ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.5

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.5

Question 1.
Using suitable identities, find the following products:
(i) (3x + 5) (3x + 5)
(ii) (9y – 5) (9y-5)
(iii) (4x + 11y) (4x – 11y)
(iv) \(\left(\frac{3}{2} m+\frac{2}{3} n\right)\left(\frac{3}{2} m-\frac{2}{3} n\right)\)
(v) \(\left(\frac{2}{a}+\frac{5}{b}\right)\left(\frac{2}{a}+\frac{5}{b}\right)\)
(vi) \(\left(\frac{p^{2}}{2}+\frac{2}{q^{2}}\right)\left(\frac{p^{2}}{2}-\frac{2}{q^{2}}\right)\)
Solution:
(i) (3x + 5)(3x + 5)
= (3x + 5)²
= (3x)² + 2 × 3x × 5 + (5)²
= 9x² + 30x + 25
{(a + b)² = a² + 2zb + b²}

(ii) (9y – 5)(9y – 5)
= (9y – 5)²
= (9y)² – 2 × 9y × 5 + (5)²
= 81y² – 90y + 25

(iii) (4x + 11y)(4x – 11y)
= (4x)² – (11y)²
{(a + b)(a – b) = a² – b²}
= 16x² – 121y²


= \(\frac{p^{4}}{4}-\frac{4}{q^{4}}\)

Question 2.
Using the identities, evaluate the following:
(i) 81²
(ii) 97²
(iii) 105²
(iv) 997²
(v) 6.1²
(vi) 496 × 504
(vii) 20.5 × 19.5
(viii) 9.62
Solution:
(i) (81)² = (80 + 1)²
= (80)² + 2 × 80 × 1 + (1)² {(a + b)² = a² + 2ab + b²}
= 6400 + 160+ 1 = 6561

(ii) (97)² = (100 – 3)²
= (100)² – 2 × 100 × 3 + (3)² {(a – b)² = a² – 2ab + b²}
= 10000 – 600 + 9
= 10009 – 600 = 9409

(ii) (105)² = (100 + 5)²
= (100)² + 2 × 100 × 5 + (5)² {(a + b)² = a² + 2ab + b²}
= 10000+ 1000 + 25 = 11025

(iv) (997)² = (1000 – 3)²
= (1000)² – 2 × 1000 × 3 + (3)² {(a – b)² = a² – 2ab + b²}
= 1000000 – 6000 + 9
= 1000009 – 6000 = 994009

(v) (6.1)² = (6 + 0.1)²
= (6)² + 2 × 6 × 0.1 +(0.1)² {(a + b)² = a² + 2ab + b²}
= 36 + 1.2 + 0.01 = 37.21

(vi) 496 × 504
= (500 – 4) (500 + 4) {(a + b) (a – b) = a² – b²}
= (500)² – (4)²
= 250000 – 16 = 249984

(vii) 20.5 × 19.5
(20 + 0.5) (20 – 0.5) {(a + b) (a – b) = a² – b²}
= (20)² – (0.5)²
= 400 – 0.25 = 399.75

(viii) (9.6)² = (10 – 0.4)²
= (10)² – 2 × 10 × 0.4 + (0.4)² {(a – b)² = a² – 2ab + b²}
= 100 – 8.0 + 0.16 = 92.16

Question 3.
Find the following squares, using the identities:

Solution:
(i) (pq + 5r)²
(pq)² + 2 × pq × 5r + (5r)² {(a + b)² = a² + 2ab + b²}
= p²q² + 10pqr + 25r²

(iii) (\(\sqrt{2}\)a + \(\sqrt{3}\)b)²
= (\(\sqrt{2}\)a)² + 2 × \(\sqrt{2}\)a × \(\sqrt{3}\)b + (\(\sqrt{3}\)b) {(a – b)² = a² – 2ab + b²}
= 2a² + 2\(\sqrt{6}\)ab + 3b²

Question 4.
Using the identity, (x + a) (x + b) = x² + (a + b)x + ab, find the following products:
(i) (x + 7) (x + 3)
(ii) (3x + 4) (3x – 5)
(iii) (p² + 2q) (p² – 3q)
(iv) (abc + 3) (abc – 5)
Solution:
(i) (x + 7) (x + 3)
= (x)² + (7 + 3)x + 7 × 3
= x² + 10x + 21

(ii) (3x + 4) (3x – 5)
= (3x)² + (4 – 5) (3x) + 4 × (-5)
= 9x² – 3x – 20

(iii) (P² + 2q)(p² – 3q)
= (p²)² + (2q – 3q)p² + 2q × (-3q)
= p⁴ – p²q – 6pq

(iv) (abc + 3) (abc – 5)
= (abc)² + (3 – 5)abc + 3 × (-5)
= a²b²c² – 2abc – 15

Question 5.
Using the identity, (x + a) (x + b) = x² + (a + b)x + ab, evaluate the following:
(i) 203 × 204
(ii) 8.2 × 8.7
(iii) 107 × 93
Solution:
(i) 203 × 204
= (200 + 3) (200 + 4)
= (200)² + (3 + 4) × 200 + 3 × 4
= 40000 + 1400 + 12 = 41412

