ID : ww-8-algebraic-identities [1] Grade 8 Algebraic Identities For more such worksheets visit www.edugain.com Answer t he quest ions (1) There are two numbers such that the sum of their squares is 25 and sum of the numbers is 7. Find their product. (2) Solve the f ollowing using the standard identity (x + a) (x + b) = x 2 + (a + b)x + ab A) 105 104 B) 996 995 C) 1004 999 D) 1004 998 (3) Solve the f ollowing using the standard identity (a+b) (a-b) = a 2 - b 2 A) 109 91 B) 108 92 C) 107 93 D) 1009 991 (4) If, f ind the value of p 2 - q 2. (5) There are two numbers such that sum of the numbers is 25 and the dif f erence of their squares is 175. Find their dif f erence. (6) If 3(p 2 + q 2 + r 2 ) = (p + q + r) 2, f ind the value of p + q - 2r. (7) If xy = -8, and -x - 2y = 0, f ind value of x 2 + 4y 2. (8) If = 8, f ind the value of. (9) If (u - 5) 2 + (v - 5) 2 + (w - 3) 2 + (x - 5) 2 + (y - 3) 2 = 0, f ind the value of u + v + w + x + y. (10) Solve the f ollowing using standard identities A) C) 21 2 B) 399 2 D) 89 2 501 2 Choose correct answer(s) f rom given choice (11) If, f ind the value of a. 2 b. 1/2 c. 0 d. 1.
(12) Find the value of (209.876)2 - (189.876) 2 399.752 using standard identities. ID : ww-8-algebraic-identities [2] a. 40 b. 2 c. 20 d. 30 (13) Find the value of (35.7)2 - (24.3) 2 11.4 using standard identities. a. 60 b. 6 c. 70 d. 600 (14) Solve the f ollowing using the standard identity a 2 - b 2 = (a+b) (a-b) 96 2-4 2 a. 9218 b. 9200 c. 9189 d. 9196 Fill in the blanks (15) If x 2 + y 2 = 41 and xy = 20, the value of 5(x + y) 2 - (x - y) 2 =. 2016 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : ww-8-algebraic-identities [3] (1) 12 Let s assume the two numbers be x and y. It is given that, sum of their squares is 25. Theref ore, x 2 + y 2 = 25 -----(1) Also the sum of the numbers is 7. Theref ore, x + y = 7 On squaring both sides we get: (x + y) 2 = 49 x 2 + y 2 + 2xy = 49...[Since, (x + y) 2 = x 2 + y 2 + 2xy] 25 + 2xy = 49...[From eqution (1)] 2xy = 49-25 2xy = 24 xy = 24 2 xy = 12 Step 4 Thus, their product is 12. (2) A) 10920 We have been asked to f ind the value of 105 104 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 105. Two such simple numbers are 100 and 5. Similarly, two simple numbers whose sum is 104 are 100 and 4. Thus, 105 104 = { 100 + (5)} { 100 + (4)} = 100 2 + {(5) + (4)} 100 + (5)(4)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = 10000 + (9)(100) + (20) = 10000 + (900) + (20) = 10920 Theref ore, the result is 10920.
B) 991020 ID : ww-8-algebraic-identities [4] We have been asked to f ind the value of 996 995 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 996. Two such simple numbers are 1000 and -4. Similarly, two simple numbers whose sum is 995 are 1000 and -5. Thus, 996 995 = { 1000 + (-4)} { 1000 + (-5)} = 1000 2 + {(-4) + (-5)} 1000 + (-4)(-5)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = 1000000 + (-9)(1000) + (20) = 1000000 + (-9000) + (20) = 991020 Theref ore, the result is 991020. C) 1002996 We have been asked to f ind the value of 1004 999 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 1004. Two such simple numbers are 1000 and 4. Similarly, two simple numbers whose sum is 999 are 1000 and -1. Thus, 1004 999 = { 1000 + (4)} { 1000 + (-1)} = 1000 2 + {(4) + (-1)} 1000 + (4)(-1)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = 1000000 + (3)(1000) + (-4) = 1000000 + (3000) + (-4) = 1002996 Theref ore, the result is 1002996.
