Prasanna – Page 274

RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS

January 19, 2023

RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
The point of intersect of the co-ordiante axes is
(a) ordinate
(b) abscissa
(c) quadrant
(d) origin
Solution:
Origin (d)

Question 2.
The abscissa and ordinate of the origin are
(a) (0, 0)
(b) (1, 0)
(c) (0, 1)
(d) (1, 1)
Solution:
The abscissa and ordinate of the origin are (0, 0). (a)

Question 3.
The measure of the angle between the co-ordinate axes is
(a) 0°
(b)   90°
(c) 180°
(d)   360°
Solution:
The measure of the angle between the coordinates of axes is 90°. (b)

Question 4.
A point whose abscissa and ordinate are 2 and -5 respectively, lies in
(a) First quadrant
(b) Second quadrant
(c) Third quadrant
(d) Fourth quadrant
Solution:
The point whose abscissa is 2 and ordinate -5 will lies in fourth quadrant. (d)

Question 5.
Points (-4, 0) and (7, 0) lie
(a) on x-axis
(b)   y-axis
(c) in first quadrant
(d) in second quadrant
Solution:
∵ The ordinates of both the points are 0
∴ They lie on x-axis            (a)

Question 6.
The ordinate of any point on x-axis is
(a) 0
(b) 1
(c) -1
(d) any number
Solution:
The ordinate of any point lying on x-axis is 0.          (a)

Question 7.
The abscissa of any point on y-axis is
(a) 0
(b) 1
(c) -1
(d) any number
Solution:
The abscissa of any point on y-axis is 0. (a)

Question 8.
The abscissa of a point is positive in the
(a) First and Second quadrant
(b) Second and Third quadrant
(c) Third and Fourth quadrant
(d) Fourth and First quadrant
Solution:
The abscissa of a point is positive in the fourth and First quadrant. (d)

Question 9.
A point whose abscissa is -3 and ordinate 2 lies in
(a) First quadrant
(b) Second quadrant
(c) Third quadrant
(d) Fourth quadrant
Solution:
A point (-3, 2) will lies in second quadrant. (b)

Question 10.
Two points having same abscissae but different ordinates lie on
(a) x-axis
(b) y-axis
(c) a line parallel to y-axis
(d) a line parallel to x-axis
Solution:
Two points having same abscissae but different ordinates is a line parallel to y- axis. (c)

Question 11.
The perpendicular distance of the point P (4, 3) from x-axis is
(a) 4
(b) 3
(c) 5
(d) none   of these
Solution:
The perpendiculat distance of the point P (4, 3) from x-axis is 3.         (b)

Question 12.
The perpendicular distance of the point P (4, 3) from y-axis is
(a) 4
(b) 3
(c) 5
(d) none of these
Solution:
perpendicular distance of the point P (4, 3) from y-axis is 4.         (a)

Question 13.
The distance of the point P (4, 3) from the origin is
(a) 4
(b) 3
(c) 5
(d) 7
Solution:
The distance of the point P (4, 3) from origin is \(\sqrt { { 4 }^{ 2 }+{ 3 }^{ 3 } }\)
= \(\sqrt { 16+9 }\)
= \(\sqrt { 25 }\) = 5 (c)

Question 14.
The area of the triangle formed by the points A (2, 0), B (6, 0) and C (4, 6) is
(a) 24 sq. units
(b) 12 sq. units
(c) 10 sq. units
(d) none of these
Solution:
RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS Q14.1

Question 15.
The area of the triangle formed by the points P (0, 1), Q (0, 5) and R (3, 4) is
(a) 16 sq. units
(b) 8 sq. units
(c) 4 sq. units
(d) 6 sq. units
Solution:
RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS Q15.1

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RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2

January 19, 2023

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2

Other Exercises

Question 1.
Write two solutions for each of the following equations
(i) 3x + 4y = 7
(ii) x = 6y
(iii) x + πy = 4
(iv) \(\frac { 2 }{ 3 }\) x – y = 4
Solution:

