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,
3992 = (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)2 + (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 9x2 + 25y2 = 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, 9x2 + 25y2 = 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 9x2 + 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.