(ii) 8.2 × 8.7
= (8 + 0.2) (8 + 0.7)
= (8)² + (0.2 + 0.7) × 8 + 0.2 × 0.7
= 64 + 8 × (0.9) + 0.14
= 64 + 7.2 + 0.14 = 71.34

(iii) 107 × 93
= (100 + 7) (100 – 7)
= (100)² + (7 – 7) × 100 + 7 × (-7)
= 10000 + 0 – 49 = 9951

Question 6.
Using the identity a² – b² = (a + b) (a – b), find
(i) 53² – 47²
(ii) (2.05)² – (0.95)²
(iii) (14.3)² – (5.7)²
Solution:
(i) 53² – 47²
= (50 + 3) (50 – 3)
= (50)² – (3)²
= 2500 – 9 = 2491

(ii) (2.05)² – (0.95)²
= (2.05 + 0.95) (2.05 – 0.95)
= 3 × 1.10 = 3.3

(iii) (14.3)² – (5.7)²
= (14.3 + 5.7) (14.3 – 5.7)
= 20 × 8.6 = 172

Question 7.
Simplify the following:
(i) (2x + 5y)² + (2x – 5y)²
(ii) \(\left(\frac{7}{2} a-\frac{5}{2} b\right)^{2}-\left(\frac{5}{2} a-\frac{7}{2} b\right)^{2}\)
(iii) (p² – q²r)² + 2p²q²r
Solution:
(i) (2x + 5y)² + (2x – 5y)²
= (2x)² + 2 × 2x × 5y + (5y)² + (2x)² – 2 × 2x × 5y + (5y)²
= 4x² + 20xy + 25y² + 4x² – 20xy + 25y²
= 8x² + 50y²

(ii) \(\left(\frac{7}{2} a-\frac{5}{2} b\right)^{2}-\left(\frac{5}{2} a-\frac{7}{2} b\right)^{2}\)

(iii) (p² – q²r)² + 2p²q²r      {(a – b)² = a² – 2ab + b²}
= (p²)² – 2 × p² × q²r + (q²r)² + 2p²q²r
= p⁴ – 2p²q + q⁴r² + 2p²q²r
= p⁴ + q⁴r²

Question 8.
Show that:
(i) (4x + 7y)² – (4x – 7y)² = 112xy
(ii) \(\left(\frac{3}{7} p-\frac{7}{6} q\right)^{2}+p q=\frac{9}{49} p^{2}+\frac{49}{36} q^{2}\)
(iii) (p – q)(p + q) + (q – r)(q + r) + (r – p) (r + p) = 0
Solution:
(i) (4x + 7y)² – (4x – 7y)² = 112xy
LHS = (4x + 7y)² – (4x – 7y)²
= [(4x)² + 2 × 4x × 7y + (7y)²]
– [(4x)² – 2 × 4x + 7y + (7y)²]
{∵ (a ± b)² = a² ± 2ab + b²}
= (16x² + 56xy + 49y²) – (16x² – 56xy + 49y²)
=- l6x² + 56xy + 49y² – 16x² + 56xy – 49y²
= 112xy = RHS

(iii) (p – q) (p + q) + (q – r) (q + r) + (r – p) (r + p) = 0
LHS = (p – q) (p + q) + (q – r) (q + r) + (r – p)(r + p)
= p² – q² + q² – r² + r² – p²
{(a + b) (a – b) = a² – b²}
= 0 = RHS

Question 9.
If x + \(\frac{1}{x}\) = 2, evaluate:

Solution:
x + \(\frac{1}{x}\) = 2

Question 10.
If x = \(\frac{1}{x}\) = 7, ecaluate:

Solution:
x = \(\frac{1}{x}\) = 7

Question 11.
If x² + \(\frac{1}{x^{2}}\) = 23, evaluate:

Solution:
x² + \(\frac{1}{x^{2}}\) = 23

Question 12.
If a + b = 9 and = 10, find the value of a² + b².
Solution:
a + b = 9, ab = 10
a + b = 9
Squaring both sides,
(a + b)² = (9)
⇒ a² + b² + 2ab = 81
⇒ a² + b² + 2 × 10 = 81
⇒ a² + b² + 20 = 81
⇒ a² + b² = 81 – 20 = 61
∴ a² + b² = 61

Question 13.
If a – b = 6 and a² + b² = 42, find the value of
Solution:
a – b = 6, a² + b² = 42
a – b = 6
Squaring both sides,
(a – b)² = (6)²
⇒ a² + b² – 2ab = 36
⇒ 42 – 2ab = 36
⇒ 2ab = 42 – 36 = 6
⇒ ab = \(\frac{6}{2}\) = 3
∴ ab = 3

Question 14.
If a² + b² = 41 and ab = 4, find the values of
(i) a + b
(ii) a – b
Solution:
a² + b² = 41, ab = 4
(i) (a + b)² = a² + b² + 2ab
= 41 + 2 × 4 = 41 + 8 = 49
= (±7)²
∴ a + b = ±7

(ii) (a – b)² = a² + b² – 2ab
= 41 – 2 × 4 = 41 – 8 = 33
∴ a – b = ±\(\sqrt{33}\)

ML Aggarwal Class 8 Solutions for ICSE Maths