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2 Q1.1
(ii) x = 6y
Let y = 0, then
x = 6 x 0 = 0
∴ x = 0, y = 0
x = 0, y = 0 are the solutions of the equation
Let y= 1, then
x = 6 x 1 = 0 –
∴ x = 6, y = 1 are the solutions of the equation.
(iii) x + πy = 4
Let x = 4, then
4 + πy = 4
⇒ πy = 4- 4 = 0
∴ y = 0
∴ x = 4, y = 0 are the solutions of the equation
Let x = 0, then
0 + πy = 4 ⇒ πy = 4
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2 Q1.2

Question 2.
Check which of the following are solutions of the equations 2x – y =6 and which are not
(i) (3, 0)
(ii) (0, 6)
(iii) (2,-2)
(iv)(\(\sqrt { 3 } \) ,0)
(v) (\(\frac { 1 }{ 2 }\) ,-5)
Solution:
Equation is 2x – y = 6
(i) Solution is (3, 0) i.e. x = 3, y = 0
Substituting the value of x and y in the equation
2 x 3 – 0 = 6 ⇒ 6 – 0 = 6
6 = 6
Which is true
∴ (3, 0) is the solutions.
(ii) (0, 6) i.e. x =0, y =6
Substituting the value of x and y in the equation
2 x 0 – 6 = 6 ⇒ 0-6 = 6
⇒ -6 = 6 which is not true
∴ (0, 6) is not its solution’
(iii) (2, -2) i.e. x = 2, y = -2
Substituting the value of x and y in the equation
2 x 2 – (-2) = 6 ⇒ 4 + 2 = 6
⇒ 6 = 6 which is true.
∴ (2, -2) is the solution.
(iv) (\(\sqrt { 3 } \),0) i.e. x = \(\sqrt { 3 } \) , y = 0,
Substituting the value of x and y in the equations
2 x \(\sqrt { 3 } \)-(0) = 6
⇒ 2\(\sqrt { 3 } \)-0 = 6
⇒ 2 \(\sqrt { 3 } \) =  6 which is not true
∴ (\(\sqrt { 3 } \) > 0) is not the solution.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2 Q2.1

Question 3.
If x = -1, y = 2 is a solution of the equation 3x + 4y =k Find the value of k.
Solution:
x = -1, y = 2
The equation is 3x + 4y = k
Substituting the value of x and y in it
3 x (-1) + 4 (2) = k
⇒ -3+ 8 = k
⇒ 5 = k
∴ k = 5

Question 4.
Find the value of λ if x = -λ and y = \(\frac { 5 }{ 2 }\) is a solution of the equation x + 4y – 7 = 0.
Solution:
x = -λ, y= \(\frac { 5 }{ 2 }\)
Equation is x + 4y – 7 = 0
Substituting the value of x and y,
-λ + 4 x \(\frac { 5 }{ 2 }\) -7 = 0
⇒ -λ + 10 – 7 = 0
⇒ -λ +3 = 0
∴ -λ = -3
⇒ λ = 3
Hence λ = 3

Question 5.
If x = 2α + 1 and y = α – 1 is a solution of the equation 2x – 3y + 5 = 0, find the value of α.
Solution:
x = 2α + 1, y = α – 1
are the solution of the equation 2x – 3y + 5 – 0
Substituting the value of x and y
2(2α + 1) -3 (α – 1) + 5 = 0
⇒ 4α+ 2-3α+ 3 + 5 = 0
⇒ α+10 = 0
⇒ α = -10
Hence α = -10

