Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 4 · Chapter companion part
Explain square identities, expand signed expressions, recover values from sums/products and respect reciprocal conditions.
Class 8 English Version: part A of the chapter 4 companion. Eight sections and 30 written tasks. Selected companion coverage, not complete textbook coverage; use all four parts and compare the mapped source methods. The software checks stated numerical values or coefficients; symbolic work, equivalent forms and proofs need guardian or teacher review.
Open this topic in Skills Lab →
An identity is an equality true for every allowed value of its variables. For example (a + b)² = a² + 2ab + b² follows by multiplying (a + b)(a + b). An equation such as x + 2 = 5 instead restricts x. Trying several numerical examples can reveal a false identity, but does not by itself prove that a statement works for every value.
The four products in (a + b)(a + b) are a², ab, ba and b². Since ab = ba, the two equal cross products combine to 2ab. For (2x + 3y)², use whole terms a = 2x and b = 3y: 4x² + 12xy + 9y². In the positive-length area model, two different rectangles each have area ab; there is no missing middle rectangle.
Replace b by −b in the sum-square identity to get (a − b)² = a² − 2ab + b². For (3x − 4y)², the middle term is −24xy while the last term is +16y². Also (−2x − y)² = (2x + y)². Squaring a negative entire expression makes it non-negative at real inputs; it does not erase every internal cross-product sign.
Conjugates a + b and a − b multiply to a² − b² because the cross products cancel. Thus (5x + 2)(5x − 2) = 25x² − 4. More generally (u + r)(u + s) = u² + (r + s)u + rs. For (3x + 2)(3x − 5), use u = 3x, r = 2, s = −5 to get 9x² − 9x − 10.
Expand (a + b + c)² as a² + b² + c² + 2ab + 2bc + 2ca. There are three square cells and six off-diagonal cells in the positive-length grid. For signed terms, substitute them with signs attached: (2x + 3y − z)² = 4x² + 9y² + z² + 12xy − 4xz − 6yz. Grouping u = 2x + 3y and then squaring u − z gives the same result.
Let s = a + b, d = a − b and p = ab. Then a² + b² = s² − 2p = d² + 2p; d² = s² − 4p; s² = d² + 4p. Also 2(a² + b²) = s² + d² and 4ab = s² − d². If s = 11 and p = 24, the square sum is 73 and d² = 25. A squared difference gives only |d|; choosing d = 5 requires an ordering such as a > b.
For expressions U and V, set M = (U + V)/2 and D = (U − V)/2. Then U = M + D, V = M − D and UV = M² − D². For U = 2x + 5 and V = 4x − 1, M = 3x + 2 and D = −x + 3. Their product is (3x + 2)² − (x − 3)² = 8x² + 18x − 5. Define a temporary expression without confusing it with a new independent variable.
For x ≠ 0, x × (1/x) = 1. If s = x + 1/x, then x² + 1/x² = s² − 2 and x⁴ + 1/x⁴ = (s² − 2)² − 2. If t = x − 1/x, then (x + 1/x)² = t² + 4. Source exercise 4.4 gives x² − √3 x + 1 = 0: completing the square gives (x − √3/2)² + 1/4 = 0, so there is no real solution. Its formal complex-number consequences are different from a feasible real example; our numerical variants state feasible conditions.
For (100 + 3)², compute 10000 + 600 + 9 = 10609. The 600 is 2 × 100 × 3, not 3² or a second copy of 10000. Multiplying 103 × 103 checks the result independently.
The tile lengths a = 4 and b = 2 are original teaching values. Negative variables require algebraic reasoning; the drawing does not represent negative physical area.
The source exercise with coefficient √3 has no real root, although formal complex consequences can be computed. We explain that limitation and use feasible real conditions for automatic tasks.
These are exact identities for permitted real variables, not measured observations.
| Structure | Identity | What to notice |
|---|---|---|
| Square of a sum | (a + b)² = a² + 2ab + b² | Two cross products |
| Square of a difference | (a − b)² = a² − 2ab + b² | Last square is positive |
| Conjugates | (a + b)(a − b) = a² − b² | Cross products cancel |
| Common first term | (u + r)(u + s) = u² + (r + s)u + rs | Use signed r and s |
| Three-term square | (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca | Three paired cross products |
For (3x + 2)(3x − 5), the common first term is 3x and the signed constants are 2 and −5. The middle coefficient is (2 − 5) × 3 = −9.
Distribute first; adding and subtracting the expansions supplies all six square corollaries.
(a + b)² = a² + 2ab + b² (a − b)² = a² − 2ab + b² a² + b² = (a + b)² − 2ab = (a − b)² + 2ab (a + b)² = (a − b)² + 4ab (a − b)² = (a + b)² − 4ab 2(a² + b²) = (a + b)² + (a − b)² 4ab = (a + b)² − (a − b)²
An area diagram uses positive lengths. The algebraic derivation also handles signed values. A squared difference does not fix a signed difference.
Original positive lengths a = 4, b = 2. Each side of the complete square is 6 units.
The pieces meet without overlap. Play is optional; Previous/Next and all static stages remain available. Reduced motion disables timed playback, and changing lesson mode pauses it. Physical lengths are positive.
The mean and half-difference use complete expressions, including their signs.
M = (U + V)/2 D = (U − V)/2 UV = (M + D)(M − D) = M² − D²
Changing the sign of D does not change D². This is an identity, not a need to solve for x.
