Class 8 Algebraic Identities
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1 ID : in-8-algebraic-identities [1] Class 8 Algebraic Identities For more such worksheets visit Answer t he quest ions (1) Solve the f ollowing using the standard identity a 2 - b 2 = (a+b) (a-b) A) C) B) D) (2) Solve the f ollowing using standard identities A) 31 2 B) C) 71 2 D) (3) If, f ind the value of. (4) Solve the f ollowing using the standard identity (x + a) (x + b) = x 2 + (a + b)x + ab A) B) C) D) (5) If (a - 1) 2 + (b - 2) 2 + (c - 1) 2 = 0, f ind the value of abc. (6) Find the value of (37.65)2 - (22.35) 2 using standard identities. Choose correct answer(s) f rom given choice (7) Solve the f ollowing using the standard identity (a+b) (a-b) = a 2 - b a b c d (8) There are two numbers such that their dif f erence is 3 and the dif f erence of their squares is 39. Find sum of the numbers. a. 12 b. 13 c. 16 d. 20 (9) If 3(a 2 + b 2 + c 2 ) = (a + b + c) 2, f ind the value of a - 2b + c. a. 2 b. -1 c. 0 d. 3
2 ID : in-8-algebraic-identities [2] (10) Simplif y (1pq + 2qr) 2-4pq 2 r a. 1p 2 q 2 + 4q 2 r 2 - pq 2 r b. 1p 2 q 2 + 4q 2 r 2 + pq 2 r c. 1p 2 q 2 + 4q 2 r 2 d. 1p 2 q 2 + 4q 2 r 2-4pq 2 r (11) If x 2 + y 2 = 29 and xy = 10, f ind the value of 4(x + y) 2-3(x - y) 2 a. 164 b. 176 c. 169 d. 167 (12) If, f ind the value of p 2 - q 2. a. 2 b. 2 2 c. 3 2 d. 0 (13) If -p - 4q = -5, and pq = -6, f ind value of p q 2. a. 65 b. 73 c. 69 d. 75 (14) Find the value of (147.75)2 - (117.75) 2 using standard identities. a. 3 b. 40 c. 30 d. 60 Fill in the blanks (15) There are two numbers such that their product is 36 and sum of the numbers is 12, the sum of their squares = Edugain ( All Rights Reserved Many more such worksheets can be generated at
3 Answers ID : in-8-algebraic-identities [3] (1) A) 7600 We have been asked to f ind the value of using the f ollowing identity: a 2 - b 2 = (a + b)(a - b). Applying the identity, we can write as: ( )(88-12) = = 7600 Theref ore, the result is B) 7800 We have been asked to f ind the value of using the f ollowing identity: a 2 - b 2 = (a + b)(a - b). Applying the identity, we can write as: ( )(89-11) = = 7800 Theref ore, the result is C) We have been asked to f ind the value of using the f ollowing identity: a 2 - b 2 = (a + b)(a - b). Applying the identity, we can write as: ( )(986-14) = = Theref ore, the result is
4 D) ID : in-8-algebraic-identities [4] We have been asked to f ind the value of using the f ollowing identity: a 2 - b 2 = (a + b)(a - b). Applying the identity, we can write as: ( )(997-3) = = Theref ore, the result is (2) A) 961 Use the standard identities here For example (a+b) 2 = a 2 + b 2 +2ab Similarly,(a-b) 2 = a 2 + b 2-2ab Take the last question here, which is 31 2 Now, 31 = Theref ore, 31 2 = (30 + 1) = (2 x 30 x 1) 31 2 = = 961 B) Use the standard identities here For example (a+b) 2 = a 2 + b 2 +2ab Similarly,(a-b) 2 = a 2 + b 2-2ab Take the last question here, which is Now, 499 = Theref ore, = (500-1) = (2 x 500 x 1) = =
5 C) 5041 ID : in-8-algebraic-identities [5] Use the standard identities here For example (a+b) 2 = a 2 + b 2 +2ab Similarly,(a-b) 2 = a 2 + b 2-2ab Take the last question here, which is 71 2 Now, 71 = Theref ore, 71 2 = (70 + 1) = (2 x 70 x 1) 71 2 = = 5041 D) Use the standard identities here For example (a+b) 2 = a 2 + b 2 +2ab Similarly,(a-b) 2 = a 2 + b 2-2ab Take the last question here, which is Now, 199 = Theref ore, = (200-1) = (2 x 200 x 1) = = (3) 2 (4) A) 9888 We have been asked to f ind the value of using the f ollowing identity: (x + a) (x + b) = x 2 + (a + b)x + ab. Let us think of two simple numbers whose sum is 103. Two such simple numbers are 100 and 3. Similarly, two simple numbers whose sum is 96 are 100 and -4. Thus, = { (3)} { (-4)} = {(3) + (-4)} (3)(-4)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = (-1)(100) + (-12) = (-100) + (-12) = 9888 Theref ore, the result is 9888.
