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Chapter 4: Algebraic Identities Exercise – 4.1

Question: 1

Evaluate each of the following using identities:

(i) (2× − 1/×)2

(ii) (2× + y)(2× - y)

(iii) (a2b − ab2)2

(iv) (a - 0.1)(a + 0.1)

(v) (1.5×2 − 0.3y2)(1.5×2 + 0.3y2)

Solution:

(i) Given,

(2× − 1/×)2 = (2×)2 + (1/×)2 − 2 ∗ 2× ∗ 1/×

(2× − 1/×)2 = 4×2 + 1/×2 – 4      [∴  (a − b)2 = a2 + b2 − 2ab]

Where, a = 2×, b = 1/×

∴  (2× − 1/×)2 = 4×2 + 1/×2 − 4

(ii) Given,

(2× + y)(2× - y)

= (2×)2 − (y)2      [ ∴ (a + b)(a − b) = a2 − b2]

= 4×2 − y2

∴  (2× + y)(2× − y) = 4×2 − y2

(iii) Given, 

(a2b − ab2)2

= (a2b)2 + (ab2)2 − 2 ∗ a2b ∗ ab2            [∴ (a − b)2 = a2 + b2 − 2ab]

Where, a = a2b, b = ab2

= a4b2 + b4a2 − 2a3b3

(a2b − ab2)2 = a4b2 + b4a2 − 2a3b3

(iv)  Given,

(a - 0.1)(a + 0.1)

= a2 − (0.1)2                  [∴ (a + b)(a − b) = a2 − b2]

Where, a = a and b = 0.1

= a2 − 0.01

∴ (a − 0.1)(a + 0.1) = a2 − 0.01

(v)  Given, 

(1.5×2 − 0.3y2)(1.5×2 + 0.3y2)

= (1.5×2)2 − (0.3y2)2      [∴ (a + b)(a − b) = a2 − b2]

Where, a = 1.5×2, b = 0.3y2

= 2.25×4 − 0.09y4

∴ (1.5×2 − 0.3y2)(1.5×2 + 0.3y2) = 2.25×4 − 0.09y4

 

Question: 2

Evaluate each of the following using identities:

(i) (399)2

(ii) (0.98)2

(iii) 991 × 1009

(iv) 117 × 83

Solution:

(i) We have,

399= (400-1)2

= (400)2 + (1)2 - 2 × 400 ×1     [(a - b)2 = a2 + b2 - 2ab]

Where, a = 400 and b = 1

= 160000 + 1 - 8000

= 159201

Therefore, (399)2 = 159201.

(ii)  We have,

(0.98)2 = (1-0.02)2

= (1)+ (0.02)2 - 2 × 1 × 0.02

= 1 + 0.0004 - 0.04   [Where, a = 1 and b = 0.02]

= 1.0004 - 0.04

= 0.9604

Therefore, (0.98)2 = 0.9604

(iii) 991 × 1009

Solution:

We have,

991 × 1009

= (1000 - 9)(1000 + 9)

= (1000)2 - (9)2      [(a + b)(a - b) = a2 - b2]

= 1000000 - 81        [Where a = 1000 and b = 9]

= 999919

Therefore, 991 × 1009 = 999919

(iv)  We have,

117 × 83

= (100 + 17)(100 - 17)

= (100)2 - (17)2     [(a + b)(a - b) = a2 - b2]

= 10000 - 289        [Where a = 100 and b = 17]

= 9711

Therefore, 117 × 83 = 9711

 

Question: 3

Simplify each of the following:

(i) 175 × 175 + 2 × 175 × 25 + 25 × 25

(ii) 322 × 322 - 2 × 322 × 22 + 22 × 22

(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24

Solution:

(i) We have,

175 × 175 + 2 × 175 × 25 + 25 × 25 = (175)2 + 2 (175) (25) + (25)2

= (175 + 25)2        [a2 + b2 + 2ab = (a + b)2]

= (200)2               [Where a = 175 and b = 25]

= 40000

Therefore, 175 × 175 + 2 × 175 × 25 + 25 × 25 = 40000.

(ii)  We have,

322 × 322 - 2 × 322 × 22 + 22 × 22

= (322-22)2       [a2+ b2- 2ab = (a – b)2]

= (300)2            [ Where a = 322 and b = 22]

= 90000

Therefore, 322 × 322 - 2 × 322 × 22 + 22 × 22 = 90000.