Question 6.
If x = 1, and y = 6 is a solution of the equation 8x – ay + a² = 0, find the values of a.
Solution:
x = 1, y = 6 is a solution of the equation
8x – ay + a² = 0
Substituting the value of x and y,
8 x 1-a x 6 + a² = o
⇒ 8 – 6a + a² = 0
⇒ a² – 6a + 8 = 0
⇒ a² – 2a -4a + 8 = 0
⇒ a (a – 2) – 4 (a – 2) = 0
⇒ (a – 2) (a – 4) = 0
Either a – 2 = 0, then a = 2
or a – 4 = 0, then a = 4
Hence a = 2, 4

Question 7.
Write two solutions of the form x = 0, y = a and x = b, y = 0 for each of the following equations.
(i) 5x – 2y = 10
(ii) -4x + 3y = 12
(iii) 2x + 3y = 24
Solution:
(i) 5x – 2y = 10
Let x = 0, then
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2 Q7.1

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RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1

January 18, 2023

RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1

Other Exercises

Question 1.
Which of the following numbers are perfect squares ?
(i)484
(ii) 625
(iii) 576
(iv) 941
(v) 961
(vi) 2500
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 1
Grouping the factors in pairs, we have left no factor unpaired
∴ 484 is a perfect square of 22

∴ Grouping the factors in pairs, we have left no factor unpaired
∴ 625 is a perfect square of 25.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 576 is a perfect square of 24
(iv) 941 has no prime factors
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 4
∴ 941 is not a perfect square.
(v) 961 =31 x 31
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 961 is a perfect square of 31

Grouping the factors in pairs, we see that no factor is left impaired
∴ 2500 is a perfect square of 50 .

Question 2.
Show that each of the following* numbers is a perfect square. Also find the number whose square is the given number in each case :
(i) 1156
(ii) 2025
(iii) 14641
(iv) 4761
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 6
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 1156 is a perfect square of 2 x 17 = 34

Grouping the factors in pairs, we see that no factor is left unpaired
2025 is a perfect square of 3 x 3 x 5 =45
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 8
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 14641 is a perfect square of 11×11 = 121

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 4761 is a perfect square of 3 x 23 = 69

Question 3.
Find the smallest number by which the given number must be multiplied so that the product is a perfect square.
(i) 23805
(ii) 12150
(iii) 7688
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 10
Grouping the factors in pairs of equal factors, we see that 5 is left unpaird
∴ In order to complete the pairs, we have to multiply 23805 by 5, then the product will be the perfect square.
Requid smallest number = 5
(ii) 12150 = 2 x 3 x 3×3 x 3×3 x 5×5
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 11
Grouping the factors in pairs of equal factors, we see that factors 2 and 3 are left unpaired
∴ In order to complete the pairs, we have to multiply 12150 by 2 x 3 =6 i.e., then the product will be the complete square.
∴ Required smallest number = 6
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 12
Grouping the factors in pairs of equal factors, we see that factor 2 is left unpaired
∴ In order to complete the pairs we have to multiply 7688 by 2, then the product will be the complete square
∴ Required smallest number = 2

Question 4.
Find the smallest number by which the given number must be divided so that the resulting number is a perfect square.
(i) 14283
(ii) 1800
(iii) 2904
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 13
Grouping the factors in pairs of equal factors, we see that factors we see that 3 is left unpaired
Deviding by 3, the quotient will the perfect square.

Grouping the factors in pair of equal factors, we see that 2 is left unpaired.
∴ Dividing by 2, the quotient will be the perfect square.
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 15
Grouping the factors in pairs of equal factors, we see that 2 x 3 we left unpaired
∴ Dividing by 2 x 3 = 6, the quotient will be the perfect square.

Question 5.
Which of the following numbers are perfect squares ?
11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121
Solution:
11 is not a perfect square as 11 = 1 x 11
12 is not a perfect square as 12 = 2×2 x 3
16 is a perfect square as 16 = 2×2 x 2×2
32 is not a perfect square as 32 = 2×2 x 2×2 x 2
36 is a perfect square as 36 = 2×2 x 3×3
50 is not a perfect square as 50 = 2 x 5×5
64 is a perfect square as 64 = 2×2 x 2×2 x 2×2
79 is not a perfect square as 79 = 1 x 79
81 is a perfect square as 81 = 3×3 x 3×3
111 is not a perfect square as 111 = 3 x 37
121 is a perfect square as 121 = 11 x 11
Hence 16, 36, 64, 81 and 121 are perfect squares.

