NCERT Solutions Class 9 Maths Chapter 4 Think and Reflect — Exploring Algebraic Identities

NCERT Class 9 Mathematics — Ganita Manjari Part I, Exploring Algebraic Identities. All 16 Think and Reflect prompts answered.

Also here. Every exercise in this chapter is solved, one page per exercise.

Chapter 4 sets eleven Think and Reflect boxes into its text, holding sixteen prompts. They do more work here than in most chapters: several of them are where an identity is actually discovered, with the main text only writing down the result afterwards.

Three in particular are worth attempting before reading on — the comparison of $(a+b)^2$ with $a^2 + b^2$, the identity hidden in Fig. 4.6, and the disagreement between James and Reshma.

Key insight. Almost every prompt here turns on one fact: $(a+b)^2$ and $a^2 + b^2$ differ by exactly $2ab$. That single term decides which is larger, explains what happens when $b$ is replaced by $-b$, and is the term the algebra-tile rectangles are made of. When a prompt looks unfamiliar, ask where the $2ab$ has gone.

Prompt 1 comes from the box on page 69, after Example 1 has shown that for any three consecutive square numbers, (smallest $+$ largest) $-$ $2\times$(middle) always equals $2$.

Question 1

Try and find other patterns like this one. For example, you could consider $4$ consecutive squares and see if you can find a pattern.

Solution. First, see why the three-square pattern works, because that shows what to look for. Writing the three consecutive squares as $n^2$, $(n+1)^2$, $(n+2)^2$:

$$n^2 + (n+2)^2 – 2(n+1)^2 = n^2 + n^2 + 4n + 4 – 2n^2 – 4n – 2 = 2$$

Everything with an $n$ in it cancels, leaving the constant $2$ — which is why the answer never depends on which squares you start from.

Now four consecutive squares. Take $n^2$, $(n+1)^2$, $(n+2)^2$, $(n+3)^2$ and try the natural analogue: add the outer two and subtract the inner two.

$$n^2 + (n+3)^2 – (n+1)^2 – (n+2)^2$$ $$= n^2 + (n^2 + 6n + 9) – (n^2 + 2n + 1) – (n^2 + 4n + 4)$$ $$= (n^2 + n^2 – n^2 – n^2) + (6n – 2n – 4n) + (9 – 1 – 4) = 4$$

Always $4$. Test it: with $1, 4, 9, 16$ we get $1 + 16 – 4 – 9 = 4$ ✓, and with $25, 36, 49, 64$ we get $25 + 64 – 36 – 49 = 4$ ✓

Another pattern worth finding. The difference between consecutive squares:

$$(n+1)^2 – n^2 = 2n + 1$$

always an odd number, and every odd number appears exactly once. That is why the squares are the running totals of the odd numbers: $1$, $1+3 = 4$, $1+3+5 = 9$, $1+3+5+7 = 16$.

Prompts 2 to 4 come from the box on page 71, after the text observes that $(a+b)^2 \neq a^2 + b^2$.

Question 2

What can you say about $a$ and $b$ if $(a + b)^2 < a^2 + b^2$?

Solution. Compare the two by subtracting. Expanding the left-hand side:

$$(a+b)^2 – \left( a^2 + b^2 \right) = a^2 + 2ab + b^2 – a^2 – b^2 = 2ab$$

So the whole comparison reduces to the sign of $2ab$ — that is, the sign of $ab$.

For $(a+b)^2$ to be smaller, we need $2ab < 0$, so $ab < 0$: the product of $a$ and $b$ is negative, which happens exactly when one is positive and the other negative.

Test it: $a = 5$, $b = -2$ gives $(a+b)^2 = 9$ and $a^2 + b^2 = 29$, and indeed $9 < 29$.

Question 3

What can you say about $a$ and $b$ if $(a + b)^2 > a^2 + b^2$?

Solution. By the same subtraction, this needs $2ab > 0$, so $ab > 0$ — $a$ and $b$ have the same sign. Either both are positive or both are negative.

Test both cases: $a = 10$, $b = 2$ gives $144 > 104$ ✓; and $a = -10$, $b = -2$ gives $144 > 104$ as well, since $2ab = 40$ is still positive.

Question 4

When will $(a + b)^2$ be equal to $a^2 + b^2$?

Solution. Equality needs the difference to vanish:

$$2ab = 0$$

A product is zero only when one of its factors is zero, so

$$a = 0 \qquad \text{or} \qquad b = 0$$

(or both). In that case the “identity” $(a+b)^2 = a^2 + b^2$ does hold — but only for those values, which is exactly why it is not an identity in general.

