Grade 8 Algebraic Identities In

ID : in-8-Algebraic-Identities [1]

Class 8 Algebraic Identit Id entities ies For more such worksheets visit www.edugain.com

Answer Answer t he quest ions ions (1)

(2)

Solve the following using the standard identity a2 - b 2 = (a+b) (a-b) A)

882 - 122

B)

892 - 112

C)

986 2 - 14 2

D)

997 2 - 32

Solve the following using standard identities A)

312

B)

4 992

C)

7 12

D)

199 2

(3)

If

(4 )

Solve the following using the standard identity (x + a) (x + b) = x 2 + (a + b)x + ab

, f ind the value value of

.

A)

103 × 96

B)

1004 × 995

C)

995 × 1004

D)

1001 × 998

(5)

If (a - 1) 2 + (b - 2)2 + (c - 1) 2 = 0 , find f ind the value value of abc .

(6 )

(37.65)2 - (22.35) (22.35)2

Find the value of

using standard st andard identities.

15.3

Choose correct answer(s) from given choice (7 )

(8)

(9 )

Solve the following using the standard identity (a+b) (a-b) = a 2 - b 2 1007 × 993 a. 999966

b. 999951

c. 999944

d. 999939

There are two numbers numbers such that t heir heir diff erence erence is 3 and the dif dif f erence erence of their squares squares is 39. Find sum of the numbers. a. 12

b. 13

c. 16

d. 20

If 3(a2 + b 2 + c 2) = (a + b + c)2, f ind the value value of a - 2b + c. a. 2

- 1 b. -1

c. 0

d. 3

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ID : in-8-Algebraic-Identities [2] (10) Simplify (1pq + 2qr)2 - 4 pq2r a. 1p 2q2 + 4 q2r 2 - pq 2r

b. 1p 2q2 + 4 q2r 2 + pq2r

c. 1p 2q2 + 4 q2r 2

d. 1p 2q2 + 4 q2r 2 - 4 pq2r

(11) If x2 + y 2 = 29 and xy = 10 , find the value of 4(x + y) 2 - 3(x - y) 2 a. 164

b. 176

c. 169

d. 167

, f ind the value of p2 - q2.

(12) If a. √2

b. 2√2

c. 3√2

d. 0

(13) If -p - 4q = - 5, and pq = -6, find value of p2 + 16q2. a. 65

b. 73

c. 69

d. 75

(14)

Find the value of

(147.75)2 - (117.75) 2

using standard identities.

265.5 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 =

.

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ID : in-8-Algebraic-Identities [3]

Answers (1)

A)

7600 Step 1

We have been asked to f ind the value of 882 - 12 2 using the f ollowing identity: a2 - b2 = (a + b)(a - b) . Step 2

Applying t he identity, we can write 882 - 122 as: (88 + 12)(88 - 12) = 100 × 76 = 7600 Step 3

Theref ore, the result is 7600 .

B)

7800 Step 1


Applying t he identity, we can write 892 - 112 as: (89 + 11)(89 - 11) = 100 × 78 = 7800 Step 3

Theref ore, the result is 7800 . C)

972000 Step 1


Applying t he identity, we can write 9862 - 14 2 as: (986 + 14)(986 - 14 ) = 1000 × 972 = 972000 Step 3




ID : in-8-Algebraic-Identities [4] D)

994000 Step 1

We have been asked to f ind the value of 9972 - 32 using the f ollowing identity: a2 - b2 = (a + b)(a - b) . Step 2

Applying t he identity, we can write 997 2 - 32 as: (997 + 3)(997 - 3) = 1000 × 994 = 994000 Step 3


(2)

A)

961 Use the standard identities here For example (a+b) 2 = a2 + b2 +2ab Similarly, (a-b) 2 = a2 + b2 - 2ab Take the last quest ion here, which is 312 Now, 31 = 30 + 1 Therefo re, 312 = (30 + 1) 2 312 = 302 + 1 2 + (2 x 30 x 1) 312 = 900 + 1 + 60 312 = 961

B)

249001 Use the standard identities here For example (a+b) 2 = a2 + b2 +2ab Similarly, (a-b) 2 = a2 + b2 - 2ab Take the last question here, which is 4992 Now, 499 = 500 - 1 Therefo re, 4992 = (500 - 1)2 4992 = 5002 + 1 2 - (2 x 500 x 1) 4992 = 250000 + 1 - 1000 4992 = 249001



