An identity holds for every value; an equation need not

What to assume they know

Multiplying two brackets by the distributive property, and collecting like terms afterwards
Squaring a negative number, and that a product of two negatives is positive (Why a debt times a debt is a fortune)
Adding and multiplying rational numbers, including finding a common denominator (What "rational" means, and why the denominator cannot be zero)
Solving a simple equation such as x² = 25 and knowing it has two solutions
Terms, coefficients and variables in an algebraic expression (Turning a situation into an expression: terms, variables, coefficients)

What they should be able to do

Carry out the chapter's opening trick on three sets of three consecutive square numbers and report that the outcome is 2 each time
State why "it came out 2 three times" is not yet a reason, and name what is missing
Substitute a negative pair and a rational pair into a² + 2ab + b² and into (a + b)², and compare the two results
Derive (a + b)² = a² + 2ab + b² by multiplying (a + b)(a + b) with the distributive property, and identify which step made the argument general
Decide, for a given equation, whether it is an identity or an equation with particular solutions, and justify the verdict
Use the identity to expand the square of a binomial in which a and b are themselves products, such as 5x and 2y
Use the identity to square a two-digit number by splitting it at a round number
Explain which single term decides whether (a + b)² exceeds a² + b², and state the three cases that term produces
Represent three consecutive whole numbers as (n − 1), n and (n + 1) and complete the algebraic explanation of the opening pattern

Where it usually goes wrong

"It worked, so it is true." This is the misconception the chapter is built around. It runs the pattern three times, tests the identity on negatives, tests it on fractions, and then says in print that it is not yet sure. Show a student that a rule which survives four trials can still fail on the fifth, and that no finite number of trials removes that risk.
"But a proof is just a very careful check." No. The distributive argument on p. 70 checks nothing. It contains no numbers, which is the reason it covers the numbers nobody tried.
"One counterexample is not enough to reject a rule." It is. The asymmetry is the point of the section: trials can never confirm an identity, and a single failure refutes it outright.
"(a + b)² = a² + b²." The chapter states the inequality flatly and then spends a whole box on which side is bigger. The missing 2ab is not a correction bolted on; it is two of the four products you get when you open the brackets.
"(a + b)² is always the bigger one." Only when a and b share a sign. The chapter poses exactly this question and does not answer it.
"The middle term is always added, because there is a plus in front." With a = −2/3 and b = 3/4 the middle term is −1. The plus sign in the printed identity belongs to the notation, not to the value.
"If the letters are lengths, the picture settles everything." The page says the drawn square establishes the rule for a and b that are lengths of segments, and then immediately asks what happens for numbers that are not lengths. Negative lengths do not exist; negative numbers do.
"The trick works because 1, 4, 9 are small." The algebra on p. 73 shows the answer never depended on which three squares you picked, and it also shows the answer is 2 and not "about 2".

Questions to check understanding

Expand a square of a binomial in which each part is itself a product or a fraction, e.g. ((7/5)x + (3/2)y)²
Evaluate a two- or three-digit square by splitting the number at a round value, showing the three intermediate terms
Given an equation, state with reasons whether it is an identity, and if not, find the values that satisfy it
Given a claimed identity, either prove it by expansion or produce one set of values that breaks it
Explain, in terms of signs, when (a + b)² is greater than a² + b²
Prove a stated numerical pattern by writing the consecutive numbers as (n − 1), n, (n + 1) — the competency-based form of this chapter's opening

Examples worth working on the board

Inputs, not answers. Values marked Verified are worked out here on the chapter's stated inputs; this book prints no answer key anywhere.