Check: $a = 7$, $b = 0$ gives $49 = 49 + 0$ ✓

The box’s closing hint asks which term decides the comparison, and the answer to all three prompts is the same: the middle term $2ab$ of the expansion. It is present in $(a+b)^2$ and absent from $a^2 + b^2$, and that single term accounts for the whole difference.

Prompt 5 comes from the box on page 73, before the identity for $(a-b)^2$ is derived.

Question 5

What if we replace $b$ by $-b$ in $(a + b)^2 = a^2 + 2ab + b^2$?

Solution. An identity holds for every value of its letters, so $b$ may be replaced by $-b$ throughout and the result is still true. Substitute carefully:

$$\left( a + (-b) \right)^2 = a^2 + 2a(-b) + (-b)^2$$

Simplify each piece. On the left, $a + (-b) = a – b$. On the right, $2a(-b) = -2ab$, and $(-b)^2 = +b^2$ since squaring removes the sign. So

$$(a – b)^2 = a^2 – 2ab + b^2$$

which is the second identity of the chapter, obtained without any new work.

This is worth noticing as a method, not just a result. Substituting into an identity you already have is one of the cheapest ways to produce a new one — and it is how the chapter derives $(a+b+c)^2$ as well, by replacing $b$ with $b+c$.

Prompt 6 comes from the box on page 76, beside Fig. 4.4.

Question 6

Label the squares and rectangles in Fig. 4.4 so that it represents the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

Solution. Fig. 4.4 is a square of side $a + b + c$, cut by two horizontal and two vertical lines into a $3 \times 3$ grid. The columns have widths $a$, $b$, $c$ and the rows have heights $a$, $b$, $c$, so each of the nine pieces is a rectangle whose area is the product of its row and column labels.

	width $a$	width $b$	width $c$
height $a$	$a^2$	$ab$	$ac$
height $b$	$ab$	$b^2$	$bc$
height $c$	$ac$	$bc$	$c^2$

Now add the nine areas. Three are squares, sitting on the diagonal:

$$a^2 + b^2 + c^2$$

The other six come in matching pairs, one on each side of that diagonal:

$$ab + ab = 2ab, \qquad bc + bc = 2bc, \qquad ac + ac = 2ca$$

Total area:

$$a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

And the same square, measured as a whole, has area $(a+b+c)^2$. The two must agree, which is the identity.

The picture also explains the coefficient $2$. Each cross term appears twice because each pair of lengths meets in two different rectangles — once above the diagonal and once below.

Prompts 7 and 8 come from the box on page 78.

Question 7

Try to evaluate the following using a suitable identity: (i) $35^2$ (ii) $65^2$ (iii) $85^2$ (iv) $105^2$. Do you observe any interesting pattern?

Solution. Each number ends in $5$, so write it as (a multiple of ten) $+\,5$ and use $(a+b)^2$.

$$35^2 = (30+5)^2 = 900 + 300 + 25 = 1225$$ $$65^2 = (60+5)^2 = 3600 + 600 + 25 = 4225$$ $$85^2 = (80+5)^2 = 6400 + 800 + 25 = 7225$$ $$105^2 = (100+5)^2 = 10000 + 1000 + 25 = 11025$$

The pattern. Every answer ends in $25$, and the part before it is:

$$12, \qquad 42, \qquad 72, \qquad 110$$

Compare with the tens digits $3$, $6$, $8$, $10$:

$$3 \times 4 = 12, \qquad 6 \times 7 = 42, \qquad 8 \times 9 = 72, \qquad 10 \times 11 = 110$$

Each is the tens part multiplied by one more than itself. So the rule is: to square a number ending in $5$, multiply the part before the $5$ by the next whole number, and write $25$ after it.

Why it works, with $n$ for the part before the $5$ — the number is $10n + 5$:

$$(10n + 5)^2 = 100n^2 + 100n + 25 = 100\,n(n+1) + 25$$

The $100$ is what shifts $n(n+1)$ two places left, leaving room for the $25$.

Question 8

Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.

Solution. Read the side lengths off Fig. 4.6.

The first row holds four squares, of sides

$$a + b + c, \qquad a + b – c, \qquad a – b + c, \qquad a – b – c$$

The second row holds three squares, of sides $2a$, $2b$ and $2c$.

