Individualized Education Programme IEP

Learning is not a hard science. There is no single formula that can be applied to all learners to achieve the desired learning outcome.Furthermore, each learner is unique and so is the facilitator of the learning process and as a consequence the learning process should be unique to an individual learner. That complicates the learning process in a school setup where the availability of qualified and passionate facilitators of learning are scarce, to allow a 1:1 ratio. Hence an individualized education programme will then be tailored made for the so called special cases only. Unfortunately this contradicts the assumption that we are all unique and we all need individualized education. In most cases we target the marginalized minority on the negative side i.e. those who have poor handwritings, get all sums wrong, the easily agitated, those who have many spelling errors etc. What about those who are excellent as per our standards? How do we ensure they exploit their potential to the fullest? What about the normal student? are they just normal or conformist, or have they mastered  what the society want and follow what the society advocates to be the right standards? Despite the fact we all complain of a poor education system that cannot address the societal challenges, not forgetting we are part of the system.

Reflection on what kind of learners i have and what kind of teacher i am, can have great benefits. Analysing critical incidences in and out of our classrooms can help discover the uniqueness in each one of us. Of course reflection per se can be hard and bias. The assumption is, if you are highly exposed to the right knowledge, skills and attitudes, you can be in a better position to utilize these abilities - reflecting and CIA, to their fullest.

This post welcomes replies on what potential you discovered, what insights you had, when you reflected on your IEP and took whatever steps you undertook. Sharing of experiences can be beneficial to any reflective individual.

Rewriting an Exam Question

When answering exam questions some weak students normally rewrite the question in the Answers booklet before providing the solution.why do they rewrite it?it does not add any value. Moreover it is a waste of precious time. No matter how many times you tell them they always repeat.Some explain this behaviour by saying the tutor may confuse the solution and its respective item.

Basic techniques of differentiation

Techniques of Differentiating

The following are basic techniques of differentiating. 

(a) Power rule

So far we have seen that given; y=f(x)=kx^{n}

Then the gradient function is; dy/dx=knx^{n-1}

Example 1: Differentiate the function y=3x^{5} with respect to x

solution: dy/dx=15x^{4}

Example 2: Find the derivative of the function y=2x^{3}-6x+3

solution : dy/dx=6x^{2}-6

Example 3: Differentiate the function y= [6x^{3}-7x]/[x^{2}]

solution: we need to first simplify the function by rewriting it as;

y=6x^{3}/x^{2}-7x/x^{2}=6x-7x^{-1}

Hence we have;

          dy/dx =6+7x^{-2}

(b) Product rule

Let y=f(x)=uv where u and v are both differentiable functions of x, then the gradient function is given as;

dy/dx= u dv/dx+v du/dx

Alternatively;

 (uv)^/=uv^/ +vu^/

This rule is appropriate when it comes to integrating product functions.

Example 1: Given y=(3x^{4}+7x)(5x^{7}-3x^{2}+9x) determine the derivative.

Solution: by definition dy/dx=u dv/dx+ v du/dx

we let,

u=3x^{4}+7x then du/dx=12x^{3}+7  and we let v=5x^{7}-3x^{2}+9x  then dv/dx =35x^{6}-6x+9

Therefore; 

dy/dx=u dv/dx+ v du/dx=(3x^{4}+7x)(35x^{6}-6x+9)+(5x^{7}-3x^{2}+9x)(12x^{3}+7)

Example 2: Given y=3x^{5}sinx , determine the gradient function.

Solution: by definition dy/dx =u dv/dx+v du/dx

We let, u=3x^{5} then du/dx =15x^{4}

Also let, v = sinx    then  dv/dx =cosx

Therefore; 

dy/dx=u dv/dx+v du/dx =3x^{5}cos x+15x^{4}sin x=3x^{4}(xcos x+5sin x)

Example 3: Let y=3x^{2}e^{x}, find dy/dx

Solution: We let u=3x^{2} then du/dx=6x

Also v=e^{x} then dv/dx=e^{x}

Hence;

dy/dx=(3x^{2})(e^{x})+(6x)(e^{x})=3x^{2}e^{x}+6xe^{x}

(c)  Quotient rule

Let y=f(x)= u/v  where u and  v are both differentiable functions of  x, then the gradient function is given as;

dy/dx=[v du/dv –u dv/dx]/v^{2}

Example 1:  Given y= 4x^{3}/[3x^{2}-4x]  find dy/dx;

