Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths

RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

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

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

Solution:

Given,

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

RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics [?(a−b)2 = a2+b2−2ab]

 

Where, a = 2x, b = 1/x

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

(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) RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

We have,

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

RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics   [?(a−b)= (a+b)(a−b)]

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

= 9

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

Q.4. If RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics ,find the value ofRD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

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

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

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


Q.5. If RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics ,find the value ofRD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution: We have,

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

Now,  RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics


Q. 6. If RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics find the value of  RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

We have,

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

Q. 7. If RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics find the value ofRD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

We have,

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

Q. 8. If RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics find the value of RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

We have,

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

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) RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

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

RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 MathematicsRD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 MathematicsRD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 MathematicsRD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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

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

Solution:

We have,

RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics   [?(a+b)(a+b) = (a+b)2 and (a+b)(a−b) = a2−b2]

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

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

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

Solution:

We have,

RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics   [?(a−b)(a−b) = (a−b)2]
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(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’

RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 MathematicsRD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 MathematicsRD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

?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.

The document RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on RD Sharma Solutions Ex-4.1, Algebraic Identities, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What are algebraic identities?
Ans. Algebraic identities are mathematical expressions that hold true for any value of the variables involved. These identities help in simplifying and solving algebraic equations. Some examples of algebraic identities are the distributive property, the commutative property, and the associative property.
2. How can algebraic identities be used to solve equations?
Ans. Algebraic identities can be used to simplify and solve equations by manipulating the expressions using the properties of algebraic identities. By applying these identities, we can transform complex expressions into simpler forms, making it easier to solve the equations and find the values of the variables.
3. Can you provide an example of how to use algebraic identities to solve an equation?
Ans. Sure! Let's consider the equation 2(x + 3) - 5x = 4x - 7. We can use the distributive property to expand 2(x + 3) as 2x + 6. Rewriting the equation, we have 2x + 6 - 5x = 4x - 7. Now, we can simplify the equation by combining like terms. -3x + 6 = 4x - 7. By rearranging the equation and solving for x, we can find the value of x.
4. Are there any specific rules to remember while using algebraic identities?
Ans. Yes, there are a few rules to keep in mind when using algebraic identities: 1. Always apply the distributive property when there is a multiplication or division outside the parentheses. 2. Pay attention to the signs of the terms when combining like terms. 3. Remember the order of operations (PEMDAS) to simplify expressions correctly.
5. Can algebraic identities be used in real-life applications?
Ans. Yes, algebraic identities have various applications in real-life situations. They are used in fields such as engineering, physics, and computer science to model and solve problems. For example, in physics, algebraic identities are used to derive equations for motion, electricity, and magnetism. In computer science, algebraic identities play a crucial role in designing algorithms and optimizing code for efficiency.
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