MathematicsClass 9ICSE

Square Any Number Near a Hundred Without Writing Anything Down

Expand the square of a binomial and use it to square 103 and 97 mentally, square a three-term expression with the correct signs, expand a product of two binomials, and combine two expansions.

How can you square 103 in your head in two seconds?

Multiplying

103 \times 103

by long multiplication takes a while and offers several places to go wrong.

Now do it this way.

103

100 + 3

, and

(100 + 3)^{2} = 10 0^{2} + 2 (100) (3) + 3^{2} = 10000 + 600 + 9 = 10609

Three terms, each of them trivial, and the answer arrives before you have finished writing the question down.

The same trick works below a round number.

97 = 100 - 3

, so

(100 - 3)^{2} = 10000 - 600 + 9 = 9409

What makes it work is an identity — a statement true for every value of the letters, not an equation to be solved.

(a + b)^{2} = a^{2} + 2 ab + b^{2}

holds whatever

a

and

b

are, so you are free to choose them to make the arithmetic easy.

That is the real content of this chapter. An identity is a pattern you install once and then apply wherever you recognise the shape — in algebra to expand brackets, and in arithmetic to avoid long multiplication.

This page covers the first part of the ICSE Class 9 Mathematics chapter on expansions: the square of a binomial, the square of a three-term expression, the product of two binomials, and expressions built by combining more than one expansion.

Formula

How do you expand the square of a binomial?

Square the first term, square the second, and put twice their product in between — with the sign of the middle term matching the sign in the bracket.

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(a - b)^{2} = a^{2} - 2 ab + b^{2}

Notice that the last term is

+ b^{2}

in both, because a square is never negative. Only the middle term changes sign.

Worked numerical example 1. Evaluate

10 3^{2}

.

Choose

a = 100

and

b = 3

10 3^{2} = (100 + 3)^{2} = 10 0^{2} + 2 (100) (3) + 3^{2} = 10000 + 600 + 9 = 10609

Worked numerical example 2. Evaluate

9 7^{2}

.

Choose

a = 100

and

b = 3

, with a minus:

9 7^{2} = (100 - 3)^{2} = 10000 - 600 + 9 = 9409

Worked algebraic examples.

(2 x + 3 y)^{2} = (2 x)^{2} + 2 (2 x) (3 y) + (3 y)^{2} = 4 x^{2} + 12 x y + 9 y^{2}

(3 a - 5 b)^{2} = (3 a)^{2} - 2 (3 a) (5 b) + (5 b)^{2} = 9 a^{2} - 30 ab + 25 b^{2}

Check the second by substituting numbers, which is the fastest way to catch a slip. Put

a = 1

and

b = 1

: the bracket is

(3 - 5)^{2} = 4

, and the expansion gives

9 - 30 + 25 = 4

. Agreed.

Now the error this identity exists to prevent.

(a + b)^{2}

is not

a^{2} + b^{2}

. The middle term

2 ab

is not optional.

Test it on numbers:

(3 + 4)^{2} = 49

, while

3^{2} + 4^{2} = 25

. The gap of

24

is exactly

2 ab = 2 (3) (4)

.

So squaring does not distribute over addition, and the

2 ab

is the whole reason. **This is the same mistake as writing

a + b = a + b

in the surds chapter — an operation being treated as distributive when it is not.

Choosing

a

and

b

well is the whole skill in numerical questions.** For

10 3^{2}

you could take

a = 103

and

b = 0

, which is useless, or

a = 50

and

b = 53

, which is worse than the original problem. **Pick the round number as

a

and the small remainder as

b

**, and the identity does the rest.

How do you square an expression with three terms?

Square each of the three terms, then add twice each of the three possible products.

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)

There are three squares and three cross products, so six terms in all.

When some signs are negative, the squares stay positive and each cross product takes the product of the two signs involved:

(a + b - c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab) - 2 (b c) - 2 (c a)

(a - b - c)^{2} = a^{2} + b^{2} + c^{2} - 2 (ab) + 2 (b c) - 2 (c a)

The rule for each cross term: it is positive if the two signs are the same and negative if they differ. In

(a - b - c)^{2}

the product of

- b

and

- c

is positive, which is why

+ 2 (b c)

appears there.

Worked algebraic example. Expand

(2 x + 3 y - 4 z)^{2}

.