Solution: Let u=4x^{3} then du/dx =12x^{2} and v =3x^{2}-4x then dv/dx = 6x-4

dy/dx=[12x^{2}(3x^{2}-4x)-4x^{3}(6x-4)}]/[(3x^{2}-4x)^{2}]

=[36x^{4}-48x^{3}-24x^{4}+16x^{3}]/[{(3x^{2}-4x)^{2}}]

=[12x^{4}-32x^{3}]/[(3x^{2}-4x)^{2}]

=[4x^{3}(3x-8)]/[(3x^{2}-4x)^{2}]

Example 2: Find the derivative of the function;

y= 2x/sin x

Solution: Let;

u = 2x then du/dx  =2

Let; v= sin x then dv/dx=cos x

Hence;

dy/dx=[2sin x -2xcos x]/sin^2x=2[sin x – x cos x]/ sin^{2}x

(d) Chain rule

Let y be a differentiable function of  u  i.e.  y=f(u)  and that u  is a differentiable function of x i.e. u=g(x), then y is a differentiable function of x i.e. y=f(g(x)), and;

dy/dx=(dy/du )(du/dx)

Example 1: Suppose y=(3x^{5}+7x^{4})^{4}, find dy/dx

Solution:

Let u=3x^{5}+7x^{4} then  du/dx=15x^{4}+28x^3

Also y=u^{4} then  dy/du=4u^{3}

By definition

dy/dx=(dy/du)(du/dx)=(4u^{3})(15x^{4}+28x^3)

=4(15x^{4}+28x^3)(3x^{5}+7x^{4})^{3}

=4x^{3}(15x+28)(3x^{5}+7x^{4})^{3}

Introduction to Differentiation

Definition 1:
To understand the concept of basic differentiation, we need to consider the gradient or slope or steepness of a straight line of the form y=mx+c where y and x are variables, m the gradient of the line and c is the y-intercept. Now the gradient of a straight line is the change in the vertical distance over the change in the horizontal distance. That is;
Gradient, m=(change in vertical distance)/(change in the horizontal distance)

Note that the gradient of a straight line is constant, but the gradient of a curve keeps on varying.

Definition 2:
To determine the gradient of a curve at point we need to use a gradient function. Given a function;
y=f(x)=kx^{n}
Then the gradient function;
dy/dx=knx^{n-1}

Example 1: Let, y=3x^{5},then dy/dx=15x^{4}

Example 2: Let; y=2 , then dy/dx =0

Remark 1: Note that, y=2 can also be written as y=2x^{0}. Recall that x^{0}=1 where x is not 0

Example 3: Differentiate the following function with respect to x; y=(2x-3)(4x+5)

Solution: Note that another term that is used to mean, 'finding the gradient function' is 'differentiate' with respect to a certain variable. One can also refer differentiation as 'determining the derived function'.
In this example, we need to expand the RHS to get;
y=8x^{2}-2x-15 then dy/dx =16x-2

Remark 1: The rule so far discussed is the most basic rule of differentiating. Some functions may not be differentiated at all over certain intervals or may not be differentiated using the above rule. Some differentiating techniques are discussed in the next section while others are beyond the scope of this blog.

P.S.
1) Next post will be on techniques of differentiating
2) This is an excerpt from the textbook, Basic Mathematics; by Kahenya NP

Incident 4 The Missing Girls

It was on Thursday mid morning at around 10.25am. This was immediately after a short break and learners were expected to attend science lesson. Due to the harshness of their class teacher who was also their science teacher twenty pupils did not attend science lesson and therefore only twenty three learners were in the classroom ready to be taught. The other twenty hid behind the classroom and others in the latrine.

This was the day I was expecting to be assessed for the second time because we had already communicated by the assessor that morning and even gave her direction to my school. I was expecting her to be in the classroom for assessment in the same class. When the teacher realized that some of the learners were not in the classroom she decided to report the matter to the deputy headteacher who also reported the same matter to the headteacher. The headteacher did not take his time to know what could be the root cause of the misconduct instead ordered boys from standard seven and eight to go and look for the standard three pupils who hid within the school compound. They were assumed to be within the school compound because the school is fenced and has lockable gate and a gate keeper, later on the learners were found in the latrines except two girls who were missing for the whole day.

On Friday very early in the morning the parents of the two girls came in tears (sic) asking for their beloved daughters, None of the teachers could explain the where-about of the girls. The Headteacher asked the teachers on duty to assemble learners in the assembly ground so that he could make an announcement concerning the two girls. During the announcement one of the boys raised up his hand and informed the headteacher that the previous evening he saw the two girls in one of the herdmen house and he expected that they slept there.