Division by x requires x ≠ 0; check that a stated real condition is possible.
s = x + 1/x; t = x − 1/x x² + 1/x² = s² − 2 = t² + 2 x⁴ + 1/x⁴ = (s² − 2)² − 2 (x + 1/x)² = t² + 4
The source’s formal complex consequences do not establish a real solution. Never divide by a variable before checking the condition.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original positive-length model, drawn with one scale for every region. It illustrates an identity; negative physical lengths or negative-area tiles are not assumed.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact positive-length grid. The general signed identity is established by algebra, rather than by negative physical areas.
Show your working on paper. Check your answer, then choose Next.
10404
(100 + 2)² = 10000 + 400 + 4 = 10404.
9604
(100 − 2)² = 10000 − 400 + 4 = 9604.
12
2 × 2 × 3 = 12.
9
(3x)² gives 9x².
-24
2 × 3 × (−4) = −24.
73
11² − 2 × 24 = 73.
53
5² + 2 × 14 = 53.
16
10² − 4 × 21 = 16.
20
4² + 4 = 20.
7
3² − 2 = 7.
47
The square sum is 7, so the fourth-power sum is 7² − 2 = 47.
25
The product is 25x² − 4.
1
4 + (−3) = 1.
-6
2 × 3 × (−1) = −6.
20
4ab = 9² − 1² = 80, so ab = 20.
Write in your own words. A guardian or teacher should review this response.
4x² + 12xy + 9y². Multiplication produces two 6xy rectangles, whose sum is 12xy.
Show all four products and combine like terms.
Write in your own words. A guardian or teacher should review this response.
9x² − 24xy + 16y². The cross products are −12xy each; (−4y)² is positive.
Check the middle and last signs separately.
Write in your own words. A guardian or teacher should review this response.
The whole expression is −(2x + y). Its square equals (2x + y)² = 4x² + 4xy + y².
Do not replace the middle plus with a minus simply because the original terms were negative.
Write in your own words. A guardian or teacher should review this response.
4x² + 9y² + z² + 12xy − 4xz − 6yz. The three cross terms are 2(2x)(3y), 2(2x)(−z) and 2(3y)(−z).
Require all three squares and all three signed pair terms.
Write in your own words. A guardian or teacher should review this response.
x⁶ − 4x⁴y + 4x²y². The middle product is −2(x³)(2xy), and powers of x add in a product.
Do not square only the coefficient of a compound term.
Write in your own words. A guardian or teacher should review this response.
Let U = x + 2y. Product = U² − (3z)² = x² + 4xy + 4y² − 9z².
Name the complete conjugate terms before expanding.
Write in your own words. A guardian or teacher should review this response.
9x² − 9x − 10, using (r + s)u = (2 − 5)(3x).
Check the linear coefficient and constant product.
Write in your own words. A guardian or teacher should review this response.
Mean M = 3x + 2; half-difference D = −x + 3. Product = (3x + 2)² − (x − 3)² = 8x² + 18x − 5.
Either sign for the squared half-difference is equivalent; show its derivation.
Write in your own words. A guardian or teacher should review this response.
The expression is (U + V)² = (3a + 2b)² = 9a² + 12ab + 4b².
Substitute whole expressions and collect before expanding.
Write in your own words. A guardian or teacher should review this response.
The square sum is 121 − 48 = 73. The squared difference is 121 − 96 = 25.
Explain why knowing the squared difference alone does not fix its sign.
Write in your own words. A guardian or teacher should review this response.
Divide by x to obtain x + 1/x = 3. The square sum is 7 and the fourth-power sum is 47. This quadratic has real nonzero solutions.
State the division condition and both identity steps.
Write in your own words. A guardian or teacher should review this response.
Over real values a − b may be 4 or −4. Under the added condition a > b it must be 4.
Never infer a positive sign from a square without the ordering condition.
Write in your own words. A guardian or teacher should review this response.
For a > b > 0, cut a b-by-b square from an a-by-a square and rearrange the remaining pieces into a rectangle with sides a + b and a − b. Algebraic multiplication proves the identity for all real a,b; the physical-area drawing needs positive lengths.
Explain the two different domains and preserve area during rearrangement.
Write in your own words. A guardian or teacher should review this response.
Add the two expansions to get 2(a² + b²) = (a + b)² + (a − b)². Subtract to get 4ab = (a + b)² − (a − b)². Rearranging either original expansion gives the two square-sum forms and the two sum/difference-square forms.
Write the rearrangements with their signs; no division by a variable is needed.
Write in your own words. A guardian or teacher should review this response.
The expression is a² − b² + 4ab = (a + b)(a − b) + 4ab. Since (a − b)² = 100 − 84 = 16 and a > b, the difference is 4. The value is 10 × 4 + 84 = 124. Values a = 7, b = 3 give 13² − 45 = 124.
Derive the transformation for all allowed a,b. Checking one pair alone is not a proof of the identity.
Build the positive-length square model, then write an identity page with signed and reciprocal examples.
Keep your work on paper or in a local file. This page does not upload it or automatically grade the project.
This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 4, printed pages 48–76. It does not replace the complete chapter or its practical teaching.
Official book catalogue and source →
Teacher-led lessons, schedules and prices for this outline are upcoming. You can use any starter practice above now without booking.
Solo practice works without an account. Sign in to save a practice record to your learner profile.
Exact lesson links and related activities are labelled separately. Return-to-course links retain your selected learning path.
Play this exact lesson in Skills Lab →