Take

a = 2 x

b = 3 y

c = - 4 z

.

- The three squares:

(2 x)^{2} = 4 x^{2}

(3 y)^{2} = 9 y^{2}

(- 4 z)^{2} = 16 z^{2}

2 \times (2 x) \times (3 y) = 12 x y

2 \times (3 y) \times (- 4 z) = - 24 y z

2 \times (- 4 z) \times (2 x) = - 16 z x

(2 x + 3 y - 4 z)^{2} = 4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 z x

Worked numerical check of the identity itself. Take

a = 10

b = 5

c = 2

. Then

a + b + c = 17

and

1 7^{2} = 289

.

Now through the identity:

- Squares:

100 + 25 + 4 = 129

2 \times 10 \times 5 = 100

2 \times 5 \times 2 = 20

2 \times 2 \times 10 = 40

129 + 100 + 20 + 40 = 289

The identity checks out.

Now the part that is most often got wrong. Students remember the three squares and lose one of the three cross products — usually the one joining the first term to the last, because it is the pair you do not read consecutively.

The safe method is to write the three pairs down before evaluating any of them: first with second, second with third, third with first. Three pairs, always, however many minus signs are involved.

And the count is worth remembering as a check. A three-term square has six terms. If your answer has five, a cross product is missing; if it has seven, something has been written twice. Counting the terms before you move on is a two-second check that catches the commonest error in this section.

How do you expand the product of two different binomials?

**The coefficient of

x

is the sum of the two numbers, and the constant is their product.**

(x + a) (x + b) = x^{2} + (a + b) x + ab

With minus signs, the same identity works if you carry the signs into

a

and

b

(x + a) (x - b) = x^{2} + (a - b) x - ab

(x - a) (x - b) = x^{2} - (a + b) x + ab

Worked algebraic example. Expand

(x + 5) (x - 3)

.

Here

a = 5

and

b = - 3

, so the sum is

5 + (- 3) = 2

and the product is

5 \times (- 3) = - 15

(x + 5) (x - 3) = x^{2} + 2 x - 15

Worked numerical example 1. Evaluate

103 \times 107

.

Take

x = 100

a = 3

b = 7

103 \times 107 = 10 0^{2} + (3 + 7) (100) + (3) (7) = 10000 + 1000 + 21 = 11021

Worked numerical example 2, with one number below the round figure. Evaluate

98 \times 103

.

Take

x = 100

a = - 2

b = 3

98 \times 103 = 10000 + (- 2 + 3) (100) + (- 2) (3) = 10000 + 100 - 6 = 10094

Worked numerical example 3, with both below. Evaluate

96 \times 97

.

Take

x = 100

a = - 4

b = - 3

96 \times 97 = 10000 + (- 4 - 3) (100) + (- 4) (- 3) = 10000 - 700 + 12 = 9312

Check that one by ordinary arithmetic:

96 \times 97 = 96 \times 100 - 96 \times 3 = 9600 - 288 = 9312

. Agreed.

Notice how this identity contains the previous one. Put

b = a

and it becomes

(x + a) (x + a) = x^{2} + 2 a x + a^{2}

which is exactly

(x + a)^{2}

. So the square of a binomial is the special case of this product where the two numbers are equal, and that is worth seeing once, because it means there are fewer identities to remember than the textbook suggests.

And notice how it contains the difference of two squares as well. Put

b = - a

(x + a) (x - a) = x^{2} + (a - a) x - a^{2} = x^{2} - a^{2}

The middle term vanishes because the two numbers cancel.

So one identity generates three, depending on whether the two numbers are equal, opposite, or neither — and recognising which case you are looking at is faster than recalling three separate formulae.

How do you simplify an expression made of two squares?

Expand each square separately, then collect like terms — and watch for cross terms that cancel or double.

The two most useful combinations are worth knowing as results in their own right.

(a + b)^{2} + (a - b)^{2} = 2 (a^{2} + b^{2}) (a + b)^{2} - (a - b)^{2} = 4 ab

Why they come out so neatly. Write both expansions one above the other:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(a - b)^{2} = a^{2} - 2 ab + b^{2}

Add them and the

2 ab

terms cancel, leaving

2 a^{2} + 2 b^{2}

. Subtract them and the squares cancel instead, leaving

2 ab - (- 2 ab) = 4 ab

.

