to explain without using drawings but let me try )

First of all, the eqn you have is a quadratic eqn (ie eqn in the form of ax^2 + bx + c) which, if you were to draw a graph of the eqn, it's either a U shape curve (if a is +ve), or an inverted U (if a is -ve).

Next, if we look at your eqn x^2 -4x +3 >=0, or (x-3)(x-1)>=0
since a (ie coef of x^2) is = +ve, it's a U shape curve which cuts the x-axis (ie y=0) at x=3 and x=1

What it means is that when x is between 1 and 3 (ie 1<x<3) the curve falls below the x-axis (ie eqn is -ve), and when x is <1 and when x is >3, the curve rises above the x-axis (ie eqn is +ve).

Hence for eqn to be >= 0 (ie +ve), x<1 or x>3.

If you can't understand the above explanation, alternatively, just bear in mind that for quadratic inequality eqn, say (x-3)(x-1)>0, ans is always either in the form

1<x<3, or

x<1, or x>3

To find out which is the correct ans, one way is to pick a number between 1 and 3, say x=2, and insert into the eqn, and you'll get (2-3)(2-1) = -1, which is <0. Since question ask for >=0, correct ans has to be the second one, ie x<1 or x>3.

[bran36]

mum_sugoku:

bran36:

Hi, as the exams are near, I'm preparing for my papers. However, while doing my assessment book, there's this question I do not understand.

It's an inequality question. I don't understand why must the sign change. Perhaps anyone can enlighten me? Many thanks!!

My work:
http://i44.tinypic.com/2gxfqdv.jpg

The Answer:
http://i44.tinypic.com/35irdbp.jpg

(Actually it's quite hard to explain without using drawings but let me try )

First of all, the eqn you have is a quadratic eqn (ie eqn in the form of ax^2 + bx + c) which, if you were to draw a graph of the eqn, it's either a U shape curve (if a is +ve), or an inverted U (if a is -ve).

Next, if we look at your eqn x^2 -4x +3 >=0, or (x-3)(x-1)>=0
since a (ie coef of x^2) is = +ve, it's a U shape curve which cuts the x-axis (ie y=0) at x=3 and x=1

What it means is that when x is between 1 and 3 (ie 1<x<3) the curve falls below the x-axis (ie eqn is -ve), and when x is <1 and when x is >3, the curve rises above the x-axis (ie eqn is +ve).

Hence for eqn to be >= 0 (ie +ve), x<1 or x>3.

If you can't understand the above explanation, alternatively, just bear in mind that for quadratic inequality eqn, say (x-3)(x-1)>0, ans is always either in the form

1<x<3, or

x<1, or x>3

To find out which is the correct ans, one way is to pick a number between 1 and 3, say x=2, and insert into the eqn, and you'll get (2-3)(2-1) = -1, which is <0. Since question ask for >=0, correct ans has to be the second one, ie x<1 or x>3.

Hi thanks a lot for your help! However, how do you which value of x is > or <? I don't really understand this part. Thanks again!