What is an Absolute Value?

Part 1:

Absolute Value & The Opposite of a Number

Nomenclature

Part 2:

Solving Absolute Value Equations

Part 3:

Absolute value function / graphicals

Part 4:

Solving Absolute Value Equations Graph Calculator

Using TI 84Plus and simple programme

INTRODUCTION TO TI84 Plus

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

QD Question 2 (Kai Chek, Yi De, Dylaine)

Sorry for the late post.

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 

               Jemaimah

              Sherman

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

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