Numerical check. Take

a = 7

and

b = 4

. Then

(11)^{2} + (3)^{2} = 121 + 9 = 130

, and

2 (49 + 16) = 2 (65) = 130

. And

(11)^{2} - (3)^{2} = 121 - 9 = 112

, while

4 ab = 4 (28) = 112

. Both agree.

Worked example 1. Simplify

(2 x + 3)^{2} + (2 x - 3)^{2}

= (4 x^{2} + 12 x + 9) + (4 x^{2} - 12 x + 9) = 8 x^{2} + 18

The

x

terms cancel, as the identity predicts:

2 (a^{2} + b^{2})

with

a = 2 x

and

b = 3

gives

2 (4 x^{2} + 9) = 8 x^{2} + 18

.

Worked example 2. Simplify

(3 x - 2 y)^{2} - (3 x + 2 y)^{2} + 24 x y

= (9 x^{2} - 12 x y + 4 y^{2}) - (9 x^{2} + 12 x y + 4 y^{2}) + 24 x y

= - 24 x y + 24 x y = 0

The whole expression is zero for every value of

x

and

y

— which the identity shows at once, since the difference of the two squares is

- 4 ab = - 4 (3 x) (2 y) = - 24 x y

.

Worked example 3 — the identities used backwards. If

a + b = 9

and

ab = 20

, find

a^{2} + b^{2}

and

a - b

.

From

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a^{2} + b^{2} = (a + b)^{2} - 2 ab = 81 - 40 = 41

And from

(a - b)^{2} = (a + b)^{2} - 4 ab

(a - b)^{2} = 81 - 80 = 1 \Rightarrow a - b = \pm 1

Check it. The two numbers with sum

9

and product

20

are

4

and

5

. Indeed

16 + 25 = 41

, and

5 - 4 = 1

.

Now the point of example 3, which is the most useful idea in the chapter. You were never told what

a

and

b

were, and you did not need to find out.

An identity lets you compute a combination of two unknowns from another combination of them, without ever solving for the unknowns themselves. That is why these questions are set — not to test expansion, but to test whether you can see which identity connects what you are given to what is wanted. The next part of this chapter is almost entirely that skill, applied to cubes.

Exam tip

Exam tip: count the terms and check with small numbers

(a \pm b)^{2}

has THREE terms** and

(a + b + c)^{2}

has SIX. Count them before moving on — a missing term is the commonest error here.

**In

(a - b)^{2}

only the middle term is negative.** The last term is

+ b^{2}

, because a square cannot be negative.

**

(a + b)^{2} \neq = a^{2} + b^{2}

.** Check on numbers:

(3 + 4)^{2} = 49

while

9 + 16 = 25

, and the gap

24

is exactly

2 ab

.

For a three-term square, list the three PAIRS first — first with second, second with third, third with first. The last one is the one people forget.

Each cross term is positive if the two signs agree and negative if they differ. So in

(a - b - c)^{2}

the term joining

- b

and

- c

is positive.

**For

(x + a) (x + b)

, the coefficient of

x

is the SUM and the constant is the PRODUCT.** Carry minus signs into

a

and

b

rather than remembering separate versions.

For a numerical square, pick the round number and the small remainder.

103 = 100 + 3

97 = 100 - 3

9812 = 9800 + 12

.

For a numerical product, use the same round number for both:

96 \times 97 = (100 - 4) (100 - 3)

.

Learn the two combinations:

(a + b)^{2} + (a - b)^{2} = 2 (a^{2} + b^{2})

and

(a + b)^{2} - (a - b)^{2} = 4 ab

.

And their backwards forms:

a^{2} + b^{2} = (a + b)^{2} - 2 ab

and

(a - b)^{2} = (a + b)^{2} - 4 ab

.

**When asked for

a - b

from a square, give BOTH signs** —

(a - b)^{2} = 1

means

a - b = \pm 1

.

And **always check an algebraic expansion by substituting

a = 1, b = 1

** or another easy pair. It takes five seconds and catches a wrong sign every time.

Did you know

Why a mental arithmetic trick and an algebraic identity are the same thing

There is a pleasing unity in this chapter that is easy to miss because the two halves look so different.