(iii)  We have,

0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24

= (0.76 + 0.24)2          [a2 + b2 + 2ab = (a + b)2]

= (1.00)2                       [Where a = 0.76 and b = 0.24]

= 1

Therefore, 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24 = 1.

(iv) We have,

 

Question: 4

If x + 1/x = 11, find the value of x2 + 1/x2

Solution:

We have, x + 1/x = 11

Now, (x + 1/x)2 = x2 + (1/x)2 + 2 ∗ x∗1/x

⇒ (x + 1/x)2 = x/2 + 1/x2 + 2

⇒(11)2 = x2 + 1/x2 + 2[?x + 1/x = 11]

⇒ 121 = x2 + 1/x2 + 2

⇒ x2 + 1/x2 = 119

 

Question: 5

If x − 1/x = −1, find the value of x2 + 1/x2

Solution:

We have, x − 1/x = −1

Now, (x − 1/x)2 = x2 + (1/x)2 − 2 ∗ x ∗ 1/x

⇒ (x − 1/x)2 = x2 + 1/x2 − 2

⇒ (−1)2 = x2 + 1/x2 − 2[ ∴ x − 1/x = −1]

⇒ 2 + 1 = x2 + 1/x2

⇒ x2 + 1/x2 = 3

 

Question: 6

Solution:

We have,

(x + 1/x)2 = x2 + (1/x)2 + 2 ∗ x ∗ 1/x

⇒ (x + 1/x)2 = x2 + 1/x2 + 2

⇒ 5 = x2 + 1/x2 + 2

⇒ x2 + 1/x2 = 3  ...(1)

Now, (x2 + 1/x2)2 = x4 + 1/x4 + 2 ∗ x2 ∗ 1/x2

⇒ (x2 + 1 × 2)2 = x4 + 1/x4 + 2

⇒ 9 = x4 + 1/x4 + 2 [∴ x2 + 1/x2 = 3]

⇒ x4 + 1/x4 = 7

Hence, x2 + 1/x2 = 3; x4 + 1/x4 = 7.

 

Question: 7

If x2 + 1/x2 = 66, find the value of x − 1/x

Solution:

We have,

(x − 1/x)2 = x2 + (1/x)2 − 2 ∗ x ∗ 1/x

⇒ (x − 1/x)2 = x2 + 1/x2 − 2

⇒ (x − 1/x)2 = 66 – 2   [∴ x2 +1/x2 = 66]

⇒ (x − 1/x)2 = 64

⇒ (x − 1/x)2 = (± 8)2

⇒ x − 1/x = ± 8

 

Question: 8

If x2 + 1/x2 = 79, find the value of x + 1/x

Solution:

We have,

(x + 1/x)2 = x2 + (1/x)2 + 2 ∗ x ∗ 1/x

⇒ (x + 1/x)2 = x2 + 1/x2 + 2

⇒ (x + 1/x)2 = 79 + 2      [∴ x2 + 1/x2 = 79]

⇒ (x + 1/x)2 = 81

⇒ (x + 1/x)2 = (± 9)2

⇒ x +1/x = ± 9

 

Question: 10

If 9x+ 25y= 181 and xy = -6, find the value of 3x + 5y.

Solution:

We have,

(3x + 5y)2 = (3x)2 + (5y)2 + 2 * 3x * 5y

⇒ (3x + 5y)2 = 9×2 + 25y2 + 30xy

= 181 + 30(-6)             [Since, 9x+ 25y= 181 and xy = - 6]

⇒ (3x + 5y)2 = 1

⇒ (3x + 5y)2 = (± 1)2

⇒ 3x + 5y = ± 1

 

Question: 11

If 3x - 7y = 10 and xy = -1, find the value of 9x+ 49y2.

