RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths

# RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

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Q. 1.  Evaluate each of the following using identities:

Solution:

Given,

[?(a−b)2 = a2+b2−2ab]

Where, a = 2x, b = 1/x

(ii) (2x+y) (2x-y)

Solution:

Given, (2x+y) (2x-y)

= (2x)2−(y)2[?(a+b)(a−b) = a2−b2]

= 4x2−y2

?(2x+y)(2x−y) = 4x2−y2

(iii) (a2b−ab2)2

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) (a-0.1) (a+0.1)

Solution:

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) (1.5x2−0.3y2)(1.5x2+0.3y2)

Solution:

Given, (1.5x2−0.3y2)(1.5x2+0.3y2)

= (1.5x2)2−(0.3y2)2         [?(a+b)(a−b) = a2−b2]

Where, a = 1.5x2,b = 0.3y2

= 2.25x4−0.09y4

?(1.5x2−0.3y2)(1.5x2+0.3y2) = 2.25x4−0.09y4

Q.2. Evaluate each of the following using identities:

(i) (399)2

Solution:

We have,

399= (400-1)2

= (400)2+(1)2 – 2x400x1               [ (a-b)2 = a2+ b2-2ab ]

Where, a = 400 and b = 1

= 160000 + 1 – 8000

= 159201

Therefore, (399)2 = 159201.

(ii) (0.98)2

Solution:

We have,

(0.98)2 = (1-0.02)2

= (1)+ (0.02)2 – 2x1x0.02

= 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) 991x1009

Solution:

We have,

991x1009

= (1000-9) (1000+9)

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

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

= 999919

Therefore, 991x1009 = 999919

(iv) 117x83

Solution:

We have,

117x83

= (100+17) (100-17)

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

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

= 9711

Therefore, 117x83 = 9711

Q.3. Simplify each of the following:

(i) 175 x 175 +2 x 175 x 25 + 25 x 25

Solution:

We have,

175 x 175 +2 x 175 x 25 + 25 x 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 x 175 +2 x 175 x 25 + 25 x 25 = 40000.

(ii) 322 x 322 – 2 x 322 x 22 + 22 x 22

Solution:

We have,

322 x 322 – 2 x 322 x 22 + 22 x 22

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

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

= 90000

Therefore, 322 x 322 – 2 x 322 x 22 + 22 x 22= 90000.

(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24

Solution:

We have,

0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 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 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24 = 1.

(iv)

Solution:

We have,

[?(a−b)= (a+b)(a−b)]

= 9

Q.4. If  ,find the value of

Solution:

Q.5. If  ,find the value of

Solution: We have,

Now,

Q. 6. If  find the value of

Solution:

We have,

Q. 7. If  find the value of

Solution:

We have,

Q. 8. If  find the value of

Solution:

We have,

Q. 9. 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 = 9x2 + 25y2 + 30xy

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

⇒ (3x+5y)2 = 1

⇒ (3x+5y)2 = (±1)2

⇒ 3x+5y = ±1

Q.10. If 2x + 3y = 8 and xy = 2, find the value of 4x+ 9y2.

Solution:

We have,

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

⇒ (2x + 3y)2 = 4x2 + 9y2 + 12xy                   [Since, 2x + 3y = 8 and xy = 24 ]

⇒ (8)2 = 4x2 + 9y2 + 24

⇒ 64 – 24 = 4x2 + 9y2

⇒ 4x2 + 9y2 = 40

Q. 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

⇒ 9x2 + 49y2 = 58

Q.12. Simplify each of the following products:

(i)

Solution:

Solution:

We have,

[?(a+b)(a+b) = (a+b)2 and (a+b)(a−b) = a2−b2]

Solution:

We have,

[?(a−b)(a−b) = (a−b)2]

(iv) (x2+x−2)(x2−x+2)

Solution:

(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) (x3−3x−x)(x2−3x+1)

Solution:

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) (2x4−4x2+1)(2x4−4x2−1)

Solution:

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)= a2+b2−2ab]

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

Q.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’

?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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