In addition the boy declared (sic) to the Headteacher infront of the two parents that the herdman had been luring school girls using money and therefore he had a chain of girl friends from our school. Above all, the herdman had been sickly and some weeks ago he was diagnosed H.I.V. positive. When the two parents heard that, one of them fainted and the other jumped on the headteacher in front of his office (sic). The situation was unbearable teachers tried to bring calm but it was in vain. The parent who fainted later recovered and went with the other parent and the deputy headteacher and two other teachers to where the girls were. Then girls were found still sleeping. To them it was not a big deal; one of the them declared to the deputy headteacher that even her parents knew she was befriending the guy (herdman). And many are the occasions she has ever taken shopping to the mother from the guy.

The long story ended in the police station whereby the herdsman was jailed because of child abuse. The girls were  later transferred to another school.
P.S.
(1) This is unedited version of the incident.
(2) Give reflective comments on this incident. Reflect on so what... and now what situation of the incident in respect to the teaching/learning process.

Incident 3: Std 3 CRE lesson

The following is unedited version of the incident as reported by Mary (not her real name):

It was a standard 3 CRE (christian religious education) lesson. The topic was 'Jesus Resurrection'. We learnt the purpose of Jesus being born, dying and resurrection. I explained that when we die our soul goes to heaven (sic) and our body remains here on earth (sic). Therefore in the case of Jesus, it seems God put the soul back into Jesus' body after three days.

After the lesson, it was questions time to check whether i had achieved my objectives. Question 3: which part of our body goes to heaven when we die? Answers: Child 1 gave: brain. Child 2 gave: mind. Child 3 gave: Heart. Child 4 gave: legs. I got curious and i further inquired why he said legs. The boy replied,'Teacher! it is because the other day my dad was struggling (sic) my mom while lying on her. Her legs were raised up and she was screaming in pain while saying, ooh my god, am coming'.I was extremely shocked, breathless...a moment of silent followed. I hoped other learners had not heard this. I didn't know what to say or do.

I recollected myself very fast and advised the learners that it was not the legs but the soul that goes to heaven. Afterwards i called the boy and inquired further. He disclosed that they live in single room. I was a bit mixed up because i felt i needed to talk to his mother but i did not know how she would respond or react. I prayed about the issue, gathered courage and called her. On hearing the story, she was honest enough and apologized. she also promised to check on the situation. Being a very private matter i left it at that.

How would you interpret the above incident?
Now what... next?

Incident 2: Class 2 pupil’s Traditional Brew Present

It was during the second week of my teaching practice when my assessor Mr Achi* came for assessment during teaching (sic). Earlier I had prepared my learner’s psychologically (sic) that the external visitor (sic) would come to class. The learners are always excited about visitors.

Prior to the day (sic), I informed them t0 clean their uniform and to come to class early. When the assessor came, learners were quite cooperative. Learning took place well as intended but during the lesson one of the learners had brought a present to the visitor. The boy stood and asks for permission to present the present to the visitor but I told him to wait.

At the end of the lesson, the boy stood and walked steadily to the visitor and presented a bottle of traditional brew (traditional brewed alcohol) to my assessor. The rest of the class laughed and tried to run out of the class but I controlled them.

Immediately, I allowed the class to go for the long break and called the boy and asked why he did so. The boy told me that it is a normal drink at their home. I advised him not be using the brew and later I called the parent and talked with her on the dangers of brew to young kids and the health effect in general.

NOTE:

·         No editing has been done.

·         Names have been changed to protect privacy.

What was most interesting or bad aspect about the incident?

What new lessons did the teacher learn?

What is the way forward as far as the teaching/learning is concerned, in regard to this incident?

Incident 1: Class Two Incident

On the morning of 14th March 2015, i woke up early than usual since i was the teacher on duty and i was also expecting external supervisor from the university. I was able to arrive at school at about 7:00 a.m. Pupils began tickling in one by one and soon learning started.

At about 10:00 a.m our school chairman came to school to inform us that the county governor would be visiting our school in the afternoon. We hurriedly held a brief staff-meeting and shared duties amongst ourselves. I was charged with responsibility of ensuring that the environment (sic) becomes clean and tidy. i asked pupils to collect litter and some to sweep their classrooms.

