Lesson Summary - 20th Feb (Wednesday)

We started a bit on Logarithm and we learnt how to convert from Index to Log form, and Log to index form.

An example of converting index to log form:
10^x=50 (Index form)
Log1050=x (Log form)

We also learnt the relationship between Log and Indices:
Firstly, Log and Indices are inverse of each other (n^7=x / Lognx=7).

It can be seen from this graph:

To solve Log, sometimes it is easier to convert it to Index form then solve because it is easier. 
An example question:

Or in a Log equation: 
By Comparing Coefficient,

For the paper this coming Tuesday, remember to work smart and do not get stuck at one question as time is a factor in the paper. DO NOT over-read the question and over-answer the question as well. 

For questions that have the word 'Hence', you MUST use the solution from the previous part of the same question.  'Otherwise' means you can either resolve the question or use the solution from the previous part of the same question. 

Plotting of Quadratic Graphs (Revision):
General form of Quadratic: ax^2+bx+c

Take a look at this graph:

At the blue line (x^2), there is no coefficient. 
-When the coefficient is bigger than 1 it will make the graph narrower, and if the coefficient is less than 1 it will make the graph wider. When the coefficient is negative, it will invert the graph. 

Example question:
The two x intercepts are 4 and -2, and the y intercept is -8. 

Sample Questions for Quadratic Graphs:
What is given in the question? 
1. Table of values (DON'T spend too much time trying to draw a nice nice table)
2. Scale

What must you show? 
1. Crosses on points
2. Table
3. Axis
4. Labeling of equation 

1. Line of symmetry (Turning point) 
2. Gradient of Tangent (Tangent at x=1) ( Use the formula (y1-y2)/(x1-x2) )
3. Plotting of the graph
4. Find the value of x when y = ____. 
5. Plotting of a linear graph and find the intersection. 

Because of the late posting (Sorry), I will post up a clearer summary of Surds

Surds are bascially numbers left in 'square root form' or 'cube root form', etcetc, and are irrational. We leave them it this form because writing them out in decimal form would not be a very good way of expressing. 

Firstly, there are a few operations and rules that we must know of, just like in Indices. 

Operations of Surds:
1. √a x b = b√a
2. √a (divide) b = √a (over) b

Rules: (When a rational number is multiplied or divided by a surd, the result will be a surd.)
3. √a x √b = √ab (√3 x √4 = √12)
4. √a x √a = √a^2 = a (√8 x √8 = 8)
5. √a (divide) √b = √a (over) √b = √a/b (√20 / √4 = √5)

NOTE: √(a+b) ≠ √a + √b and √(a-b) ≠ √a - √b
6. x√a + z√a = (x+z)√a (IT'S NOT (x+z) ROOT OF A) (5√2 + 3√2 = 8√2)
7. x√a - z√a = (x-z)√a (IT'S NOT (x-z) ROOT OF A) (5√2 - 3√2 = 2√2)

Rationalizing of Surds:

√a + √b and √a - √b are conjugate surds. This means that the product of them is a rational number. 

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

Note: a-b is a rational number, thus we can use conjugate surds to rationalize the denominator. 

Example question: 

Solving equations in Surds:

1-Equations for surds can be solved by squaring both sides, to remove the square root. 
2-Sometimes we have to square both sides TWICE when the equation involves sum or difference of 2 or more surds. 

Example Question: 

Once again sorry for the late posting .

Friday 22 February - Lesson summary

Today's lesson was basically on laws of logarithm lol. Nothing much else, we used the graphic calculator to prove them and stuff. So well here it is~

(Ps: Sorry for my lousy handwriting)

Applications OF Exponential and Logarithm










TI84plus : Logarithms - Change of Base

TI84plus : Logarithms - Change of Base

How to use TI84 to Calculate Log of a different Base

How to Programme Change of Base

TI84 Solving Quadratic

Solving Quadratic Equation Using TI8plus programme

18/2/2013 Summery of the lesson



A-Math Question types: 
1.“Polynomial and identities”
2.“Factor and remainder theorm” and “Division of Polynomials”
3.Cubic equation and solving cubic equation.
4.Partial Fraction
5. Combinations of the topics.

