Test (elaborations) Maths

Access SAT Maths Solutions for Algebraic Identities
Exercise 4.1
Question 1: Evaluate each of the following using identities:
(i) (2x – 1/x)2
(ii) (2x + y) (2x – y)
(iii) (a2b – b2a)2
(iv) (a – 0.1) (a + 0.1)
(v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2)
Solution:
(i) (2x – 1/x)2
[Use identity: (a – b)2 = a2 + b2 – 2ab ]
(2x – 1/x)2 = (2x)2 + (1/x)2 – 2 (2x)(1/x)
= 4x2 + 1/x2 – 4
(ii) (2x + y) (2x – y)
[Use identity: (a – b)(a + b) = a2 – b2 ]
(2x + y) (2x – y) = (2x )2 – (y)2
= 4x2 – y2
(iii) (a2b – b2a)2
[Use identity: (a – b)2 = a2 + b2 – 2ab ]
(a2b – b2a)2 = (a2b) 2 + (b2a)2 – 2 (a2b)( b2a)
= a4b 2 + b4a2 – 2 a3b3
(iv) (a – 0.1) (a + 0.1)
[Use identity: (a – b)(a + b) = a2 – b2 ]
(a – 0.1) (a + 0.1) = (a)2 – (0.1)2
= (a)2 – 0.01
(v) (1.5 x2 – 0.3y2) (1.5 x2 + 0.3y2)
[Use identity: (a – b)(a + b) = a2 – b2 ]
(1.5 x2 – 0.3y2) (1.5x2 + 0.3y2) = (1.5 x2 ) 2 – (0.3y2)2
= 2.25 x4 – 0.09y4
Question 2: Evaluate each of the following using identities:
(i) (399)2
(ii) (0.98)2

,(iii) 991 x 1009
(iv) 117 x 83
Solution:
(i)




(ii)




(iii)

, (iv)




Question 3: Simplify each of the following:
(i) 175 x 175 +2 x 175 x 25 + 25 x 25
(ii) 322 x 322 – 2 x 322 x 22 + 22 x 22
(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24
(iv)




Solution:
(i) 175 x 175 +2 x 175 x 25 + 25 x 25 = (175)2 + 2 (175) (25) + (25)2
= (175 + 25)2
[Because a2+ b2+2ab = (a+b)2 ]