Two rows of figures representing an identity means their total areas are equal:

$$(a+b+c)^2 + (a+b-c)^2 + (a-b+c)^2 + (a-b-c)^2 = (2a)^2 + (2b)^2 + (2c)^2$$

Check it by expanding. Each of the four squares on the left contributes $a^2 + b^2 + c^2$, giving $4a^2 + 4b^2 + 4c^2$ in total. The cross terms cancel in threes:

$$ab: \; +2 +2 -2 -2 = 0 \qquad bc: \; +2 -2 +2 -2 = 0 \qquad ca: \; +2 -2 -2 +2 = 0$$

So the left-hand side is $4a^2 + 4b^2 + 4c^2$, and the right-hand side is $4a^2 + 4b^2 + 4c^2$ as well ✓

$$\boxed{(a+b+c)^2 + (a+b-c)^2 + (a-b+c)^2 + (a-b-c)^2 = 4a^2 + 4b^2 + 4c^2}$$

This is worth comparing with Question 4 of Exercise Set 4.3, which asks whether the three-bracket version is an identity. It is not — and this figure shows why: the four brackets here form a complete symmetric set whose cross terms cancel, and dropping any one of them spoils that balance.

Prompt 9 comes from the box on page 79, after $(x+3)(x+4) = x^2 + 7x + 12$ has been laid out with algebra tiles.

Question 9

Suppose $7x$ is split as $2x + 5x$; can a similar rectangular arrangement be formed? Consider other possibilities and check.

Solution. The tiles are one $x^2$ tile, seven $x$ tiles and twelve unit tiles, and they must fill a rectangle with no gaps. Splitting $7x$ as $mx + nx$ makes the sides $x + m$ and $x + n$, whose product is

$$(x + m)(x + n) = x^2 + (m + n)x + mn$$

For this to be the expression we have, two conditions must hold: $m + n = 7$, and $mn = 12$.

Try $2x + 5x$: here $m = 2$, $n = 5$. The sum is $7$ ✓, but the product is $2 \times 5 = 10$, not $12$ ✗. A rectangle of sides $x+2$ and $x+5$ would need only ten unit tiles, leaving two over. No rectangle can be formed.

The other splits of $7$:

Split	$m \times n$	Works?
$1x + 6x$	$6$	no
$2x + 5x$	$10$	no
$3x + 4x$	$12$	yes

Only $3x + 4x$ gives both the right number of $x$ tiles and the right number of unit tiles, producing the rectangle $(x+3)(x+4)$.

That is the whole method of splitting the middle term: the split must satisfy two conditions, not one, and it is the product that usually decides it.

Prompts 10 to 12 come from the second box on page 79.

Question 10

Algebra tiles can be used to represent products and find factors.

Solution. This is a statement to be understood rather than a question, and the idea behind it is worth stating clearly.

Three kinds of tile are used:

• a large square of side $x$, with area $x^2$;
• a rectangular strip of sides $x$ and $1$, with area $x$;
• a small square of side $1$, with area $1$.

An expression such as $x^2 + 7x + 12$ says which tiles you have: one large square, seven strips, twelve small squares.

Multiplying is building: lay the tiles out as a rectangle of known sides and read off the total area. Factorising is the reverse: given the tiles, arrange them into a rectangle and read off the sides. Those sides are the factors.

This works because a rectangle’s area is the product of its sides, which is exactly what a factorisation asserts — so an arrangement with no gaps or overlaps is a proof of the factorisation.

Question 11

Figure out the product of $x + 2$ and $x + 3$ using algebra tiles.

Solution. Build a rectangle with one side $x + 2$ and the other $x + 3$, then count what fills it.

Split each side into its parts and take the four resulting blocks:

• $x \times x$ — one $x^2$ tile
• $x \times 3$ — three strips
• $2 \times x$ — two strips
• $2 \times 3$ — six unit squares

Counting up: one $x^2$ tile, $3 + 2 = 5$ strips, and $6$ unit squares. So

$$(x + 2)(x + 3) = x^2 + 5x + 6$$

Check by the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$ with $a = 2$, $b = 3$: the middle coefficient is $5$ and the constant is $6$ ✓

Notice where the middle term comes from in the picture: it is the two strip regions, $3x$ along one edge and $2x$ along the other. That is the geometric reason the middle coefficient is a sum.

Question 12

Lay out algebra tiles for $x^2 + 11x + 30$ in such a way that you will see its factors.

Solution. The tiles available are one $x^2$ tile, eleven strips and thirty unit squares. Arranging them into a rectangle means splitting the eleven strips between the two edges, say $m$ and $n$ with $m + n = 11$, in such a way that the corner block of unit squares is exactly $m \times n = 30$.

