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O-Level Additional Math

PSLE marks the graduation of Primary school students and their entry into Secondary schools as teenagers. Discuss all issues about Secondary schooling here.
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PiggyLalala
KiasuGrandMaster
KiasuGrandMaster
Posts: 4053
Joined: Tue May 24,
Total Likes:23

Re: Secondary School Mathematics

Post by PiggyLalala » Sat Jun 18, 2011 11:35 am

suiyuan wrote:Secondary Two Maths

Hi

Solve this equation
(x-2)^2 = 3(2x-3)(2-x)

(x-2)^2 means (x-2) to the power of 2.

Please help.

Thank you.
(x-2)^2 = 3(2x-3)(2-x)
(x-2)^2 = -3(2x-3)(x-2)
(x-2)^2 + 3(2x-3)(x-2) = 0 ( note: (2-x) = -(x-2)
(x-2)[(x-2) + 3(2x-3)] = 0
(x-2)[ x- 2 + 6x -9] = 0
(x-2)( 7x - 11) = 0
x- 2 = 0 or 7x - 11 = 0
x = 2 or 11/7
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FrekiWang
GreenBelt
GreenBelt
Posts: 119
Joined: Tue Mar 08,

Re: Secondary School Mathematics

Post by FrekiWang » Sun Jun 19, 2011 2:04 pm

suiyuan wrote:Secondary Two Maths

Hi

Solve this equation
(x-2)^2 = 3(2x-3)(2-x)

(x-2)^2 means (x-2) to the power of 2.

Please help.

Thank you.
(x-2)^2-3(2x-3)(2-x)=0
(x-2)^2+3(2x-3)(x-2)=0 (note that -(2-x)=+(x-2))
(x-2)[(x-2)+3(2x-3)]=0
(x-2)(7x-11)=0
x=2 or x=11/7
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JadeDry
OrangeBelt
OrangeBelt
Posts: 30
Joined: Thu Oct 28,

Re: Secondary School Mathematics

Post by JadeDry » Sun Jun 19, 2011 10:36 pm

hometutors.sg wrote:
JadeDry wrote:I currently use "New Syllabus Mathematics" (8th Grade) books, and would appreciate if you could recommend additional good quality publications.

Thanks in advance.
A good tutor / mentor would be able to select what is good for the student.
Thank you for your time.

I have just finished:

Secondary one Mathematics Tutor: 1A and 1B.

Shinglee New Syllabus Mathematics 6th edition 1

and Singlee Mathematics Workbook 1.

I am homeschooled and have been studying singapore math from 1st grade, since my parents are my teachers I would like to see what you people think I should get for 8th Grade.

Thank you in advance for your assistance.
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suiyuan
OrangeBelt
OrangeBelt
Posts: 40
Joined: Fri Sep 12,

Re: Secondary School Mathematics

Post by suiyuan » Wed Jun 22, 2011 8:33 am

Sec Two Maths

1) Given that x^2 = 5x – 1, find the value of 2x^4 + 2/x^4

2) a)Factorise 2x^10 – 13x^5 -15 completely.
b)Factorise 10x^10y – 65x^5y-75y

Note a) ^2 means to the power of 2 b) 2/x^4 means 2 divided by x^4

Please help.

Thank you
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CoffeeCat
BrownBelt
BrownBelt
Posts: 512
Joined: Sat Feb 27,

Re: Secondary School Mathematics

Post by CoffeeCat » Wed Jun 22, 2011 1:03 pm

suiyuan wrote:Sec Two Maths

1) Given that x^2 = 5x – 1, find the value of 2x^4 + 2/x^4

2) a)Factorise 2x^10 – 13x^5 -15 completely.
b)Factorise 10x^10y – 65x^5y-75y

Note a) ^2 means to the power of 2 b) 2/x^4 means 2 divided by x^4

Please help.

Thank you
Do you happen to be from some IP school or sth?
Your teacher probably koped qns 1 from SMO.

Qns 1)
Lousy Method : solve for x, substitute and torture yourself in the process.
Proper Method : divide the quadratic equation by x throughout to get
x + 1/x = 5
You will get x^2 + 1/x^2 = 23 by squaring the above.
Repeat the process one more time.
Final ans should be 1054 if I have no calculation error.