ID : in-8-Algebraic-Identities [5] C)

5041 Use the standard identities here For example (a+b) 2 = a2 + b2 +2ab Similarly, (a-b) 2 = a2 + b2 - 2ab Take the last question here, which is 712 Now, 71 = 7 0 + 1 Therefo re, 712 = (70 + 1) 2 712 = 7 02 + 1 2 + (2 x 7 0 x 1) 712 = 4 900 + 1 + 140 712 = 5041

D)

39601 Use the standard identities here For example (a+b) 2 = a2 + b2 +2ab Similarly, (a-b) 2 = a2 + b2 - 2ab Take the last quest ion here, which is 1992 Now, 199 = 200 - 1 Therefore, 1992 = (200 - 1)2 1992 = 2002 + 1 2 - (2 x 200 x 1) 1992 = 4 0000 + 1 - 4 00 1992 = 39601

(3)

2

(4 )

A)

9888 Step 1

We have been asked to f ind the value of 103 × 96 using the f ollowing identity: (x + a) (x + b) = x 2 + (a + b)x + ab.

Let us think of two simple numbers whos e sum is 103. Two such simple numbers are 100 and 3. Similarly, two s imple numbers who se s um is 96 are 100 and - 4. Step 2

Thus, 103 × 96 = { 100 + (3)} { 100 + (- 4)} = 1002 + {(3) + (-4)} 100 + (3)(-4 ) ...[Using the identit y (x + a) (x + b) = x 2 + (a + b)x + ab]

= 10000 + (-1)(100) + (-12) = 10000 + (-100) + (- 12) = 9888 Step 3

Theref ore, the result is 9888.



ID : in-8-Algebraic-Identities [6] B)

998980 Step 1

We have been asked to f ind the value of 1004 × 995 using the f ollowing identity: (x + a) (x + b) = x2 + (a + b)x + ab.

Let us t hink of two simple numbers whos e sum is 1004. Two such simple numbers are 1000 and 4 . Similarly, two simple numbers whos e sum is 995 are 1000 and - 5. Step 2

Thus, 1004 × 995 = { 1000 + (4)} { 1000 + (- 5)} = 10002 + {(4) + (-5)} 1000 + (4)(-5) ...[Using the identit y (x + a) (x + b) = x 2 + (a + b)x + ab]

= 1000000 + (-1)(1000) + (-20) = 1000000 + (-1000) + (- 20) = 998980 Step 3


C)

998980 Step 1


Let us think of two simple numbers whos e sum is 995. Two such simple numbers are 1000 and -5. Similarly, two s imple numbers who se s um is 1004 are 1000 and 4 . Step 2

Thus, 995 × 1004 = { 1000 + (-5)} { 1000 + (4)} = 10002 + {(-5) + (4)} 1000 + (-5)(4) ...[Using the identit y (x + a) (x + b) = x 2 + (a + b)x + ab]

= 1000000 + (-1)(1000) + (-20) = 1000000 + (-1000) + (- 20) = 998980 Step 3




ID : in-8-Algebraic-Identities [7] D)

998998 Step 1


Let us think of two simple numbers whos e sum is 1001. Two such simple numbers are 1000 and 1. Similarly, two simple numbers whos e sum is 998 are 1000 and - 2. Step 2

Thus, 1001 × 998 = { 1000 + (1)} { 1000 + (- 2)} = 10002 + {(1) + (- 2)} 1000 + (1)(-2) ...[Using the ide nt ity (x + a) (x + b) = x 2 + (a + b)x + ab]

= 1000000 + (-1)(1000) + (-2) = 1000000 + (- 1000) + (- 2) = 998998 Step 3


(5)

2 Step 1

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. Step 2

We know that t he 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. Step 3

Now, (a - 1)2 = 0 a - 1 = 0 a = 1 Similarly, b = 2, c = 1. ⇒ ⇒

Step 4

Thus, the value of abc = 1 × 2 × 1 = 2



ID : in-8-Algebraic-Identities [8] (6 )

60 Step 1

We have been asked to f ind the value of

(37.65)2 - (22.35)2

using st andard identities.

15.3 Step 2

Now, (37.65)2 - (22.35)2

=

(37.65 + 22.35)(37.65 - 22.35) 15.3

15.3

[By using the identity a2 - b 2 = (a +

b)(a - b) in the numerato r]

=

60 × 15.3 15.3

= 60 Step 3

Therefo re, the value of

(37.65)2 - (22.35)2

is 60 .

