PrepShorts · Study sheet · Class 9 Mathematics · Chapter 4, Exploring Algebraic Identities
Chapter 4 · Exploring Algebraic Identities
Reading (a + b)² off a partitioned square
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Put one mark on a segment and build the square on it. Two cuts, four pieces — and the four pieces are the identity, already assembled.
The idea
The four pieces of the partitioned square are not a memory aid for four terms. They are the reason there are exactly two ab rectangles and not one: cutting the horizontal side at a and cutting the vertical side at the same place are two independent decisions, so the corner pieces come in a matched pair. That is what 2ab means. And the companion picture for (a − b)² shows why a diagram cannot simply be mirrored to get a subtraction: there the identity has to be built by taking rectangles away, and the +b² at the end is not decoration — it is the amount by which the second rectangle you remove falls short of the first.
What you should be able to do
- Draw a segment of length a + b as a segment of length a followed by one of length b, and mark all three lengths
- Construct a square on that segment and account for every one of its four pieces by area
- State the area of the whole square in two ways and read the identity (a + b)² = a² + 2ab + b² off the two statements
- Explain why the two rectangles are congruent and why there are two of them rather than one
- State the restriction under which the area argument is valid, and say what the chapter does next because of that restriction
- Use the identity to square a two-digit or three-digit number by splitting it at a round value
- Obtain (a − b)² = a² − 2ab + b² from (a + b)² = a² + 2ab + b² by replacing b with −b, and check the signs of all four terms
- Account for (a − b)² inside a square of side a by subtracting two rectangles, and explain where the +b² comes from
- Use (a − b)² to square a number that sits just below a round value
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| line segment | a straight piece of a line with two endpoints, so it has a definite length | printed in this chapter (§4.2, p. 69) |
| partition | a cutting of a region into pieces that between them use it up exactly once | printed in this chapter (§4.2, p. 69) |
| geometrical model | a drawing whose areas or volumes stand for the terms of an algebraic statement | printed in this chapter (§4.4, Fig. 4.4 caption, p. 76) |
| square of side | the square drawn on a stated length, whose area is that length squared | printed in this chapter (§4.2, Fig. 4.2 caption, p. 69) |
| area | the amount of surface a region covers, measured in square units | printed in this chapter (§4.2, p. 69) |
| square units | the unit area is used in — the square built on the unit length | printed in this chapter (§4.3, p. 74) |
| rectangle | a four-sided figure with all angles right angles | printed in this chapter (§4.2, p. 69) |
| identity | an equality that holds whatever the letters stand for | printed in this chapter (§4.2, p. 70) |
| distributive property | the rule a(b + c) = ab + ac, which the printed derivation of Fig. 4.3 uses | printed in this chapter (§4.2, p. 70) |
| binomial | a two-term expression, the thing being squared here | printed in this chapter (§4.2, p. 71) |
| cross term | the 2ab piece — the part of the expansion that is neither a² nor b² | an added term; the chapter draws the two rectangles and gives them no collective name |
| dissection argument | a proof that rearranges pieces of a figure without changing total area | an added phrasing; not printed in this chapter |
Where people slip up
- "a² + 2ab + b² has four terms because the square has four pieces, and that is the whole story." The count is a consequence, not a reason. The reason there are two ab pieces is that the horizontal cut and the vertical cut are made independently, so each of a and b pairs with the other exactly once.
- "The two ab rectangles are the same rectangle." They are congruent but differently oriented: one is b wide and a tall, the other a wide and b tall. The chapter's own artwork gives them the same fill, which is a hint about equal area, not about being one object.
- "The picture proves the identity for all numbers." The page itself limits the claim to lengths, and lengths are positive. The general claim is earned three paragraphs later by distributivity.
- "(a − b)² = a² − b²." This is the commonest error in the whole chapter, and Fig. 4.3 kills it visually: a² − b² is not even the right shape — removing a b-by-b corner from a square of side a leaves an L, not a square.
- "Replacing b by −b flips every sign." It flips one. a² has no b in it, and (−b)² is positive. Walk the three terms separately.
- "The +b² in (a − b)² is a fudge to make the numbers work." It is the overlap you would otherwise remove twice. Show the movement of the two removals and let the corner get taken away, then given back.
- "You can only use the identity when a and b are whole numbers." Exercise Set 4.1 puts fractions and reciprocals into both slots; nothing in the derivation cared.
- "Splitting 193 as 200 − 7 is harder than just multiplying." Have the explanation race the two methods once. The point of the identity here is not elegance, it is that two of the three pieces are trivial to compute.
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Worked answers to this chapter’s exercises · this video explains Exercise Set 4.1 Q1, Exercise Set 4.1 Q2, Exercise Set 4.2 Q2, Exercise Set 4.4 Q2, End-of-Chapter Exercises Q7
Transcript1,324 words
Draw a straight segment, and put one mark on it. Call the piece to the left of the mark a, and the piece to the right b. The whole segment is a plus b long. That is the entire setup. There is no arithmetic in it, and there are no numbers in it at all. Everything that follows comes from building a square on that segment. So build it. A square whose side is a plus b, on all four sides.
Now carry that mark around the edge, and cut. Once across, once down, both at the same place. Two cuts. Four pieces. The bottom left piece is a square of side a, so its area is a squared. The top right is a square of side b, area b squared. And the two pieces left over are rectangles. Not squares. Look hard at those two rectangles, because they are the whole point.
