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MathematicsClass 9Exploring Algebraic

Exploring Algebraic | Class 9 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 4 min read

Exploring Algebraic | Class 9 Mathematics Notes

Exploring Algebraic – this guide gives you a concise, exam-ready overview of Exploring Algebraic from Class 9 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

4.8 SIMPLIFYING RATIONAL EXPRESSIONS

This section applies the factorisation techniques and algebraic identities learned earlier to simplify rational algebraic expressions. Simplification involves factoring numerator and denominator expressions and cancelling common factors, provided the denominator is not zero. An example simplifies (x² - 7x + 12) / (5x² + 5x - 100) by first factorising numerator as (x - 3)(x - 4) and denominator by taking out common factor 5 and factorising the quadratic as 5(x - 4)(x + 5). The common factor (x - 4) is cancelled to yield (x - 3) / [5(x + 5)]. The section encourages students to try simplifying similar rational expressions by factoring and cancelling common terms. Further, examples involving geometric arrangements illustrate how algebraic expressions represent areas and how factorisation helps find dimensions. For instance, a rectangle composed of squares and rectangular strips arranged by Saira is analysed to find its length and breadth from the total area expression. Another example involves solving a quadratic equation representing the area of a pool to find its length and breadth. The section concludes with exercises to practice simplifying rational expressions, factorisation, and applying identities in real-world contexts.

🧪 Activity: Students are asked to simplify given rational expressions by factoring and cancelling common terms.

🔗 Connection: Prepares for end-of-chapter exercises and summary.

Frequently asked questions

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (i) 117^2 (ii) 78^2 (iii) 198^2 (iv) 214^2 (v) 1104^2 (vi) 1120^2

Solutions:

(i) 117^2 Use (a + b)^2 = a^2 + 2ab + b^2 117 = 100 + 17 So, 117^2 = 100^2 + 210017 + 17^2 = 10000 + 3400 + 289 = 13689

(ii) 78^2 Use (a - b)^2 = a^2 - 2ab + b^2 78 = 80 - 2 So, 78^2 = 80^2 - 2802 + 2^2 = 6400 - 320 + 4 = 6084

(iii) 198^2 Use (a - b)^2 198 = 200 - 2 198^2 = 200^2 - 22002 + 2^2 = 40000 - 800 + 4 = 39204

(iv) 214^2 Use (a + b)^2 214 = 200 + 14 214^2 = 200^2 + 220014 + 14^2 = 40000 + 5600 + 196 = 45796

(v) 1104^2 Use (a + b)^2 1104 = 1100 + 4 1104^2 = 1100^2

2. Factor using suitable identities: (i) 16y^{2} - 24y + 9 (ii) (9/4) s^2 + 6st + 4t^2 (iii) (m^2)/9 + (mk)/3 + (k^2)/4 + 3nk + 2mn + 9n^2 (iv) (p^2)/16 - 2 + 16/(p^2) (v) 9a^{2} + 4b^{2} + c^{2} - 12ab + 6ac - 4bc

Solutions:

(i) 16y^2 - 24y + 9 = (4y)^2 - 24y3 + 3^2 = (4y - 3)^2

(ii) (9/4)s^2 + 6st + 4t^2 = (3s/2)^2 + 2(3s/2)2t + (2t)^2 = (3s/2 + 2t)^2

(iii) (m^2)/9 + (mk)/3 + (k^2)/4 + 3nk + 2mn + 9n^2 Group terms: = (m^2)/9 + (mk)/3 + (k^2)/4 + 2mn + 3nk + 9n^2 Try to write as square of (m/3 + k/2 + 3n)^2 Calculate: (m/3)^2 + (k/2)^2 + (3n)^2 + 2(m/3)(k/2) + 2(k/2)(3n) + 2(m/3)(3n) = (m^2)/9 + (k^2)/4 + 9n^2 + (mk)/3 + 3nk + 2mn Matches given expression exactly. So factorization is (m/3 + k

3. Expand the following using the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca: (i) (p + 3q + 7r)^2 (ii) (3x - 2y + 4z)^2

Solutions:

(i) (p + 3q + 7r)^2 = p^2 + (3q)^2 + (7r)^2 + 2p3q + 23q7r + 27rp = p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp

(ii) (3x - 2y + 4z)^2 = (3x)^2 + (-2y)^2 + (4z)^2 + 23x(-2y) + 2(-2y)4z + 24z3x = 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx

4. Is this an identity? (a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2.

Yes, this is an identity.

Proof: Expand each square:

(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2ac - 2bc (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc (a - b - c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc

Add all three: = (a^2 + b^2 + c^2 + 2ab - 2ac - 2bc) + (a^2 + b^2 + c^2 - 2ab + 2ac - 2bc) + (a^2 + b^2 + c^2 - 2ab - 2ac + 2bc) = 3a^2 + 3b^2 + 3c^2 + (2ab - 2ab - 2ab) + (-2ac + 2ac - 2ac) + (-2bc - 2bc + 2bc) = 3a^2 + 3b^2 + 3c^2 - 2ab - 2ac - 2bc But carefully summing the cross terms: 2ab

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