It all begun when it started raining in the afternoon. It was a heavy downpour and word came around that the governor was not going to make it. Kamau* (not his real name), one of our trusted class three prefect came running to the office and reported that something weird was happening in class two. Being the teacher on duty  i hurried to investigate. As i neared their class i could hear the whole class singing loudly ....'kila mtu na demu wake.. '(everybody hold your girl). I could clearly recognized the voice of the most notorious boy nicknamed 'Jangili' (Thug) leading the singing. When i entered the class, i was shocked to find some of the kids were half naked.

We were able to guide and counsel the kids who led others into this ugly incident. Jangili and his close associates (sic) confessed  that they have been doing thus on their way home, the reason being that, they have been seeing their parents doing 'jig jig' at night. The kids vowed not to repeat it again.

It came to our understanding that class two pupils had not being doing their assignments and homework since their class teacher has been reluctant to mark the classwork. We also learnt that Jangili and his close associates (sic) had joined a group (sic) that had been abusing drugs.

I learnt that each child has a unique characteristics/traits and should be handled differently. Learners have individual differences.
NOTE:
The name of the university, school, author, teacher, pupil, governor and county has been omitted to protect privacy.

Critical incidences in our classrooms

In the next posts I will share some of the incidences that my students experienced when they were undertaking their teaching practice. They analyzed the incidents from a reflective point of view. They were guided by reflective models. The goals of these posts is to hear your comments on theories/opinions on why such incidents happen in our classrooms. Theories to explain the why. I would also like to hear what feelings you could have experienced if you were in such a situation. And finally the lessons we have learnt or can learn from such incidences and the way forward.

sequences and series

A sequence is a set of terms which are written in a definite order obeying certain rules e.g. 2,4,6,8,10,... is an infinite (since it goes on and on forever) sequence obeying a certain rule which is; for you to get the next term in the sequence you add 2 to the preceding one.In particular this sequence is made of multiples of two or even better the even numbers.

A sequence with definite terms is said to be a finite sequence e.g. 3,6,9,12. Note that they are no three dots after 12 i.e. the three dots normally denote that the sequence goes on and on forever. Hence this sequence has 4 terms. Hence  it is a finite sequence.

A series is the sum of the terms of a sequence e.g. 2+4+6+8+10+... is a series. In general a finite series is the sum of terms of a finite sequence i.e. a1+a2+a3+...+an. While an infinite series is the sum of terms of an infinite sequence i.e. a1+a2+a3+...+an+....

The most common sequences are the arithmetic sequences and the geometric sequences. An arithmetic sequence is a sequence that proceeds with a common difference which is normally denoted by d.In general an arithmetic sequence is a sequence of the form; a, [a+d], [a+2d],[a+3d],...,[a+(n-1)d] where a is the first term, n  is the number of terms in the sequence and [a+(n-1)d] is the last term which is either denoted as l or Tn (i.e. the nth term). For example; 2,4,6,8,10,... is an arithmetic sequence with first term a=2 and common difference d=2.

An Arithmetic series also referred to as an Arithmetic progression is the sum of terms of an arithmetic sequence i.e. a+[a+d]+[a+2d]+[a+3d]+...+[a+(n-1)d] e.g. 4+7+10+13+...is an infinite arithmetic series with first term a=4 and common difference d=3


On the other hand a geometric sequence is a sequence that proceeds with a common ratio normally denoted by r .In general a geometric sequence is a sequence of the form a,ar,ar^2,ar^3,...,ar^(n-1),where a is the first term, r is the common ratio,n is the number of terms in the sequence and ar^(n-1) is the last term or the nth term of the sequence. For example; 2,4,8,16,32,... is a geometric sequence with first term a=2, and common ratio r=2

A geometric series or the geometric progression is the sum of terms of a geometric sequence i.e. a+ar+ar^2+ar^3+...+ar^(n-1) e.g. 3+9+27+81+...is an infinite geometric series with first term a=3 and common ratio r=3.