The opening trick, three runs (§4.1, p. 68). Take three consecutive square numbers, add the smallest and the largest, then subtract twice the middle one. Run 1: the squares 1, 4, 9 — the page does this one in full. Run 2: the squares 9, 16, 25, which the printed text sets as a task. Run 3: the squares 25, 36, 49, which the page then works. Verified: run 1 gives 1 + 9 = 10 and 2 × 4 = 8, so 10 − 8 = 2; run 2 gives 9 + 25 = 34 and 2 × 16 = 32, so 34 − 32 = 2; run 3 gives 25 + 49 = 74 and 2 × 36 = 72, so 74 − 72 = 2.
The answer to run 2 is inside the artwork (p. 68). A drawn schoolgirl at a desk carries a thought bubble that does run 2 in hand lettering — 34, minus 2 × 16, then 34 − 32 — and ends with a claim that the outcome is always 2, set large. None of this lettering is in the page's text layer. So the chapter both poses run 2 and answers it, but only in the picture. The explanation can hand the girl's calculation to the student as a check rather than as an authority, because the girl asserts the always that the chapter has not yet earned.
Attempt to break it with negatives (§4.2, p. 70, Example 2). Inputs a = −2 and b = −3. The page computes a + b, its square, then a², b² and 2ab separately. Verified: the sum is −5, so its square is 25; on the other side a² = 4, then b² = 9, then 2ab = 2 × (−2) × (−3) = 12, and the three total 25. The instructive part is that 2ab came out positive here — the middle term of the expansion is not automatically an addition of something positive, it is 2ab carrying 2ab's own sign.
Attempt to break it with fractions (§4.2, p. 70). Inputs a = −2/3 and b = 3/4. Verified: a + b = 1/12, so (a + b)² = 1/144. On the other side, a² = 4/9, 2ab = −1 exactly, b² = 9/16; over the common denominator 144 those are 64/144, −144/144 and 81/144, and 64 − 144 + 81 = 1, giving 1/144. The printed page shows this reduction over 144 step by step; the arithmetic is worth working through because the middle term is the only one that is negative and it is also the only one that is a whole number.
The line that settles it (§4.2, p. 70). (a + b)² is (a + b)(a + b); apply the distributive property once to get a(a + b) + b(a + b), again to get a² + ab + ba + b², and collect ab + ba into 2ab. Note: the step that makes the argument general is ab = ba, commutativity, which the page uses silently. Nothing in the chain names a particular number, which is exactly why it covers every number the chapter has not tried.
Identity against equation (§4.2–§4.3, pp. 70–71). Two statements to put side by side. x² − 1 = 24 is satisfied by x = 5 and by x = −5 and by nothing else. (x + y)² = x² + 2xy + y² is satisfied by every pair (x, y). Both are written with an equals sign; only the second is an identity.
Squaring a binomial (§4.2, p. 71, Example 3). Expand (5x + 2y)² by reading a = 5x and b = 2y. Inputs only. Verified: 25x² + 20xy + 4y².
Squaring a number (§4.2, p. 71, Example 4). Compute 43² by splitting 43 as 40 + 3. Verified: 1600 + 240 + 9 = 1849.
Exercise Set 4.1 (pp. 71–72). Q1 asks for six expansions: (7x + 4y)², ((7/5)x + (3/2)y)², (2.5p + 1.5q)², ((3/4)s + 8t)², (x + 1/(2y))² and (1/x + 1/y)². Q2 asks for three squares by the identity: 64², 105², 205². Verified: 64² = (60 + 4)² = 3600 + 480 + 16 = 4096; 105² = (100 + 5)² = 10000 + 1000 + 25 = 11025; 205² = (200 + 5)² = 40000 + 2000 + 25 = 42025. Note that Q1's last two items put a reciprocal in the b slot, so the "square" of b is 1/(4y²) and 1/y² — the identity does not care that the pieces are fractions.
Which is larger (§4.2, p. 71). The page fixes a = 10, b = 2 and gets (a + b)² = 144 against a² + b² = 104, then asks whether the same ordering holds for every a and b, and leaves three questions in a Think and Reflect box: when is (a + b)² smaller, when larger, when equal. Verified (an added algebra, not the book's): the two sides differ by exactly 2ab, so (a + b)² > a² + b² precisely when ab > 0, i.e. a and b share a sign; (a + b)² < a² + b² precisely when ab < 0, i.e. their signs differ; and the two are equal precisely when ab = 0, i.e. at least one of them is zero. The chapter's own Example 2 (a = −2, b = −3) is a case of the first, and it is worth pointing out that "both negative" lands on the greater side.
Closing the opening trick (§4.3, p. 73). Write three consecutive whole numbers as (n − 1), n, (n + 1). Their squares are (n − 1)², n² and (n + 1)². Inputs: expand the outer two and add. Verified: the −2n and +2n cancel, leaving 2n² + 2; subtracting 2n² leaves 2. The 2 is not a leftover from n at all — it is the two isolated 1s, one from each of the two outer squares, and n never had a chance to appear in the answer.

Figures to have open

The partitioned square of side (a + b), the chapter's Fig. 4.2 (p. 69). Needed here only as a one-panel reminder with its restriction stated; the full treatment belongs to Reading (a + b)² off a partitioned square. Redraw as a schematic.
A sign chart or three-panel comparison for the term 2ab. Standard schematic; the chapter has no figure for this and the argument needs one.
The p. 68 artwork is not required. Its content — run 2 of the trick, hand lettered — should be rebuilt as clean type, because the drawn bubble also asserts an "always" that this topic is deliberately withholding until section 11.
No photograph is needed anywhere in this topic.

Where this sits in the book

NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 4, "Exploring Algebraic Identities": §4.1 Introduction (p. 68), §4.2 Visualising Identities (pp. 69–72), and the closing paragraphs of the pattern argument in §4.3 (p. 73).
Exercise Set 4.1, pp. 71–72.
Think and Reflect boxes at p. 69 (four consecutive squares, an open invitation) and p. 71 (the three ordering questions).
Chapter summary, p. 90, first bullet, which states the "true for all values" property; the full list of the chapter's identities is on p. 91.
Forward pointer: the same identity is used for factorisation in §4.3 (Recognising an expression as an identity in disguise) and re-derived geometrically in Reading (a + b)² off a partitioned square.