PRACTICE QUESTIONS IN ALGEBRAIC IDENTITIES

Problem 1 :

If 2x + 2/x = 3, what is the value of x2 + 1/x2 ?

Solution :

2x + 2/x = 3

Square both sides.

(2x + 2/x)2 = 32

(2x + 2/x)(2x + 2/x) = 9

(2x)2 + (2x)(2/x) + (2/x)(2x) + (2/x)2 = 9

4x2 + 4 + 4 + 4/x2 = 9

4x2 + 8 + 4/x2 = 9

Subtract 8 from both sides.

4x2 + 4/x2 = 1

Factor.

4(x+ 1/x2) = 1

Divide both sides by 4.

x+ 1/x2 = 1/4

Problem 2 :

If a + b + c = 6 and a+ b+ c2 = 14, what is the value of

(a - b)2 + (b - c)2 + (c - a)2 ?

Solution :

Consider the following algebraic identity.

(a + b + c)2 = a+ b+ c+ 2ab + 2bc + 2ca

(a + b + c)2 = a+ b+ c+ 2(ab + bc + ca)

Substitute a + b + c = 6 and a+ b+ c2 = 14.

(6)2 = 14 + 2(ab + bc + ca)

36 = 14 + 2(ab + bc + ca)

Subtract 14 from both sides.

12 = 2(ab + bc + ca)

Divide both sides by 2.

6 = ab + bc + ca

The value of (a - b)2 + (b - c)2 + (c - a)2 :

= (a - b)2 + (b - c)2 + (c - a)2

a- 2ab + b2 + b2- 2bc + c+ c- 2ca + a2

= a2 + a2 + b2 + b+ c2 + c2 - 2ab - 2bc - 2ca

= 2a2 + 2b+ 2c2- 2ab - 2bc - 2ca

= 2(a+ b+ c2) - 2(ab + bc + ca)

Substitute a+ b+ c2 = 14 and ab + bc + ca = 6.

= 2(14) - 2(6)

= 28 - 12

= 16

Problem 3 :

If x - y = 8 and xy = 5, what is the value of

x- y+ 8(x + y)2

Solution :

Consider the square of a binomial given below.

(x - y)= x+ y- 2xy

Substitute x - y = 8 and xy = 5.

82 = x+ y- 2(5)

64 + 10 = x+ y2

74 = x+ y2

Consider the square of a binomial given below.

(x + y)= x+ y+ 2xy

Substitute x+ y2 = 74 and xy = 5.

(x + y)= 74 + 2(5)

(x + y)= 74 + 10

(x + y)= 84

The value of x- y+ 8(x + y)2 :

= x- y+ 8(x + y)2

Use the identity a3 - b3 = (a - b)(a2 + ab + b2).

= (x - y)(x2 + xy + y2) + 8(x + y)2

Substitute.

= (8)(74 + 5) + 8(84)

=  8(79) + 672

= 632 + 672

= 1304

Problem 4 :

If x + y = 5 and xy = 6 and x > y, then find 2(x+ y2).

Solution :

(x + y)2 = (x + y)(x + y)

(x + y)2 = x2 + xy + xy + y2

(x + y)2 = x+ 2xy + y2

or

x+ 2xy + y(x + y)2

Subtract 2xy from both sides.

x+ y= (x + y)- 2xy

Substitute x + y = 5 and xy = 6.

x+ y= 5- 2(6)

x+ y= 25 - 12

x+ y= 13

Multiply both sides by 2.

2(x+ y2) = 2(13)

2(x+ y2) = 26

Problem 5 :

If a- b3 = 513 and a - b = 3, what is the value of ab?

Solution :

a- b3 = 513

(a - b)+ 3ab(a - b) = 513

Substitute a - b = 3.

3+ 3ab(3) = 513

27 + 9ab = 513

Subtract 27 from both sides.

9ab = 486

Divide both sides by 9.

ab = 54

Problem 6 :

If a+ b+ c2 = 9 and ab + bc + ca = 8, find the value of 

(a + b + c)2

Solution :

(a + b + c)2 = a+ b+ c+ 2ab + 2bc + 2ca

(a + b + c )2 = a+ b+ c+ 2(ab + bc + ca)

Substitute a+ b+ c2 = 9 and ab + bc + ca = 8.

(a + b + c)2 = 9 + 2(8)

= 9 + 16

= 25

Problem 7 :

If a + b = √7 and a - b = √5, then find the value of

8ab(a+ b2)

Solution :

a + b = √7 ----(1)

a - b = √5 ----(2)

(1) + (2) :

2a = √7 + √5

a = (√7 + √5)/2

Substitute a = (√7 + √5)/2 in (1).

(√7 + √5)/2 + b = √7

b = √7 - [(√7 + √5)/2]

b = (√7 - √5)/2

ab = [(√7 + √5)/2] ⋅ [(√7 - √5)/2]

ab = (7 - 5)/4

ab = 1/2

a+ b= (a + b)- 2ab

a+ b= (√7)- 2(1/2)

a+ b= 7 - 1

a+ b= 6

The value of 8ab(a+ b2) :

8ab(a+ b2) = 8(1/2)(6)

8ab(a+ b2) = 24

Problem 8 :

If a - b = 4 and ab = 60, what is the value of a + b?

Solution :

(a - b)2(a + b)- 4ab

Substitute a - b = 4 and ab = 60.

42(a + b)- 4(60)

16 = (a + b)- 240

Add 240 to both sides.

256 = (a + b)2

or

(a + b)= 256

Take square root on both sides.

a + b = √256

a + b = ±16

Problem 9 :

If a + b = 9m and ab = 18m2, what is the value of a - b?

Solution :

(a + b)2(a - b)+ 4ab

Substitute a + b = 9m and ab = 18m2.

(9m)2(a - b)+ 4(18m2)

81m2(a - b)+ 72m2

Subtract 72m2 from both sides.

9m2(a - b)2

or

(a - b)2 = 9m2

Take square root on both sides.

a - b = √(9m2)

a - b = ±9m

Problem 10 :

Simplify :

(2345 x 2345 - 759 x 759)/(2345 - 759)

Solution :

= (2345 x 2345 - 759 x 759)/(2345 - 759)

= (23452 - 7592)/(2345 - 759)

Use the identity a- b2 = (a + b)(a - b).

= (2345 + 759)(2345 - 759)/(2345 - 759)

= (3104)(1586)/(1586)

= 3104

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