D) 1001992 ID : ww-8-algebraic-identities [5] We have been asked to f ind the value of 1004 998 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 1004. Two such simple numbers are 1000 and 4. Similarly, two simple numbers whose sum is 998 are 1000 and -2. Thus, 1004 998 = { 1000 + (4)} { 1000 + (-2)} = 1000 2 + {(4) + (-2)} 1000 + (4)(-2)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = 1000000 + (2)(1000) + (-8) = 1000000 + (2000) + (-8) = 1001992 Theref ore, the result is 1001992. (3) A) 9919 We have been asked to f ind the value of 109 91 using the f ollowing identity: (a+b) (a-b) = a 2 - b 2. Let us try to think of two numbers whose sum is 109 and dif f erence is 91. Two such numbers are 100 and 9. Thus, 109 91 = (100 + 9) (100-9) = 100 2-9 2 [Using the identity (a+b) (a-b) = a 2 - b 2 ] = 10000-81 = 9919 Theref ore, the result is 9919.
B) 9936 ID : ww-8-algebraic-identities [6] We have been asked to f ind the value of 108 92 using the f ollowing identity: (a+b) (a-b) = a 2 - b 2. Let us try to think of two numbers whose sum is 108 and dif f erence is 92. Two such numbers are 100 and 8. Thus, 108 92 = (100 + 8) (100-8) = 100 2-8 2 [Using the identity (a+b) (a-b) = a 2 - b 2 ] = 10000-64 = 9936 Theref ore, the result is 9936. C) 9951 We have been asked to f ind the value of 107 93 using the f ollowing identity: (a+b) (a-b) = a 2 - b 2. Let us try to think of two numbers whose sum is 107 and dif f erence is 93. Two such numbers are 100 and 7. Thus, 107 93 = (100 + 7) (100-7) = 100 2-7 2 [Using the identity (a+b) (a-b) = a 2 - b 2 ] = 10000-49 = 9951 Theref ore, the result is 9951.
D) 999919 ID : ww-8-algebraic-identities [7] We have been asked to f ind the value of 1009 991 using the f ollowing identity: (a+b) (a-b) = a 2 - b 2. Let us try to think of two numbers whose sum is 1009 and dif f erence is 991. Two such numbers are 1000 and 9. Thus, 1009 991 = (1000 + 9) (1000-9) = 1000 2-9 2 [Using the identity (a+b) (a-b) = a 2 - b 2 ] = 1000000-81 = 999919 Theref ore, the result is 999919. (4) 8 3 (5) 7 Let s assume the two numbers be x and y. It is given that sum of the numbers is 25. Theref ore, x + y = 25 -----(1) Also the dif f erence of their squares is 175. Theref ore, x 2 - y 2 = 175 -----(2) Step 4 Now, their dif f erence = x - y = x2 - y 2...[Since, (x - y)(x + y) = x 2 - y 2 ] x + y = 175 = 7 25...[From equation (1) and (2)] Step 5 Thus, their dif f erence is 7. (6) 0
(7) 32 ID : ww-8-algebraic-identities [8] It is given that: xy = -8 -----(1) It is also given that: -x - 2y = 0 On squaring both sides we get: ( -x - 2y) 2 = 0 (-1x) 2 + (-2y) 2 + 2 (-1x) (-2y) = 0...[Since, (a + b) 2 = a 2 + b 2 + 2ab] 1x 2 + 4y 2 + (4)xy = 0 x 2 + 4y 2 + (4)(-8) = 0...[From equation (1)] x 2 + 4y 2 = 32 Thus, the value of x 2 + 4y 2 is 32. (8) 6 If we assume, a = x, and b = 1/x, we can use standard algebraic identities which specif ies relation between a - b and a 2 + b 2 By using identity, (a - b) 2 = a 2 + b 2-2ab ( ) 2 = x 2 + ( 1 x ) 2-2 x 1 x ( ) 2 = 8-2 = 6 Theref ore, the value of is 6.