E-math question types:
*refer to the post about the tested topics.

*Find the equation

  1. Quadratic
    1. 2 real and different roots
      1.   x = -4 or x = 2
      2. (x=4)(x-2) = 0
        1. y = A(x+4)(x-2)
        2. in y-intercept = -3, x = 0
          1. -3 = A(0+4)(0-2)
            1. -3 = -8A
              1. A = 3/8
              2. Therefore, y = 3/8 (x+4)(x-2)

Drawing Graph

Given : Scale, Table

  • Draw graph using pencil
  • Label the graph
  • Values in ink
  • if exceeds the range given, (e.g. -3< x <6) marks will be penalized 
  • Plotting x  ------ (2-3marks)
  • if x = ____, find y / y = _____, find x ------- (2marks)
  • Line of symmetry - (1mark), Equation of line of symmetry - (1mark)
  • Tangent at x=1 

y= x^2 +x -2
hence, solve x^2 +x = 7.
x^2 +x-2 = 5
y = 5

y = x^2 =x -2  ----- (1)
y=5 ------ (2)
x^2 +x = 7


Using TI 84Plus and simple programme


Programming QUADRATIC INTO TI84 

Study the video provided for clarity of concept and greater appreciation of TI84plus.

Task 1: Simulate the programme into your TI84plus. Call it QUAD
Task 2: Extend your learning by considering the case when the answers are not Real ie. Imaginary

Logarithm in the Real World

Take a Break - Math Exponent

Lesson Summary 15/2/2013





♥♥ Valentines Day Math Lesson Summary ♥♥

Hi 305. ♥♥ Happy Valentines Day ♥♥
Enjoy your lesson summary! ☃

Todays notes(Darryl Lam)

This is the scribe for today's lesson. 


QD Question 1 (Hao En)

1) Expansion of 2(2x^2+3) to make life easier for yourself (at least for me). Or you could choose not to. 

2) Since 4x^2+6 is a quadratic expression, the accompanying "unknowns" should be in linear form. Hence the Ax+B. As x-1 is linear, the unknown is a constant, giving us C. 

3) Multiplying both sides by the common denominator in order for us to be able to solve the question.

4) The substitution method is used here, as it is much more easier than the Comparing Coefficient method. As the expression on the left side is 5, at one glance it should be easy to tell that if we were to use the comparing coefficient method, solving the unknowns using simultaneous equations would be required. 

In this case, by substituting x=1, we are able to remove A and B from the equation, leaving us with only C.

5) Looking at the equation, as A is the only one with a x attached to it, we are able to tell that by substituting x=0 we would be able to remove the A. By removing A and using the answer for C we found out earlier, it would effectively leave us with only B. 

6) As A is the only one left, any value of x besides the ones that remove A from the equation will work. For simplicity's sake, -1 is used here.

7) Subbing back the unknowns into the partial fractions, and then simplifying them.

8) Rewriting of the answer to make life easier for Mr Johari.

QD Question 5 (Aziel, Jerome and Kashyap)

Done by Aziel, Jerome and Kashyap.

QD Question 7 - (Balram, Tobias and Guo Jin)

Solution to Question 7:

QD Question 4 (Enoch, Jemaimah, Sherman)

Question 4:

Done by Enoch 

QD Question 1 (Charlene and Jun Jie)

Before we start on the solution for question 1, the answer that we are to find would be the A, B and C not the X.

The first step in solving a partial fraction is to use the formula given. In this case it will be the RHS of the first line.
The second step is to get rid of the denominator by multiplying it. This is shown in green.
The third step is to substitute X, always start with the roots and use simple numbers. This is shown in purple, green and red.
The last statement is how to present your answer. Put 3 dots aligned in a triangular formation and insert the right values for A, B and C. If the answer can be simplified, then simplify it.

QD Question 3 (Esther, Darryl, Sheares)

QD Question 8 (Hao Xian and Luke)

QD Question 6 solution (Jee Hoon, Bowen)

Q : Express

into partial fraction.

