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Std 7 Mathematics · Chapter 14

14. Algebraic Formulae – Expansion of Squares

Step-by-step solutions to Practice Sets 50 to 53: expanding squares of binomials, (a + b)(a − b), and factorising. Maharashtra State Board (Balbharati), Grade Seven Mathematics.

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Formulae: (a + b)2 = a2 + 2ab + b2; (a − b)2 = a2 − 2ab + b2; (a + b)(a − b) = a2 − b2.

Practice Set 50

1Expand.

(i) (5a + 6b)2
= (5a)2 + 2(5a)(6b) + (6b)2
= 25a2 + 60ab + 36b2

(ii) (a2 + b3)2
= (a2)2 + 2 × a2 × b3 + (b3)2
= a24 + ab3 + b29

(iii) (2p − 3q)2
= (2p)2 − 2(2p)(3q) + (3q)2
= 4p2 − 12pq + 9q2

(iv) (x − 2x)2
= x2 − 2 × x × 2x + (2x)2
= x2 − 4 + 4x2

(v) (ax + by)2
= (ax)2 + 2(ax)(by) + (by)2
= a2x2 + 2abxy + b2y2

(vi) (7m − 4)2
= (7m)2 − 2(7m)(4) + 42
= 49m2 − 56m + 16

(vii) (x + 12)2
= x2 + 2 × x × 12 + (12)2
= x2 + x + 14

(viii) (a − 1a)2
= a2 − 2 × a × 1a + (1a)2
= a2 − 2 + 1a2

2Which of the options given below is the square of the binomial (8 − 1x) ? (i) 64 − 1x2 (ii) 64 + 1x2 (iii) 64 − 16x + 1x2 (iv) 64 + 16x + 1x2

(8 − 1x)2
= 82 − 2 × 8 × 1x + (1x)2
= 64 − 16x + 1x2

Answer: option (iii)

3Of which of the binomials given below is m2n2 + 14mnpq + 49p2q2 the expansion? (i) (m + n)(p + q) (ii) (mn − pq) (iii) (7mn + pq) (iv) (mn + 7pq)

m2n2 = (mn)2, 49p2q2 = (7pq)2 and 14mnpq = 2 × mn × 7pq
∴ m2n2 + 14mnpq + 49p2q2 = (mn + 7pq)2

Answer: (iv) (mn + 7pq)

4Use an expansion formula to find the values. (i) (997)2 (ii) (102)2 (iii) (97)2 (iv) (1005)2

(i) (997)2 = (1000 − 3)2 = 10002 − 2 × 1000 × 3 + 32 = 1000000 − 6000 + 9 = 994009

(ii) (102)2 = (100 + 2)2 = 1002 + 2 × 100 × 2 + 22 = 10000 + 400 + 4 = 10404

(iii) (97)2 = (100 − 3)2 = 1002 − 2 × 100 × 3 + 32 = 10000 − 600 + 9 = 9409

(iv) (1005)2 = (1000 + 5)2 = 10002 + 2 × 1000 × 5 + 52 = 1000000 + 10000 + 25 = 1010025

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Practice Set 51

1Use the formula to multiply the following.

(i) (x + y)(x − y) = x2 − y2

(ii) (3x − 5)(3x + 5) = (3x)2 − 52 = 9x2 − 25

(iii) (a + 6)(a − 6) = a2 − 62 = a2 − 36

(iv) (x5 + 6)(x5 − 6) = (x5)2 − 62 = x225 − 36

2Use the formula to find the values. (i) 502 × 498 (ii) 97 × 103 (iii) 54 × 46 (iv) 98 × 102

(i) 502 × 498 = (500 + 2)(500 − 2) = 5002 − 22 = 250000 − 4 = 249996

(ii) 97 × 103 = (100 − 3)(100 + 3) = 1002 − 32 = 10000 − 9 = 9991

(iii) 54 × 46 = (50 + 4)(50 − 4) = 502 − 42 = 2500 − 16 = 2484

(iv) 98 × 102 = (100 − 2)(100 + 2) = 1002 − 22 = 10000 − 4 = 9996

Practice Set 52

1Factorise the following expressions and write them in the product form. (i) 201a3b2 (ii) 91xyt2 (iii) 24a2b2 (iv) tr2s3

(i) 201a3b2 = 3 × 67 × a × a × a × b × b
(ii) 91xyt2 = 7 × 13 × x × y × t × t
(iii) 24a2b2 = 2 × 2 × 2 × 3 × a × a × b × b
(iv) tr2s3 = t × r × r × s × s × s

Factorise the coefficient into primes first (201 = 3 × 67, 91 = 7 × 13), then write each variable as many times as its index.

Practice Set 53

1Factorise the following expressions.

Use a2 − b2 = (a + b)(a − b) or take out the common factor.

(i) p2 − q2
= (p + q)(p − q)

(ii) 4x2 − 25y2
= (2x)2 − (5y)2
= (2x + 5y)(2x − 5y)

(iii) y2 − 4
= y2 − 22
= (y + 2)(y − 2)

(iv) p2 − 125
= p2 − (15)2
= (p + 15)(p − 15)

(v) 9x2 − 116y2
= (3x)2 − (y4)2
= (3x + y4)(3x − y4)

(vi) x2 − 1x2
= x2 − (1x)2
= (x + 1x)(x − 1x)

(vii) a2b − ab
(common factor: ab)
= ab(a − 1)

(viii) 4x2y − 6x2
(common factor: 2x2)
= 2x2(2y − 3)

(ix) 12y2 − 8z2
= 12(y2 − 16z2)
= 12[y2 − (4z)2]
= 12(y + 4z)(y − 4z)

(x) 2x2 − 8y2
= 2(x2 − 4y2)
= 2[x2 − (2y)2]
= 2(x + 2y)(x − 2y)

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