Test the factor pairs of $30$: $1 \times 30$, $2 \times 15$, $3 \times 10$, $5 \times 6$. Their sums are $31$, $17$, $13$ and $11$ — and the last is the one wanted.

So put five strips along one edge and six along the other:

• one $x^2$ tile in the corner,
• five strips along one side and six along the other ($11$ in all),
• a $5 \times 6$ block of thirty unit squares filling the opposite corner.

The rectangle formed has sides $x + 5$ and $x + 6$, so

$$x^2 + 11x + 30 = (x + 5)(x + 6)$$

Check: $(x+5)(x+6) = x^2 + 11x + 30$ ✓

Prompt 13 comes from the box on page 80.

Question 13

We have seen that $(x + 3)(x + 4) = x^2 + 7x + 12$. Also $(x + 6)(x + 7) = x^2 + 13x + 42$. Generalise the pattern to get an expression for $(x + a)(x + b)$.

Solution. Look at how the two given results are built from their brackets.

Product	Middle coefficient	Constant
$(x+3)(x+4)$	$7 = 3 + 4$	$12 = 3 \times 4$
$(x+6)(x+7)$	$13 = 6 + 7$	$42 = 6 \times 7$

In both, the middle coefficient is the sum of the two numbers and the constant is their product. So the general rule is

$$(x + a)(x + b) = x^2 + (a + b)x + ab$$

Why, by multiplying out:

$$(x+a)(x+b) = x \cdot x + x \cdot b + a \cdot x + a \cdot b = x^2 + bx + ax + ab$$

and collecting the two middle terms gives $(a+b)x$. The middle coefficient is a sum precisely because the middle comes from two separate products.

This identity is what makes factorising quadratics possible: to factor $x^2 + px + q$, find two numbers adding to $p$ and multiplying to $q$.

Prompt 14 comes from the box on page 82.

Question 14

James and Reshma were talking about algebraic identities. James writes $(a – b)^2(a + b) = (a^2 – 2ab + b^2)(a + b)$. Reshma writes $(a – b)^2(a + b) = (a – b)\left[ (a – b)(a + b) \right] = (a – b)(a^2 – b^2)$. According to you, who is correct and why?

Solution. Both are correct. They have not disagreed about a fact; they have chosen different ways of grouping the same product.

James expands the squared bracket first, then multiplies by $(a+b)$.

Reshma splits $(a-b)^2$ as $(a-b)(a-b)$, pairs one of those with $(a+b)$ to make a difference of squares, and is left with $(a-b)(a^2 – b^2)$.

Both are legitimate because multiplication is associative and commutative: the three factors $(a-b)$, $(a-b)$ and $(a+b)$ may be multiplied in any order.

Check that they agree. James’s version:

$$(a^2 – 2ab + b^2)(a+b) = a^3 + a^2b – 2a^2b – 2ab^2 + ab^2 + b^3 = a^3 – a^2b – ab^2 + b^3$$

Reshma’s version:

$$(a – b)(a^2 – b^2) = a^3 – ab^2 – a^2b + b^3 = a^3 – a^2b – ab^2 + b^3$$

Identical ✓

Which is better? Reshma’s, for the work involved: her route needs one multiplication of a binomial by a binomial, while James’s needs a trinomial times a binomial — six products instead of four. Spotting that two of the three factors form $a^2 – b^2$ is what saves the effort.

The box also invites you to combine identities and find new results, and this is an example: $(a-b)^2(a+b) = (a-b)(a^2-b^2)$ is itself a small identity discovered by regrouping.

Prompt 15 comes from the box on page 85.

Question 15

We already know that $x^2 – y^2 = (x-y)(x+y)$, and we have verified that $x^3 – y^3 = (x-y)(x^2 + xy + y^2)$. Observe that $x – y$ is a common factor of $x^2 – y^2$ and $x^3 – y^3$. Generalise.

Solution. The observation is correct and it continues for every power. Look at the second factor in each case:

$$x^2 – y^2 = (x – y)(x + y)$$ $$x^3 – y^3 = (x – y)(x^2 + xy + y^2)$$ $$x^4 – y^4 = (x – y)(x^3 + x^2y + xy^2 + y^3)$$ $$x^5 – y^5 = (x – y)(x^4 + x^3y + x^2y^2 + xy^3 + y^4)$$

The pattern in the long bracket: the powers of $x$ count down from $n-1$ to $0$ while the powers of $y$ count up from $0$ to $n-1$, every term has total degree $n-1$, all signs are positive, and there are $n$ terms.

$$x^n – y^n = (x – y)\left( x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + xy^{n-2} + y^{n-1} \right)$$

Why $x – y$ is always a factor. Substituting $x = y$ makes $x^n – y^n$ equal to zero, and a value that makes an expression zero corresponds to a factor — so $(x – y)$ divides $x^n – y^n$ for every $n$.