Solution:

We have,

(2 - 7y)2 = (3x)2 + (-7y)2 - 2 * 3x *7y

⇒ (3x - 7y)2 = 9x2 + 49y2 - 42xy           [Since, 3x - 7y = 10 and xy = -1]

⇒ (10)2 = 9x2 + 49y2 + 42

⇒ 100 - 42 = 9x2 + 49y2

⇒ 9×2 + 49y2 = 58

 

Question: 12

Simplify each of the following products:

(ii) (m + n/7)3(m − n/7)

(iii) (x/2 − 2/5)(2/5 − x/2) − x2 + 2x

(iv) (x2 + x − 2)(x2 − x + 2)

(v) (x3 − 3x − x)(x2 − 3x + 1)

(vi) (2x4 − 4x2 + 1)(2x4 − 4x2 − 1)

Solution:

 (i) We have,

(ii) We have,

We have,

(m + n/7)3(m − n/7)

= (m + n/7)(m + n/7)(m + n/7)(m − n/7)

= (m + n/7)2(m2 − (n/7)2)  [∴ (a + b)(a + b) = (a + b)2 and (a + b)(a − b) = a2 − b2]

= (m + n7)2(m2 − n2/49)

 ∴ (m + n/7)3(m − n/7) = (m + n/7)2(m2 − n2/49)

(iii) We have,

(x/2 − 2/5)(2/5 − x/2) − x2 + 2x

⇒ −(2/5 − x/2)(2/5 − x/2) − x2 + 2x

⇒ −(2/5 − x/2)2 − x2 + 2x    [∴ (a − b)(a − b) = (a − b)2]

⇒ −[(2/5)2 + (x/2)2 − 2(2/5)(x/2)] − x2 + 2x

⇒ −(4/25 + x2/4 − 2x/5) − x2 + 2x

⇒ −x2/4 − x2 + 2x/5 + 2x − 4/25

⇒ −5x2/4 + 12x/5 − 4/25

∴ (x/2 − 2/5)(2/5 − x/2) − x2 + 2x = −5x2/4 + 12x/5 − 4/25

(iv) We have,

(x2 + x − 2)(x2 − x + 2)

[(x)2 + (x − 2)][(x2 − (x + 2)]

⇒ (x2)2 − (x − 2)2              [(a - b)(a + b) = a2 - b2]

⇒ x4 − (x2 + 4 − 4x)        [∴ (a − b)2 = a2 + b2 − 2ab]

⇒ x4 − x2 + 4x − 4

∴ (x2 + x − 2)(x2 − x + 2) = x4 − x2 + 4x − 4

(v) we have,

(x3 − 3x − x)(x2 − 3x + 1)

⇒ x(x2 − 3x − 1)(x2 − 3x + 1)

⇒ x[(x2 − 3x)2 − (1)2]    [∴ (a + b)(a − b) = a2 − b2]

⇒ x[(x2)2 + (−3x)2 − 2(3x)(x2) − 1]

⇒ x[x4 + 9x2 − 6x3 − 1]

⇒ x5 − 6x4 + 9x3 − x

∴ (x3 − 3x − x)(x2 − 3x + 1) = x5 − 6x4 + 9x3 − x

(vi) We have,

(2x4 − 4x2 + 1)(2x4 − 4x2 − 1)

⇒ [(2x4 − 4x2)2 − (1)2]  [∴ (a + b)(a − b) = a2 − b2]

⇒ [(2x4)2 + (4x2)2 − 2(2x4)(4x2) − 1]

⇒ 4x8 − 16x6 + 16x4 − 1 [∴ (a − b)2 = a2 + b2 − 2ab]

∴ (2x4 − 4x2 + 1)(2x4 − 4x2 − 1) = 4x8 − 16x6 + 16x4 − 1

 

Question: 13

Prove that a2 + b2 + c2 − ab − bc − ca is always non-negative for all values of a, b and c.

Solution:

We have,

a2 + b2 + c2 − ab − bc − ca

Multiply and divide by ‘2’

= 2/2[a2 + b2 + c2 − ab − bc − ca]

= 1/2[2a2 + 2b2 + 2c2 − 2ab − 2bc − 2ca]

= 1/2[a2 + a2 + b2 + b2 + c2 + c2 − 2ab − 2bc − 2ca]

= 1/2[(a2 + b2 − 2ab) + (a2 + c2 − 2ca) + (b2 + c2 − 2bc)]

= 1/2[(a − b)2 + (b − c)2 + (c − a)2] [?(a − b)2 = a2 + b2 − 2ab]

∴  a2 + b2 + c2 − ab − bc − ca ≥ 0

Hence, a2 + b2 + c2 − ab − bc − ca ≥ 0 is always non-negative for all values of a, b and c.


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