Why (a+b) 2 = a2+b2+2ab ?
Why (a+b)^{ 2} = a^{2}+b^{2}+2ab ?
Ever wondered how was the above formula derived?
Probably the answer would be yes and is simple. Everybody knows it and when you multiply (a+b) with (a+b) you will get a plus b whole square.
(a+b) * (a+b) = a^{2} +ab + ba + b^{2} = a^{2 }+ 2ab + b^{2}
But how did this equation a plus b whole square became generalized.
Let’s prove this formula geometrically.( Please refer to the pictures on the side)
 Consider a line segment.
 Consider any arbitrary point on the line segment and name the first part as ‘a’ and the second part as ‘b’. Please refer to fig a.
 So the length of the line segment in fig a is now (a+b).
 Now, let’s draw a square having length (a+b). Please refer to fig b.
 Let’s extend the arbitrary point to other sides of the square and draw lines joining the points on the opposite side. Please refer to fib b.
 As we see, the square has been divided into four parts (1,2,3,4) as seen in fig b.
 The next step is to calculate the area of the square having length (a+b).
 As per fig b , to calculate the area of the square : we need to calculate the area's of parts 1,2,3,4 and sum up.
 Calculation : Please refer to fig c.
Area of part 1 :
Part 1 is a square of length a.
Therefore area of part 1 = a^{2 } (i)
Area of part 2 :
Part 2 is a rectangle of length : b and width : a
Therefore area of part 2 = length * breadth = ba (ii)
Area of part 3:
Part 3 is a rectangle of length: b and width : a
Therefore area of part 3 = length * breadth = ba (iii)
Area of part 4:
Part 4 is a square of length : b
Therefore area of part 4 = b^{2 }(iv)
So, Area of square of length (a+b) = (a+b)^{2 }= (i) + (ii) + (iii) + (iv)
Therefore :
(a+b)^{2} = a^{2} + ba + ba +b^{2}
i.e. (a+b)^{2 }= a^{2} + 2ab + b^{2}
Hence Proved.
This simple formula is also used in proving The Pythagoras Theorem.Pythagoras Theorem is one of the first proof in Mathematics.
In my view, in mathematics when a generalized formula has been framed there will be a proof to prove and and this is my small effort to exhibit one of the proof's.
Will definitely come up with some more.
Also please find the video proof.
Questions & Answers
Comments
It's been observed that committing formulae by heart is tentative and so it is of no use later on but working them out practically is of greater use as leaves an indellible mark on one's mind and stays forever.
Thank you ever so much
We tend to teach students just "formulas" without demonstrating and what they are need for
Thanks
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good sir but in hindi
Good one
Can you give
related questions
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What is the solution of (a+b)2=
Thanks a lot for help
I am not satisfied with that proof. Any other please!
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Good
minded
very nice.........and helpful
Answer in hindi
Actually good but as well as now its common.. way to prove but it will b more better if u find out some more way 2 prove ,becoz that may give u more greetings from everyone
Then also u tried ur best to help us so ..good job keep it up
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Diploma Electrical Engnnering
Mathematics =3 semester Question and Answer send me please VVI Question
How a2B 2 +aB2
A+b =a2*bxv


Tens
Vvvvv good
Could u give me this kind of geometrical explanation for (a+b)³.?
Very nice and helpful to me thanx for it..........
Very Good.
Nice sir...........""
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Thanks for this but can you give some questions related to this and its answer .please
hello bawa! today I really found out the mystry of the Universe.
Good, like it
Very help ful
thakyou
Nice explanation it is easy to learn
Thank you sir
I really liked the proof and I never knew that it can be proved in this way tooooo......
thanks for this
How many a and how many b in a+bwhole square..?
can you plz proo another formula
Very useful and interesting
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(ab)2=a2+b22ab.
Good
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How d formula z used in daily life
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(y)
very good
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explainted very well.
really helpful proof
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in the same way, do u know (ab)^2
in the same way, do u know (ab)^2
This is fantastic; I had no idea, and I have never seen it explained in this way. Great job!
Actually, it is proved by using Pascal's triangle, with the general formula: (a+b)^n = (n 0) a^n * b^0 + (n 1) a^n1 * b^1 + ..... + (n n1) a^1 b^n1 + (n n) a^0 * b^n1
Or shorter (a+b)^n = n Σ k=0 (n k) a^nk * b^k
not bad
good
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