6 B) ID : in-8-algebraic-identities [6] We have been asked to f ind the value of using the f ollowing identity: (x + a) (x + b) = x 2 + (a + b)x + ab. Let us think of two simple numbers whose sum is Two such simple numbers are 1000 and 4. Similarly, two simple numbers whose sum is 995 are 1000 and -5. Thus, = { (4)} { (-5)} = {(4) + (-5)} (4)(-5)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = (-1)(1000) + (-20) = (-1000) + (-20) = Theref ore, the result is C) We have been asked to f ind the value of using the f ollowing identity: (x + a) (x + b) = x 2 + (a + b)x + ab. Let us think of two simple numbers whose sum is 995. Two such simple numbers are 1000 and -5. Similarly, two simple numbers whose sum is 1004 are 1000 and 4. Thus, = { (-5)} { (4)} = {(-5) + (4)} (-5)(4)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = (-1)(1000) + (-20) = (-1000) + (-20) = Theref ore, the result is
7 D) ID : in-8-algebraic-identities [7] We have been asked to f ind the value of using the f ollowing identity: (x + a) (x + b) = x 2 + (a + b)x + ab. Let us think of two simple numbers whose sum is Two such simple numbers are 1000 and 1. Similarly, two simple numbers whose sum is 998 are 1000 and -2. Thus, = { (1)} { (-2)} = {(1) + (-2)} (1)(-2)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = (-1)(1000) + (-2) = (-1000) + (-2) = Theref ore, the result is (5) 2 Given (a - 1) 2 + (b - 2) 2 + (c - 1) 2 = 0 It means the sum of (a - 1) 2, (b - 2) 2 and (c - 1) 2 is equals to 0. We know that the square of a number cannot be negative. Theref ore, the sum of these non-negative numbers (a - 1) 2, (b - 2) 2 and (c - 1) 2 can be zero only if all of them are also equal to zero. Now, (a - 1) 2 = 0 a - 1 = 0 a = 1 Similarly, b = 2, c = 1. Step 4 Thus, the value of abc = = 2
8 (6) 60 ID : in-8-algebraic-identities [8] We have been asked to f ind the value of (37.65)2 - (22.35) 2 using standard identities. Now, (37.65) 2 - (22.35) 2 ( )( ) = b)(a - b) in the numerator] 60 = = 60 [By using the identity a 2 - b 2 = (a + Theref ore, the value of (37.65)2 - (22.35) 2 is 60. (7) b We have been asked to f ind the value of using the f ollowing identity: (a+b) (ab) = a 2 - b 2. Let us try to think of two numbers whose sum is 1007 and dif f erence is 993. Two such numbers are 1000 and 7. Thus, = ( ) (1000-7) = [Using the identity (a+b) (a-b) = a 2 - b 2 ] = = Theref ore, the result is
9 (8) b. 13 ID : in-8-algebraic-identities [9] Let s assume the two numbers be x and y. It is given that their dif f erence is 3. Theref ore, x - y = (1) Also the dif f erence of their squares is 39. Theref ore, x 2 - y 2 = (2) Step 4 Now, sum of the numbers = x + y = x2 - y 2...[Since, (x - y)(x + y) = x 2 - y 2 ] x - y = 39 3 = 13...[From equation (1) and (2)] Step 5 Thus, sum of the numbers is 13. (9) c. 0 (10) c. 1p 2 q 2 + 4q 2 r 2 We know that (a + b) 2 = a 2 + b 2 + 2ab. Now, let us start simplif ying (1pq + 2qr) 2-4pq 2 r by applying the identity (a + b) 2 = a 2 + b 2 + 2ab to the part (1pq + 2qr) 2 : (1pq + 2qr) 2-4pq 2 r = (1pq) 2 + (2qr) 2 + 2(1pq)(2qr) - 4pq 2 r = 1p 2 q 2 + 4q 2 r 2 + 4pq 2 r - 4pq 2 r = 1p 2 q 2 + 4q 2 r 2 Thus, the given expression can be simplif ied as 1p 2 q 2 + 4q 2 r 2.
10 (11) c. 169 ID : in-8-algebraic-identities [10] It is given that, x 2 + y 2 = 29 and xy = 10 Now, 4(x + y) 2-3(x - y) 2 = 4(x 2 + y 2 + 2xy) - 3(x 2 + y 2-2xy) = 4x 2 + 4y 2 + 8xy - 3x 2-3y 2 + 6xy = 1x 2 + 1y xy = 1(x 2 + y 2 ) + 14xy = 1(29) + 14(10) = 169 Thus, the value of 4(x + y) 2-3(x - y) 2 is 169. (12) b. 2 2 (13) b. 73 It is given that: pq = (1) It is also given that: -p - 4q = -5 On squaring both sides we get: ( -p - 4q) 2 = 25 (-1p) 2 + (-4q) (-1p) (-4q) = 25...[Since, (a + b) 2 = a 2 + b 2 + 2ab] 1p q 2 + (8)pq = 25 p q 2 + (8)(-6) = 25...[From equation (1)] p q 2 = 73 Thus, the value of p q 2 is 73.
11 (14) c. 30 ID : in-8-algebraic-identities [11] We have been asked to f ind the value of (147.75)2 - (117.75) 2 using standard identities. Now, (147.75) 2 - (117.75) 2 = ( )( ) [By using the identity a 2 - b 2 = (a + b)(a - b) in the numerator] 30 = = 30 Theref ore, the value of (147.75)2 - (117.75) 2 is 30. (15) 72 Let s assume the two numbers be x and y. It is given that, their product is 36. Theref ore, xy = (1) Also the sum of the numbers is 12. Theref ore, x + y = 12 On squaring both sides we get: (x + y) 2 = 144 x 2 + y 2 + 2xy = [Since, (x + y) 2 = x 2 + y 2 + 2xy] x 2 + y 2 + (2 36) = [From eqution (1)] x 2 + y 2 = x 2 + y 2 = 72 Step 4 Thus, the sum of their squares is 72.
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