The one above the a square is a across and b up. The one beside it is b across and a up. Same area, a times b, both of them. Different shape, unless the mark happens to sit exactly in the middle. And here is why there are two of them rather than one. Cutting the horizontal side at a, and cutting the vertical side at a, are two separate decisions.
So each of the two lengths gets to pair with the other, and there are two ways round to do it. That is what the two in two a b is counting. It is not a tidying-up step. It is a fact about the drawing. Now describe the area of the whole square twice over. Once as a whole. The side is a plus b, so the area is a plus b, all squared.
And once piece by piece. a squared, plus a b, plus a b, plus b squared. Those pieces were cut from that square, nothing was thrown away and nothing was counted twice. Rebuilt on sixteen hundred different pairs of lengths, with no overlap anywhere and nothing poking out. So the two descriptions are of the same thing, and an equals sign goes between them. a plus b, all squared, equals a squared, plus two a b, plus b squared.
Before going any further, read the fine print. Every a and every b in that argument was a length. A side you could actually draw. Lengths are positive. Nobody can draw a segment of length minus three. So what the picture has established is the identity for positive numbers, and nothing else. That is a real limitation, and not a small one, because most of the numbers you will want to use it on are not the length of anything.
Closing that gap takes a different kind of argument, made of algebra rather than of area. Meanwhile the picture is already worth something. Square forty three. Draw the square of side forty three, and put the mark at forty. The four pieces are a forty by forty tile, which is sixteen hundred; two forty by three strips, two hundred and forty between them; and a three by three corner, which is nine.
Sixteen hundred, plus two hundred and forty, plus nine. Eighteen forty nine. That is the same drawing, to scale. Sixty four splits at sixty and gives four thousand and ninety six. A hundred and five splits at a hundred, for eleven thousand and twenty five. Two hundred and five gives forty two thousand and twenty five. Now push on that fine print instead of accepting it. What if b were negative?
Take the identity as it stands, and replace b by minus b everywhere it appears. a squared has no b in it at all, so it does not move. Two a b becomes two a times minus b, which is minus two a b. That one flips. And b squared becomes minus b, all squared, which is plus b squared. That one does not flip. One sign changed out of three. So the square of a minus b is a squared, minus two a b, plus b squared.
Almost everybody flips the last sign as well. There is nothing there to flip: a minus times a minus is a plus. Can a picture do that? Drawings add areas. Subtracting is harder to draw. Here is how it goes. Start with a square of side a, the bigger of the two lengths. Take a strip of width b off the right hand side, running the full height. Its area is a b.
What is left is a rectangle, a minus b across and a tall. Cut that at height b. The bottom of it is a strip of height b, and the top is a square of side a minus b, which is exactly the thing we are after. Three pieces, and they fill the square exactly. Rebuilt on seven hundred and eighty pairs of lengths, with no overlap and nothing left over.
So take the square of a minus b out of that. It is the whole square, less the tall strip, less the bottom strip. a squared, minus a b, minus that bottom strip. And here is the thing everybody misses. The bottom strip is not a b. It has height b, but it only runs a minus b across, because the tall strip already took the rest. So its area is a b, less b squared.
Which means that if you subtract a b twice, you have taken away b squared too much. The plus b squared at the end of the identity is that corner going back on. It is not decoration and it is not a fudge. It is the piece you removed twice. Which hands you the other shortcut. Square twenty nine by treating it as thirty minus one. Nine hundred, minus sixty, plus one. Eight hundred and forty one.
Seventy nine is eighty minus one, giving six thousand two hundred and forty one. Two hundred and ninety nine is three hundred minus one, giving eighty nine thousand four hundred and one. The rule for choosing is just to go to the nearest round number, whichever side of you it is on. One warning, though. A hundred and ninety three is two hundred minus seven, and seven squared is forty nine.
Drop that last term and your answer is wrong by forty nine. Do the same on two hundred and ninety nine and you are wrong by one. Which is exactly why the last term looks ignorable right up until it is not. The second drawing also kills the commonest mistake in the whole subject. a minus b, all squared, is not a squared minus b squared. Look at what a squared minus b squared actually is. Take the square of side a, and cut a b by b corner out of it.
What is left is not a square. It is an L. Its area is a minus b, times a plus b. A rectangle, and a square only when b is nought. On a grid of one thousand six hundred and eighty one pairs, the wrong version happens to come out right eighty one times, and those eighty one are exactly the cases where b is nought or a equals b. Every other pair says no.
So what have the two drawings actually bought? The first one hands you the identity, and hands you the reason there are two middle rectangles rather than one, which no amount of expanding brackets will ever show you. The second hands you the minus version, and the reason its last term is positive. Neither of them covers a single negative number, because neither of them is able to draw one.
A picture can show you why. It cannot show you how far. For that you need the algebra. And once you have the algebra, the pictures are the part you remember.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- An identity holds for every value; an equation need notClass 9 · Ch 4, Exploring Algebraic Identities
Comes up again in
- Recognising an expression as an identity in disguiseClass 9 · Ch 4, Exploring Algebraic Identities
- (a + b + c)² by substitution, and the square that proves itClass 9 · Ch 4, Exploring Algebraic Identities
- Algebra tiles: factorising by rebuilding the rectangleClass 9 · Ch 4, Exploring Algebraic Identities
- Simplifying a rational expression, and the factor you must not cancelClass 9 · Ch 4, Exploring Algebraic Identities