Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.9

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Find the GCD for the following:
(i) p⁵, p¹¹, p⁹
(ii) 4x³, y³, z³
(iii) 9 a² b² c³, 15 a³ b² c⁴
(iv) 64x⁸, 240x⁶
(v) ab² c³, a² b³ c, a³ bc²
(vi) 35x⁵ y³ z⁴, 49x² yz³, 14xy² z²
(vii) 25 ab³ c, 100 a² bc, 125 ab
(viii) 3abc, 5xyz, 7pqr
Solution:
(i) p⁵, p¹¹, p⁹
p⁵ = p⁵
p⁹ = p⁵ × p⁴
p¹¹ = p⁵ × p⁶
G.C.D. of p⁵, p¹¹, p⁹ = p⁵

(ii) 4x³, y³, z³
G.C.D. of 4x³ = 1 × 4x³
y³ = 1 × y³
z³ = 1 × z³
∴ G.C.D. = 1

(iii) 9 a² b² c³, 15 a³ b² c⁴
9 a² b² c³ = 3 × 3a² b² c³
15 a³ b² c⁴ = 3 × 5a³ b² c⁴
G.C.D. = 9 a² b² c³, 15 a³ b² c⁴ = 3 × a² b² c³ = 3 a² b² c³

(iv) 64x⁸, 240x⁶
64x⁸ = 2 × 2 × 2 × 2 × 2 × 2 × x⁶ × x⁸
240x⁶ = 2 × 2 × 2 × 2 × 3 × 5 x⁶
G.C.D. of 64x⁸, 240x⁶ = 2 × 2 × 2 × 2 x⁶ = 16x⁶

(v) ab² c³, a² b³ c, a³ bc²
ab² c³ = a × b × b × c × c × c
a² b³ c = a × a × b × b × b × c
a³ bc² = a × a × a × b × c × c
G.C.D. of ab² c³, a² b³ c, a³ bc² = a × b × c = abc

(vi) 35x⁵ y³ z⁴, 49x² yz³, 14xy² z²
35x⁵ y³ z⁴ = 5 × 7 x⁵ y³ z⁴
49x² yz³ = 7 × 7 x² y z³
14xy² z² = 2 × 7 x y² z²
G.C.D. is = 7 x y z²

(vii) 25 ab³ c, 100 a² bc, 125 ab
25 ab³ c = 5 × 5 ab³c
100 a² bc = 2 × 5 × 2 × 5 a² b c
125 ab = 5 × 5 × 5 ab
G.C.D is = 5 × 5 ab = 25ab

(viii) 3abc, 5xyz, 7pqr
3abc = 1 × 3 a bc
5xyz = 1 × 5 x y z
7pqr = 1 × 7 p q r
G.C.D is 1

Question 2.
Find the GCD of the following
(i) (2x + 5), (5x + 2)
(ii) a^{m + 1}, a^{m + 2}, a^{m + 3}
(iii) 2a² + a, 4a² – 1
(iv) 3a², 5b³, 7c⁴
(v) x⁴ – 1, x² – 1
(vi) a³ – 9ax², (a – 3x)²
Solution:
(i) (2x + 5), (5x + 2)
(2x + 5) = 1 × (2x + 5)
(5x + 2) = 1 × (5x + 2)
∴ G.C.D = 1

(ii) a^{m + 1}, a^{m + 2}, a^{m + 3}
a^{m + 1} = a^m × a¹
a^{m + 2} = a^m × a¹ × a¹
a^{m + 3} = a^m × a¹ × a¹ × a¹
G.C.G = a^m × a¹ = a^{m + 1}

(iii) 2a² + a, 4a² – 1
2a² + a = a (2a + 1)
4a² – 1 = (2a)² – 1² = (2a + 1) (2a -1)
G.C.D = (2a + 1)

(iv) 3a², 5b³, 7c⁴
3a² = 1 × 3a × a
5b³ = 1 × 5 b × b × b
7c⁴ = 1 × 7 × c × c × c × c
∴ G.C.D = 1

(v) x⁴ – 1, x² – 1
x⁴ – 1 = (x²)² – 1² = (x² + 1) (x² – 1)
= (x² + 1) (x² – 1²) = (x² + 1) (x + 1 ) (x – 1)
x² – 1 = x² – 1² = (x + 1) (x – 1)
G.C.D = (x + 1) (x – 1) = x² – 1

(vi) a³ – 9ax², (a – 3x)²
a³ – 9ax² = a (a² – (3x)²) = a (a + 3x) (a – 3x)
(a – 3x)² = (a – 3x) (a – 3x)
G.C.D = (a – 3x)