Check $n = 4$: $(x-y)(x^3 + x^2y + xy^2 + y^3)$ expands to $x^4 – y^4$, every middle term cancelling against the next ✓ (This also factorises further, as $(x-y)(x+y)(x^2+y^2)$, because $4$ is even.)

Prompt 16 comes from the box on page 87, just before Exercise Set 4.5.

Question 16

Try to simplify the following rational expression: $\dfrac{36s^2 – 12st + t^2}{t^2 + 2ts – 48s^2} = \dfrac{(6s-t)^2}{(\underline{\phantom{x}} + \underline{\phantom{x}})(\underline{\phantom{x}} + \underline{\phantom{x}})}$.

Solution. The numerator is given in factored form already:

$$36s^2 – 12st + t^2 = (6s – t)^2$$

(a perfect square, since $2(6s)(t) = 12st$).

Factorise the denominator, as the hint says. Treat $t^2 + 2ts – 48s^2$ as a quadratic in $t$: two terms multiplying to $-48s^2$ and adding to $2s$. The factor pairs of $48$ are $1\times48$, $2\times24$, $3\times16$, $4\times12$, $6\times8$ — and $8s$ with $-6s$ gives a sum of $2s$ ✓

$$t^2 + 2ts – 48s^2 = (t + 8s)(t – 6s)$$

So the blanks are filled by $t + 8s$ and $t + (-6s)$:

$$\frac{(6s – t)^2}{(t + 8s)(t – 6s)}$$

Now cancel. The numerator involves $(6s – t)$ and the denominator $(t – 6s)$, which look different — but they are negatives of each other, and the numerator is squared, so the sign does not survive:

$$(6s – t)^2 = \left[ -(t – 6s) \right]^2 = (t – 6s)^2$$

Rewriting and cancelling one factor:

$$\frac{(t – 6s)^2}{(t + 8s)(t – 6s)} = \frac{t – 6s}{t + 8s}$$

Common mistakes

• Question 1, testing one set of four squares and stopping. Two or three examples suggest a pattern; writing the squares as $n^2$, $(n+1)^2$, $(n+2)^2$, $(n+3)^2$ and watching the $n$ terms cancel is what shows it always holds.
• Questions 2 to 4, forgetting that $a$ and $b$ may be negative. With both negative, $ab > 0$, so $(a+b)^2$ is the larger — the same as when both are positive. It is the sign of the product that matters, not of the numbers separately.
• Question 5, writing $(-b)^2 = -b^2$. Squaring removes the sign, so only the middle term changes. Getting this wrong gives $a^2 – 2ab – b^2$, which is not a square of anything.
• Question 6, labelling only the three squares. Six rectangles also have to be labelled, and it is their pairing that produces the coefficient $2$.
• Question 7, stopping at the four values. The prompt asks for the pattern, and the pattern — multiply the tens part by the next number and append $25$ — is the point.
• Question 9, checking only the sum. $2 + 5 = 7$ is correct and the split still fails, because $2 \times 5 \neq 12$. Both conditions must hold.
• Question 12, trying $x^2 + 11x + 30 = (x+2)(x+15)$. The product is right but the sum is $17$, not $11$ — so the tiles would not form a rectangle.
• Question 14, deciding one of them must be wrong. Both are correct. A disagreement about method is not a disagreement about fact, and checking is a matter of expanding both.
• Question 15, expecting alternating signs in the long bracket. For $x^n – y^n$ all the signs in the second factor are positive. (It is $x^n + y^n$, for odd $n$, whose second factor alternates.)
• Question 16, saying $(6s-t)$ and $(t-6s)$ cannot cancel. They are negatives of each other, and because the numerator is squared the minus signs pair off. Rewriting $(6s-t)^2$ as $(t-6s)^2$ makes the cancellation visible.

Practise next

• Exercise Set 4.3, Question 4 — asks whether the three-bracket version of the Question 8 identity holds. It does not, and seeing why is the best follow-up to Fig. 4.6.
• Exercise Set 4.4 — the splitting-the-middle-term method that Questions 9, 12 and 13 build up to, applied systematically.
• Exercise Set 4.5 — six rational expressions to simplify, of which Question 16 above is a worked warm-up.