RD Chapter 4 Algebraic Identities Ex 4.1 Solutions
Question - 1 : - Evaluate each of the following using identities:
(i) (2x –1)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)
Answer - 1 : -
Question - 2 : - Evaluate each of the following using identities:
(i) (399)2
(ii) (0.98)2
(iii) 991 x 1009
(iv) 117 x 83
Answer - 2 : -
Question - 3 : - Simplify each of the following:
Answer - 3 : -
(i) 175 x 175 +2 x 175 x 25 + 25 x 25 = (175)2 +2 (175) (25) + (25)2
= (175 + 25)2
[Becausea2+ b2+2ab = (a+b)2 ]
= (200)2
= 40000
So, 175 x 175 +2 x 175x 25 + 25 x 25 = 40000.
(ii) 322 x 322 – 2 x 322 x 22 + 22 x 22
= (322)2 –2 x 322 x 22 + (22)2
= (322 – 22)2
[Becausea2+ b2-2ab = (a-b)2]
= (300)2
= 90000
So, 322 x 322 – 2 x322 x 22 + 22 x 22= 90000.
(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x0.24
= (0.76) 2 +2 x 0.76 x 0.24 + (0.24) 2
= (0.76+0.24) 2
[Because a2+ b2+2ab = (a+b)2]
= (1.00)2
= 1
So, 0.76 x 0.76 + 2 x0.76 x 0.24 + 0.24 x 0.24 = 1.
(iv)
Question - 4 : - 
Answer - 4 : -
Question - 5 : - 
Answer - 5 : -
Question - 6 : - 
Answer - 6 : -
Question - 7 : - If 9x2 +25y2 = 181 and xy = -6, find thevalue of 3x + 5y.
Answer - 7 : -
9x2 + 25y2 = 181, and xy = -6
(3x + 5y)2 = (3x)2 + (5y)2 +2 x 3x + 5y
⇒ 9X2 + 25y2 +30xy
= 181 + 30 x (-6)
= 181 – 180 = 1
= (±1 )2
∴ 3x +5y = ±1
Question - 8 : - If 2x + 3y = 8 and xy = 2, find the value of 4X2 + 9y2.
Answer - 8 : -
2x + 3y = 8 and xy = 2
Now, (2x + 3y)2 = (2x)2 + (3y)2 +2 x 2x x 3y
⇒ (8)2 = 4x2 +9y2 + 12xy
⇒ 64 = 4X2 + 9y2 +12 x 2
⇒ 64 = 4x2 + 9y2 +24
⇒ 4x2 + 9y2 =64 – 24 = 40
∴ 4x2 + 9y2 =40
Question - 9 : - If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2
Answer - 9 : -
3x – 7y = 10, xy = -1
3x -7y= 10
Squaring both sides,
(3x – 7y)2 = (10)2
⇒ (3x)2 +(7y)2 – 2 x 3x x 7y = 100
⇒ 9X2 + 49y2 – 42xy = 100
⇒ 9x2 + 49y2 – 42(-l) = 100
⇒ 9x2 +49y2 + 42 = 100
∴ 9x2 +49y2 = 100 – 42 = 58
Question - 10 : - Simplify each of the following products:
Answer - 10 : -