Solving a quadratic equation by completing the square method

I will demonstrate by two examples. Example 1 has the coefficient of x^2=1 and Example 2 has the coefficient of x^2 greater than 1
Example 1
solve by completing the square method; x^2-7x+10=0 (note also the coefficient of x =-7)
solution
The first step is to take the constant 10 to the RHS to get;
x^2-7x=-10
Next we add a constant k to both sides to get;
x^2-7x+k=-10+k ....(*)
Next we find the value of k to 'complete the square' on the LHS by making it a perfect square i.e.;
k=(1/2*the coefficient of x)^2
k=(1/2*-7)^2=(-7/2)^2
Now we write equation (*) as follows;
x^2-7x+(-7/2)^2=-10+(-7/2)^2
The LHS can be easily factorized as below;
(x-7/2)^2=-10+49/4
(x-7/2)^2=9/4
We next find the square root of both sides to get;
x-7/2=±3/2
x=7/2±3/2
x=3.5±1.5
x=5 or 2
Example 2
Solve by completing the square method; 2x^2+x-3=0 (Note that the coefficient of x^2=2 unlike example 1 above)
Our first step is to make the coefficient of x^2=1 by dividing every term by 2 to get;
x^2+x/2-3/2=0
Next we take the constant to the RHS to get;
x^2+x/2=3/2
Next we add a k to both sides to get;
x^2+x/2+k=3/2+k.....(**)
Next we find the value of k to 'complete the square' on the LHS by making it a perfect square i.e.;
k=(1/2*coefficient of x)^2=(1/2*1/2)^2=(1/4)^2
Equation (**) can now be written as;
x^2+x/2+(1/4)^2=3/2+(1/4)^2
the LHS can be factorized to get;
(x+1/4)^2=3/2+1/16
(x+1/4)^2=9/16
Next find the square of both sides to get;
x+1/4=±3/4
x=/1/4±3/4= 4/4 or -2/4 
x=1 or -1/2

Solving by a quadratic equation using the Quadratic Formula

Given a quadratic equation ax^2+bx+c=0 then x=(-b±√(b^2-4ac))/2a. This is the quadratic formula. Example 1 Solve the following equation using the quadratic formula; 3x^2+7x+2=0 solution. In our case a=3, b=7, and c=2. We next need to replace a, b,and c with 3, 7, and 2 respectively in the formula to get; x=(-7±√(7^2-4*3*2))/2*3. x=(-7±√(49-24))/6 = (-7±5)/6 = -2/6 or -12/6 =-1/3 or -2.

Solving a quadratic equation by factorization

A quadratic equation is an equation of the form ax^2+bx+c=0, where a, b, and c are known constants and a  i not zero e.g. 2x^2+3x+7=0; 3x^2-4x=0; 7x^2-9=0 etc.

To solve by factorization, first factorize the LHS i.e. ax^2+bx+c, by first finding two factors which i will call T1 and T2 such that; (i) their sum is b i.e. T1+T2=b and (ii) their product is ac i.e. T1*T2=ac. Next replace bx with T1x and T2x to obtain ax^2+T1x+T2x+c=0. Then carry out group factorization of the LHS.

I will next demonstrate using two examples;
Example 1
Solve by factorization; x^2+7x+12=0
Solution
In this example our a=1, b=7 and c=12. Hence we need two factors T1 and T2 such that;
(i) T1+T2=7 and (ii) T1*T2=12. The two factors are T1=3 and T2= 4.
Next we replace 7x with 3x and 4x to get;
x^2+3x+4x+12=0 ......(***)
The next step is to factorize the LHS by group factorization i.e. we take the first two terms x^2 and 3x and factorize [i.e. x^2+3x=x(x+3)] and then factorize the next two terms i.e. [4x+12=4(x+3)]
Such that our equation (***) above becomes;
x^2+3x+4x+12=x(x+3)+4(x+3)=0
Note that (x+3) is common thus we have;
(x+3)(x+4)=0
(x+3) and (x+4) are the two factors of x^2+7x+12. Since their product is 0 it implies that either;
x+3=0 or x+4=0
Hence x=-3 or x=-4.
-3 and -4 are the solutions or the roots of the quadratic equation x^2+7x+12=0
Example 2.
solve 3x^2+10x+8=0
Solution
In this example our a=3, b=10 and c=8
Hence we need two factors T1 and T2 such that (i) T1+T2=10 and (ii) T1*T2=24
The two factors are 4 and 6.

Next we replace 10x with 6x and 4x to get;

3x^2+6x+4x+8=0 ......(******)

The next step is to factorize the LHS by group factorization i.e. we take the first two terms 3x^2 and 6x and factorize [i.e.3x^2+6x=3x(x+2)] and then factorize the next two terms i.e. [4x+8=4(x+2)] 

Such that our equation (******) above becomes;

3x^2+6x+4x+8=3x(x+2)+4(x+2)=0

Note that (x+2) is common thus we have;

(x+2)(3x+4)=0

(x+2) and (3x+4) are the two factors of 3x^2+8x+10. Since their product is 0 it implies that either;

x+2=0 or 3x+4=0

Hence x=-2 or x=-4/3. 