Qns 2a)
Notice 10 is twice that of 5. hence imagine you substitute y=x^5. You will reduce the problem to factorising a quadratic expression.

Qns 2b)
Ambiguous.......Please use brackets.
10x^10y means 10x^(10y) or 10(x^10)y

Since it's part b, i will assume the latter.
First extract common factors from all terms, which is just 5y. After that what you have should be part a.
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JadeDry
OrangeBelt
OrangeBelt
Posts: 30
Joined: Thu Oct 28,

Re: Secondary School Mathematics

Post by JadeDry » Wed Jun 22, 2011 4:23 pm

Hello again, I have a problem that I am unsure about:

Simplify the following expression:

3/4((x squared)(y)) X 5/9((x)(z cubed)) / ((20/9)(x quadrupled)(y squared)(z))

These are my workings:
3/4((x squared)(y)) X 5/9((x)(z cubed))=

15/36((x cubed)(y)(z cubed))


15/36((x cubed)(y)(z cubed)) / ((20/9)(x quadrupled)(y squared)(z))=

3(Z squared) / 16(xy)

Are my calculations correct?

On a side note, is Z cubed/ Z= Z squared or Z cubed?

Thanks in advance.
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suiyuan
OrangeBelt
OrangeBelt
Posts: 40
Joined: Fri Sep 12,

Re: Secondary School Mathematics

Post by suiyuan » Wed Jun 22, 2011 4:41 pm

Sec Two Maths

Find the smallest possible value of the expression 3x^2 + 27y^2+5z^2-18xy-30z+125.

Note a) ^2 means to the power of 2

Please help.

Thank you
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FrekiWang
GreenBelt
GreenBelt
Posts: 119
Joined: Tue Mar 08,

Re: Secondary School Mathematics

Post by FrekiWang » Wed Jun 22, 2011 7:51 pm

suiyuan wrote:Sec Two Maths

Find the smallest possible value of the expression 3x^2 + 27y^2+5z^2-18xy-30z+125.

Note a) ^2 means to the power of 2

Please help.

Thank you
Note that
3x^2-18xy+27y^2=3(x^2-6xy+9y^2)=3(x-3y)^2
5z^2-30z=5(z^2-6z)=5(z^2-6z+9-9)=5(z^2-6z+9)-45=5(z-3)^2-45

So the original express is
(3x^2-18xy+27y^2)+(5z^2-30z)+125
=3(x-3y)^2+5(z-3)^2-45+125
=3(x-3y)^2+5(z-3)^2+80

Since the square of any number is non-negative (0 is the smallest possible), the minimum value of the whole expression is 80.

PS: A bit too much for a sec 2 student, probably from one of the top schools?
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suiyuan
OrangeBelt
OrangeBelt
Posts: 40
Joined: Fri Sep 12,

Re: Secondary School Mathematics

Post by suiyuan » Wed Jun 22, 2011 8:26 pm

FrekiWang wrote: 5z^2-30z=5(z^2-6z)=5(z^2-6z+9-9)=5(z^2-6z+9)-45=5(z-3)^2-45
Hi

Please explain in more details how you get this.

5z^2-30z=5(z^2-6z)=5(z^2-6z+9-9)=5(z^2-6z+9)-45=5(z-3)^2-45

Thank you
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CoffeeCat
BrownBelt
BrownBelt
Posts: 512
Joined: Sat Feb 27,

Re: Secondary School Mathematics

Post by CoffeeCat » Thu Jun 23, 2011 1:36 am

suiyuan wrote:
FrekiWang wrote: 5z^2-30z=5(z^2-6z)=5(z^2-6z+9-9)=5(z^2-6z+9)-45=5(z-3)^2-45
Hi

Please explain in more details how you get this.

5z^2-30z=5(z^2-6z)=5(z^2-6z+9-9)=5(z^2-6z+9)-45=5(z-3)^2-45

Thank you
He is basically completing the square. If your teacher posed that question, then he is assuming that you all know how to complete the square. Completing the square is a simple technique for solving simpler "find minimum of value of a simple quadratic looking expression".

You should be able to find completing the square in ur sec2 textbook.
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