One half is algebra: expand

(2 x + 3 y)^{2}

, collect like terms, tidy the answer. It feels like a manipulation exercise.

The other half is mental arithmetic: square

103

, multiply

96

97

. It feels like a party trick.

They are the same operation. In the algebra the letters stay letters; in the arithmetic you choose

a = 100

and let the letters become numbers. Nothing else changes.

And that is what an identity is — a statement about a shape rather than about particular quantities.

(a + b)^{2} = a^{2} + 2 ab + b^{2}

is not a fact about any specific numbers. It is a fact about what happens whenever you square a sum, and it is therefore available for any sum you meet.

Which is why the arithmetic shortcut is not a separate thing to be learnt. Once you have the identity, the shortcut is the identity used well — and using it well means choosing

a

and

b

so that

a^{2}

is easy,

b^{2}

is easy and

2 ab

is easy.

That choice is the whole skill. To square

103

, take

a = 100

: then

a^{2} = 10000

needs no thought,

b = 3

gives

b^{2} = 9

, and

2 ab = 600

. To square

48

, take

a = 50

and

b = - 2

2500 - 200 + 4 = 2304

. To square

9997

, take

a = 10000

and

b = - 3

100000000 - 60000 + 9 = 99940009

.

Every one of those is the same three-term calculation with the digits shifted. The larger the number, the more the shortcut saves — which is the opposite of long multiplication, where a bigger number means more work.

So the identity is not an alternative to arithmetic; it is the reason the arithmetic can be reorganised — and a student who sees

(a + b)^{2}

as a shape rather than a formula gets both halves of this chapter for the price of one.

Exam relevance

Why does JEE Main assume you know these identities cold?

Because they are not a topic in Class 11 — they are the notation everything else is written in, and no marks are given for them.

This is the foundation for Class 11 Mathematics Complex Numbers and Quadratic Equations, Binomial Theorem, Straight Lines and Conic Sections, examined in JEE Main. Nothing in those chapters will ask you to expand

(a + b)^{2}

— they will assume you did it correctly three lines earlier while doing something else.

The binomial theorem is this chapter generalised. Class 11 gives

(a + b)^{n} = \sum (r n) a^{n - r} b^{r}

, and putting

n = 2

returns

a^{2} + 2 ab + b^{2}

exactly. **So the identity you learn here is the

n = 2

row of Pascal's triangle**, and the coefficients

1, 2, 1

are its entries. Questions on a particular term or coefficient of an expansion are among the most reliably examined items in JEE Main, and they are this pattern extended.

The backwards use is the whole technique of quadratic roots. If

α

and

β

are the roots of

a x^{2} + b x + c = 0

, then

α + β = - \frac{b}{a}

and

α β = \frac{c}{a}

— and questions then ask for

α^{2} + β^{2}

, or

\frac{1}{α} + \frac{1}{β}

, or

(α - β)^{2}

. **Every one is answered with

α^{2} + β^{2} = (α + β)^{2} - 2 α β

and

(α - β)^{2} = (α + β)^{2} - 4 α β

— the two results derived on this page, applied to roots you never compute. This is one of the most frequent JEE Main question types in algebra.

Completing the square is this identity read backwards.** Solving a quadratic, finding the vertex of a parabola and converting a circle's equation to centre-radius form all work by recognising

x^{2} + 2 a x

as the start of

(x + a)^{2}

and supplying the missing

a^{2}

. Conic Sections uses it constantly, and the three-term square appears when two variables are involved.

The three-term square appears in coordinate geometry and vectors. The expansion of

(a + b + c)^{2}

is the pattern behind

∣ u + v + w ∣^{2}

in Class 12 Vector Algebra, where the cross terms become dot products. So the rule about three pairs rather than two is the same rule there, and forgetting the third pair is the same error.

Trigonometric identities use the same algebra. Expanding

(sin θ + cos θ)^{2}

to get

1 + 2 sin θ cos θ

is this chapter with trigonometric letters, and **questions giving

sin θ + cos θ

and asking for

sin θ cos θ

are standard — answered by exactly the backwards method of the last section.

What the questions look like. For board work, expect expand a given square or product, evaluate a numerical square or product using an identity, and simplify an expression combining two expansions. The identity used should be quoted. For JEE Main, expect binomial coefficients, symmetric functions of quadratic roots, completing the square in conics, and vector magnitudes.

