TOK (Theory of Knowledge) - Algebra

Algebra

In this Chapter, we are going to recap the basic knowledge regarding “Algebra". The following descriptions are summarized for preparation of past paper practice and if you do not know something, you can contact us.

 

Negative Number - Symbol (-) - Less than zero, represented on the left side of zero. 

Directed Number -   (+1,+2,+3,...) / (-1,-2,-3,....)

Addition of directed numbers 

eg  3+4+5 = (+3)+(+4)+(+5) = +12

3+4-5 = (+3)+(+4)+(-5) = +2

12-15 = (+12)+(-15) = -3

Same sign ---- sum

different sign ---- find the difference and give the answer in sign of larger number

 

Brackets and simplifying 

- distributive law - $(a+b)\times c=ac+bc$

eg $(x+6y)3=3x+(3\times 6)y=3x+18y$

 

Order of Operation 

1. brackets ()/{}/[]
2. exponents $2^3,5^2$
3. multiply and divide 
4. add or substract

Linear equations 

When order of unknown variable is 1 i.e. (x^1,y^1,...)

Eg $y=x+2$

2x-10=0

2x=10

x=10/2

x=5

   

If Exponents or Roots contains, they are Not Linear equations.

 

 

Simultaneous Equations

A group of more than one equations.

 

Two methods to solve these equations

·        substitution method

change one equation in the form y=mx+b . Then substitute value of y as (mx+b)

Eg, Given that, y+2x=20, 2y+x=25 

y+2x=20,           
2y+x=25
2(20-2x)+x=25
40-4x+x=25
40-3x=25
-3x=-15
y=20-2x
y=20-2(5)
y=20-10
y=20-2x            
x=5
y=10 

·        elimination method

More convenient in some equations, where coefficient of a unknown in first equal is same  or easily convert the same coefficient to corresponding unknown in second equation. Then Subtract two equations to eliminate that unknown with corresponding coefficient. 

Eg, Given that x+2y=25, x+y=15

x+2y=25..........Eq(1)
x+y=15............Eq(2)

Subtract eq (2)from (1)

x+2y-(x+y)=25-15
x+2y-x-y=10
y=10
then, substitute in eq (1) or (2) to get x=5

Factorising 

What is factor? A factor is a number that divides the given number exactly, leaving NO remainder.

Factorising is the process of diving number into factors.[Reversal of expanding brackets]

To factorise, you should think of the following facts,

 

• If highest common factor contains, write it down in front of bracket, and fill the bracket. $4x+8=4(\frac {4x}{4}+\frac {8}{4}=4(x+2))$
• In quadratic factorisation, 

 

    (1)Factorise $x^2+x-6$.

    (2)multiply coefficient of $x^2$ and end number ($1 \times 6$)

    (3)Find the factors of result number ($6=1 \times 5=2 \times 3$)

    (4)Find what pair would get coefficient of x{middle number} and constant number (-6) 

i.e.(1=3-2)

-6=$3 \times (-2)$

    (5)Choose that pair (+3 and -2)

    (6)Replace original expression with chosen pair i.e.

$x^2+x-6$

=$x^2+3x-2x-6$  

=$x(x+3)-2(x+3)$

=$(x+3)(x-2)$ 

• The difference of two squares

$a^2-b^2=(a+b)(a-b)$
$2^2-9x^2=2^2-(3x)^2=(2+3x)(2-3x)$

Quadratic equations

• Solution by Factors.

Consider the fact that $a \times b=0$.

$(x+1)(x-2)=0$

$(x+1) \times 0=0$ or $(0 \times (x-2)$

$x+1=0$ or $x-2=0$

$x=-1$ or  $x=2$ 

 

• Solution by Formula

$\displaystyle x=\frac {-b \pm \sqrt{(b^2-4ac)}}{2a}$ in quadratic equation $ax^2+bx+c=0$

Use this formula only if you cannot use above method.

In above equation, $(x+1)(x-2)=x^2-x-2$, a=1, b=-1, c=-2

$x=\frac {-(-1) \pm \sqrt{((-1)^2-4(1)(-2))}}{2(1)}$

$x=\frac {1 \pm \sqrt 9}{2}$

$x=\frac {1\pm 3}{2}$

$x= \frac {1+3}{2}$ or $x= \frac {1-3}{2}$

$x= 2$ or $x= -1$

 

Congrarulations! You are ready to practice past papers. you can click Home button and choose the post you would like to practice. See you in the next chapter.