(9) 21 ID : ww-8-algebraic-identities [9] Given (u - 5) 2 + (v - 5) 2 + (w - 3) 2 + (x - 5) 2 + (y - 3) 2 = 0 It means the sum of (u - 5) 2, (v - 5) 2, (w - 3) 2, (x - 5) 2 and (y - 3) 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 (u - 5) 2, (v - 5) 2, (w - 3) 2, (x - 5) 2 and (y - 3) 2 can be zero only if all of them are also equal to zero. Now, (u - 5) 2 = 0 u - 5 = 0 u = 5 Similarly, v = 5, w = 3, x = 5, y = 3. Step 4 Thus, the value of u + v + w + x + y = 5 + 5 + 3 + 5 + 3 = 21 (10) A) 441 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 21 2 Now, 21 = 20 + 1 Theref ore, 21 2 = (20 + 1) 2 21 2 = 20 2 + 1 2 + (2 x 20 x 1) 21 2 = 400 + 1 + 40 21 2 = 441 B) 7921 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 89 2 Now, 89 = 90-1 Theref ore, 89 2 = (90-1) 2 89 2 = 90 2 + 1 2 - (2 x 90 x 1) 89 2 = 8100 + 1-180 89 2 = 7921
C) 159201 ID : ww-8-algebraic-identities [10] 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 399 2 Now, 399 = 400-1 Theref ore, 399 2 = (400-1) 2 399 2 = 400 2 + 1 2 - (2 x 400 x 1) 399 2 = 160000 + 1-800 399 2 = 159201 D) 251001 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 501 2 Now, 501 = 500 + 1 Theref ore, 501 2 = (500 + 1) 2 501 2 = 500 2 + 1 2 + (2 x 500 x 1) 501 2 = 250000 + 1 + 1000 501 2 = 251001 (11) a. 2
(12) c. 20 ID : ww-8-algebraic-identities [11] We have been asked to f ind the value of (209.876)2 - (189.876) 2 identities. 399.752 using standard Now, (209.876) 2 - (189.876) 2 399.752 = (209.876 + 189.876)(209.876-189.876) 399.752 [By using the identity a 2 - b 2 = (a + b)(a - b) in the numerator] 399.752 20 = 399.752 = 20 Theref ore, the value of (209.876)2 - (189.876) 2 399.752 is 20. (13) a. 60 We have been asked to f ind the value of (35.7)2 - (24.3) 2 11.4 using standard identities. Now, (35.7) 2 - (24.3) 2 = 11.4 b) in the numerator] 60 11.4 = 11.4 = 60 (35.7 + 24.3)(35.7-24.3) 11.4 [By using the identity a 2 - b 2 = (a + b)(a - Theref ore, the value of (35.7)2 - (24.3) 2 11.4 is 60.
(14) b. 9200 ID : ww-8-algebraic-identities [12] We have been asked to f ind the value of 96 2-4 2 using the f ollowing identity: a 2 - b 2 = (a + b)(a - b). Applying the identity, we can write 96 2-4 2 as: (96 + 4)(96-4) = 100 92 = 9200 Theref ore, the result is 9200. (15) 404 It is given that, x 2 + y 2 = 41 and xy = 20 Now, 5(x + y) 2 - (x - y) 2 = 5(x 2 + y 2 + 2xy) - (x 2 + y 2-2xy) = 5x 2 + 5y 2 + 10xy - x 2 - y 2 + 2xy = 4x 2 + 4y 2 + 12xy = 4(x 2 + y 2 ) + 12xy = 4(41) + 12(20) = 404 Thus, the value of 5(x + y) 2 - (x - y) 2 is 404.