How board and competitive emphasis differ. A board paper rewards the expansion written out with the identity named. A competitive paper never asks for the expansion — it asks for something that cannot be reached without it.

The single trap that costs the most marks. Losing one of the three cross terms in a three-term square. There are three pairs — first with second, second with third, and third with first — and the last is the one skipped, because it is the only pair you do not read consecutively. A correct three-term square has six terms**, so counting them is a complete check. The defence is to write the three pairs down as pairs before evaluating any of them, because once they are on the page in a list of three, none can go missing.

Key takeaways

Squares and products of binomials: quick revision

- **

(a + b)^{2} = a^{2} + 2 ab + b^{2}

and

(a - b)^{2} = a^{2} - 2 ab + b^{2}

. Only the middle** term changes sign; the last is always

+ b^{2}

.
- **

(a + b)^{2} \neq = a^{2} + b^{2}

** — check on numbers:

(3 + 4)^{2} = 49

against

9 + 16 = 25

, and the gap

24

2 ab

.
- **

10 3^{2} = (100 + 3)^{2} = 10000 + 600 + 9 = 10609

9 7^{2} = (100 - 3)^{2} = 10000 - 600 + 9 = 9409

.
-

(2 x + 3 y)^{2} = 4 x^{2} + 12 x y + 9 y^{2}

;

(3 a - 5 b)^{2} = 9 a^{2} - 30 ab + 25 b^{2}

.
-

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)

— three squares and three cross products, six terms in all.
- Each cross term is positive if the two signs agree and negative if they differ.** So

(a - b - c)^{2} = a^{2} + b^{2} + c^{2} - 2 (ab) + 2 (b c) - 2 (c a)

.
- **

(2 x + 3 y - 4 z)^{2} = 4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 z x

.
- Check the identity on numbers**: with

10, 5, 2

the squares give

129

and the cross terms

100 + 20 + 40

, totalling

289 = 1 7^{2}

.
- List the three PAIRS before evaluating — first with second, second with third, third with first.
- **

(x + a) (x + b) = x^{2} + (a + b) x + ab

** — the coefficient of

x

is the sum, the constant the product. Carry minus signs into

a

and

b

.
- **

(x + 5) (x - 3) = x^{2} + 2 x - 15

.
-

103 \times 107 = 10000 + 1000 + 21 = 11021

;

98 \times 103 = 10000 + 100 - 6 = 10094

;

96 \times 97 = 10000 - 700 + 12 = 9312

.
- One identity generates three**: put

b = a

to get

(x + a)^{2}

; put

b = - a

to get

x^{2} - a^{2}

.
- **

(a + b)^{2} + (a - b)^{2} = 2 (a^{2} + b^{2})

— the cross terms cancel.
-

(a + b)^{2} - (a - b)^{2} = 4 ab

— the squares cancel.
- Check with

a = 7, b = 4

**:

121 + 9 = 130 = 2 (65)

, and

121 - 9 = 112 = 4 (28)

.
- **

(2 x + 3)^{2} + (2 x - 3)^{2} = 8 x^{2} + 18

;

(3 x - 2 y)^{2} - (3 x + 2 y)^{2} + 24 x y = 0

** for all

x, y

.
- Backwards forms: **

a^{2} + b^{2} = (a + b)^{2} - 2 ab

and

(a - b)^{2} = (a + b)^{2} - 4 ab

.
- Given

a + b = 9

and

ab = 20

**:

a^{2} + b^{2} = 81 - 40 = 41

, and

(a - b)^{2} = 81 - 80 = 1

a - b = \pm 1

. The numbers are

4

and

5

.
- An identity lets you find a combination of two unknowns from another combination — without ever solving for the unknowns.

Square

48

and multiply

96

104

using nothing but these identities, then check both by ordinary multiplication — if they match, the chapter is secure.

How can you square 103 in your head in two seconds?

How do you expand the square of a binomial?

How do you square an expression with three terms?

How do you expand the product of two different binomials?

How do you simplify an expression made of two squares?

Exam tip: count the terms and check with small numbers

Why a mental arithmetic trick and an algebraic identity are the same thing

Why does JEE Main assume you know these identities cold?

Squares and products of binomials: quick revision

Related study notes

Ready to put this into practice?