Ans : x+1 + 1/x+1 - 4/x-1

Working :

Use long division to divide x^3 -x^2 +2x -6 by x^2 -1 (because it is improper fraction)

x^3 -x^2 +2x -6/x^2-1 = x+1 - 3x+5/x^2-1
3x+5 = A(x-1) + B(x+1)
3x+5 = (A+B)x - A+B
By comparing coefficients :
x^1 : A+B = 3
x^0 : -A+B = 5

A = -1
B = 4

x^3 -x^2 +2x -6/x^2-1 = x+1 + 1/x+1 - 4/x-1

On-line Quiz


This is a class individual work

Topic: Partial Fractions

To demonstrate your understanding of the concepts through clear articulation on your approach to solve a problem.

Instructions (I):
  • This is an individual task. 
  • It will also help you to seek and articulate your understanding of concepts as you attempt to explain your approach and solution to the problem.
  • You should not take more than 20 minutes to complete the entire task.

Instructions (II):
  • Choose ONE question 
    • Two questions are presented below. Read through carefully. 
    • Identify a question that you think is challenging but manageable. 
  • Plan 
    • Work out the solution on a piece of paper and think through how you would explain the solution. 
    • Explaining your solution does not mean simply read out the lines of the working, but inform the assessor how you identify the key information, the method you use, and how the method works. 
    • Allocation of Question
    • Q1 # 1,  9, 17
    • Q2 # 2, 10, 18
    • Q3 # 3, 11, 19
    • Q4 # 4, 12, 20
    • Q5 # 5, 13, 21
    • Q6 # 6, 14, 22
    • Q7 # 7, 15, 23
    • Q8 # 8, 16
  • Presentation
    • Use any application/ platform that is suitable to present your solution. 
    • Remember to keep the presentation accurate and concise.
  • Submission of work
    • Post your presentation as a Blog post.

Notes - 5th Fabulous February

As shown in the above picture, the laws of indices only applies to multiplication or division, when you see addition or subtraction, chances are, you have to factorise, one example is the question Mr Ingham set for us, shown in the picture below 

As you can see Mr Ingham 'solved' his own question by using this law a^(m+n) = a^m + a^n
From there, he was able to get a quadratic equation, thereby using the calculator to solve that problem.

TASK FOR 5th February 2013

TASK FOR 5th February 2013

Period 1

1.  Self Assessment ( please refer to Quiz C [10mins]
- complete the corrections 
- identify the errors committed be it (i) conceptual or (ii) carelessness

2. Collaborative Effort - via Maths Blog  -  Linoit exercise  [20mins]

- S3-05   complete the portion on Indices
- S3-09   complete the portion on Surds
3. Peer Assessment - via Maths Blog activity 

INDICES & SURDS - The Mistake Fragment part 1
- post the responses as comments

4. Self Assessment - via Maths Blog activity 
- INDICES & SURDS - The Mistake Fragment part 2
- post your responses as comments 
================================================================Period 2

5. Diagnostic Test 2 - Indices, Surds and Logarithms
- do be done on foolscap papers
- scribe of day to collect and past to teacher

INDICES & SURDS - The Mistake Fragment part 2

By Mr Johari

Identify the mistakes shown below and post your corrections.
For the correction identify the correct rules.

Find the algebra mistake: 
1.       mistake
2.      mistake
3.       mistake

4.      Find the algebra mistake: 

5.     Find the algebra mistake: 

INDICES & SURDS - The Mistake Fragment part 1


Below shown the solution posted by the secondary 3 students during the 4th February 2013 Maths Quiz C. 
There are errors in the solution posted. 
Error Analysis and Peer Evaluation
Individual Effort

You are required to post the following as a comment:

(i) Nature of error - conceptual or carelessness (be very specific in your description)
     example - careless error due to arithmetic (addition)
     example - conceptual error due to error in (a + b)^2 = a^2 + b^2.

(2) Correction to the error
      Ensure that your solution is concise and correct.


BY MR JOHARI This is an Interclass Collaboration on INDICES and SURDS S3-05 focuses on INDICES S3-09 focuses on SURDS TASK 1. Identify as many laws of Indices as possible. 2. Include condition, where possible such as a > 0. 3. Include also examples on how the rules are used.

Indices and Surds to access Linoit