15.3

(7)

b. 999951 Step 1

We have been asked to f ind the value of 1007 × 993 using the f ollowing identity: (a+b) (ab) = a2 - b2.

Let us t ry to think of two numbers whose s um is 1007 and dif f erence is 993. Two such numbers are 1000 and 7. Step 2

Thus, 1007 × 993 = (1000 + 7) (1000 - 7 ) = 10002 - 7 2 [Using the identity (a+b) (a- b) = a2 - b2] = 1000000 - 4 9 = 999951 Step 3

Therefore, the result is 999951 .



ID : in-8-Algebraic-Identities [9] (8)

b. 13 Step 1

Let’s as sume the two numbers be x and y. Step 2

It is given that their difference is 3. Therefo re, x - y = 3 --- -- (1) Step 3

Also the dif f erence of their s quares is 39. Therefo re, x2 - y2 = 39 -- --- (2) Step 4

Now, sum of the numbers = x + y =

x2 - y2

...[Since, (x - y)(x + y) = x 2 - y2]

x- y =

39 3

...[From equat ion (1) and (2)]

= 13 Step 5

Thus, sum of the numbers is 13. (9)

c. 0

(10) c. 1p 2q2 + 4 q2r 2 Step 1

We know that (a + b) 2 = a 2 + b 2 + 2ab . Step 2

Now, let us start simplifying (1pq + 2qr)2 - 4pq2r by applying the identity (a + b)2 = a 2 + b 2 + 2ab to the part (1pq + 2qr)2 :

(1pq + 2qr)2 - 4 pq2r = (1pq)2 + (2qr)2 + 2(1pq)(2qr) - 4 pq2r = 1p2q2 + 4 q2r 2 + 4pq2r - 4pq2r = 1p2q2 + 4 q2r 2 Step 3

Thus, the given expression can be simplified as 1p2q2 + 4 q2r 2.



ID : in-8-Algebraic-Identities [10] (11) c. 169 Step 1

It is given that, x2 + y 2 = 29 and xy = 10 Step 2

Now, 4(x + y)2 - 3(x - y)2 = 4(x2 + y 2 + 2xy) - 3(x2 + y 2 - 2xy) = 4 x2 + 4 y2 + 8xy - 3x2 - 3y2 + 6xy = 1x2 + 1y2 + 14xy = 1(x2 + y 2) + 14xy = 1(29) + 14(10) = 169 Step 3

Thus, the value of 4(x + y) 2 - 3(x - y) 2 is 169 . (12) b. 2√2 (13) b. 73 Step 1

It is given that : pq = -6 --- -- (1) Step 2

It is also given that: -p - 4q = -5 On squaring both sides we get: ( -p - 4q)2 = 25 ⇒

(-1p)2 + (-4q)2 + 2 × (-1p) × (-4 q) = 25 ...[Since, (a + b) 2 = a 2 + b 2 + 2ab]

⇒

1p 2 + 16q2 + (8)pq = 25

⇒

p2 + 16q2 + (8)(-6) = 25 ...[From equat ion (1)]

⇒

p2 + 16q2 = 7 3

Step 3

Thus, the value of p2 + 16q2 is 73.



ID : in-8-Algebraic-Identities [11] (14 ) c. 30 Step 1

We have been asked to f ind the value of

(147.75)2 - (117 .75) 2

using st andard identities.

265.5 Step 2

Now, (147.75)2 - (117 .75) 2

=

(147.75 + 117.75)(147 .75 - 117.75) 265.5

265.5

[By using the identity a2 -

b2 = (a + b)(a - b) in the numerato r]

=

265.5 × 30 265.5

= 30 Step 3

Therefo re, the value of

(147.75)2 - (117.75) 2

is 30 .

265.5

(15)

72 Step 1

Let’s as sume the two numbers be x and y. Step 2

It is given that , their product is 36. Therefo re, xy = 36 --- -- (1) Step 3

Also the sum of the numbers is 12. Theref ore, x + y = 12 On squaring both sides we get: (x + y) 2 = 144 ⇒

x2 + y 2 + 2xy = 144 ...[Since, (x + y) 2 = x 2 + y 2 + 2xy]

⇒

x2 + y 2 + (2 × 36) = 144 ...[From equt ion (1)]

⇒

x2 + y 2 = 144 - 72

⇒

x2 + y 2 = 7 2

Step 4

Thus, t he sum of their squares is 72.



Grade 8 Algebraic Identities In

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