-2 and -4/3 are the solutions or the roots of the quadratic equation 3x^2+8x+10=0

self-drive

When it comes to maths,your own initiative plays a key role in learning new concepts.A self-driven person who is curious and adventurous and more so an inquisitive person can gain a lot from the world wide web.Though skill of research is crucial.The world will be at the tips of your fingers.

Difference between permutation and combination

The difference is that permutation is an arrangement where order is important while combination is a selection and order is not important. Think of a sandwich made from a slice of bread, an omelette and salad. You can arrange the three ingredients in the following order (assume you are laying the sandwich on the plate); a) bread, omelette,salad b) bread, salad, omelette c) salad,bread, omelette d) salad, omelette, bread e) omelette, bread, salad f) omelette,salad, bread The above is 6 arrangements (permutations). Now, assume 6 people ordered (a) to (f) sandwiches for breakfast, would they be right to say they ate 6 different types of sandwiches? of course NOT!It is just one COMBINATION (selection) of sandwich.

Solving system of linear equations

One can solve a system of linear equations using substitution method, elimination method, graphical method, crammer's rule among others. Some websites offer applications/solver that show you how to work out the sum.  One can get a variety of resourceful sites if you are good in research skills. As long as you know exactly what you are searching for,and you are patient, inquisitive and curious enough,and also you do not have a phobia for computers, then you have no problem as long as you can make use of the google search engine.For the time being, it basically has enough for you.

'Stop passing the buck'

'Do you like what you are doing now? If not, then stop doing it now. It is said that, ‘if you find yourself digging a hole, then the best thing you should do, is simply stop digging’. If you find yourself complaining about a situation you are in and expecting somebody else to come and better the situation for you, then brace yourself for future frustrations. Learn to manage your environment. Your environment is both intrinsic and extrinsic. You are like a plant. For you to thrive, you need the right nutrients and the right environment. Unlike the plant, you can control all these. Your thoughts reflect you. The thoughts you entertain form your environment that influence you. It is you to act on the right environment'. An excerpt from, 'Reflections'by kahenya,np

Conics

Circles, parabolas, ellipses and Hyperbolas are collectively called conic sections.The general equation of conic sections is Ax^2+By^2+Cxy+Dx+Ey+F=0. You can google and get valuable links on conic sections.

Solving Linear Systems

In your previous classes at primary and high school levels, you came across linear equations i.e. equations whose functions if plotted on a plane or graph paper you will end up getting straight lines. Linear equations may have one or more unknowns. For instance; x=3;y=-5;x+y=5;x+y+2z=6;2a+3b+3c-6d=7 and so on. Solving linear equations with 1 unknown and 4 or more unknowns will not be discussed here. Our interest is to solve linear equations with 2 and 3 unknowns. Some quantities in real life may be expressed in linear form, but in reality it is rare to have linear relationships because of the many parameters that are in play. Linear equations are used in real life to explain situations while holding some variables constant. For instance price p of a commodity say a phone may partly depend on demand q of the phones and partly constant. Note, the price cannot go down to zero even if the demand is zero. This may be written as p=bq+k.

Solving Quadratic Equations by factoring

Example
solve x2+7x+12=0
The above equation is of the form ax2+bx+c=0 i.e. the general form of a quadratic equation.
what you need to do is find two numbers or factors, which we shall call T1 and T2. These two
terms are such that; i) their sum is b i.e. T1+T2=b , and ii) their product is ac i.e. T1×T
2=ac .
After gettoing the two terms, replace the term bx in the equation, with T1x and T2x ,to get
ax2+T1x +T2x+ c=0 . Then carry out group factorization i.e. you first factorize ax2+ T1x and
then +T2x+c.
So in our case, T1+T2=7 and T1×T
2=12. clearly our two terms are 3 and 4.
hence we replace 7x with 3x and 4x to get x2+3x+4x+12=0
then we apply group factorization to get x(x+3)+4(x+3)=0
note that (x+3) is a common factor.
so we have, (x+3)(x+4)=0
for the next step think of two numbers A and B and their product is zero i.e.AB=0. This imply
that either A or B is zero.
Hence in our case (x + 3)(x + 4) = 0, it implies either (x + 3) = 0 .(i) or (x + 4) =
0 (ii)
From (i) x=−3 and from (ii) x=−4
-3 and - 4 are called the roots of the equation x2+7x+12=0 or the solutins of the equation.
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