Question 6.
Using prime factorization method, find which of the following numbers are perfect squares ?
∴ 189,225,2048,343,441,2916,11025,3549
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 16
Grouping the factors in pairs, we see that are 3 and 7 are left unpaired
∴ 189 is not a perfect square

Grouping the factors in pairs, we see no factor left unpaired
∴ 225 is a perfect square

Grouping the factors in pairs, we see no factor left unpaired
∴ 2048 is a perfect square
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 19
Grouping the factors in pairs, we see that one 7 is left unpaired
∴ 343 is not a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 441 is a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 2916 is a perfect square.
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 22
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 11025 is a perfect square.

Grouping the factors in pairs, we see that 3, no factor 7 are left unpaired
∴ 3549 is a perfect square.

Question 7.
By what number should each of the following numbers be multiplied to get a perfect square in each case ? Also, find the number whose square is the new number.
(i) 8820
(ii) 3675
(iii) 605
(iv) 2880
(v) 4056
(vi) 3468
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 24
Grouping the factors in pairs, we see that 5 is left unpaired
∴ By multiplying 8820 by 5, we get the perfect square and square root of product will be
= 2 x 3 x 5 x 7 = 210

Grouping the factors in pairs, we see that 3 is left unpaired
∴ Multiplying 3675 by 3, we get a perfect square and square of the product will be
= 3 x 5 x 7 = 105
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 26
Grouping the factors in pairs, we see that 5 is left unpaired
∴ Multiplying 605 by 5, we get a perfect square and square root of the product will be
= 5 x 11 =55

Grouping the factors in pairs, we see that 5 is left unpaired
∴ Multiplying 2880 by 5, we get the perfect square.
Square rooi of product will be = 2 x 2 * 2 – 3 x 5 = 120
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 28
Grouping the factors in pairs, we see that 2 and 3 are left unpaired
∴ Multiplying 4056 by 2 x 3 i.e., 6, we get the perfect square.
and square root of the product will be
= 2 x 2 x 3 x 13 = 156

Grouping the factors in pairs, we see that 3 is left unpaired
∴ Multiplying 3468 by 3 we get a perfect square, and square root of the product will be 2 x 3 x 17 = 102
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 30
Grouping the factors in pairs, we see that 2 and 3 are left unpaired
∴ Multiplying 7776 by 2 x 3 or 6 We get a perfect square and square root of the product will be
= 2 x 2 x 2 x 3 x 3 x 3 = 216

Question 8.
By what numbers should each of the following be .divided to get a perfect square in each case ? Also find the number whose square is the new number.
(i) 16562
(ii) 3698
(iii) 5103
(iv) 3174
(v) 1575
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 31
Grouping the factors in pairs, we see that 2 is left unpaired
∴ Dividing by 2, we get the perfect square and square root of the quotient will be 7 x 13 = 91

Grouping the factors in pairs, we see that 2 is left unpaired,
∴ Dividing 3698 by 2, the quotient is a perfect square
and square of quotient will be = 43
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 33
Grouping the factors in pairs, we see that 7 is left unpaired
∴ Dividing 5103 by 7, we get the quotient a perfect square.
and square root of the quotient will be 3 x 3 x 3 = 27

Grouping the factors iq pairs, we see that 2 and 3 are left unpaired
∴ Dividing 3174 by 2 x 3 i.e. 6, the quotient will be a perfect square and square root of the quotient will be = 23
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 35
Grouping the factors in pairs, we find that 7 is left unpaired i
∴ Dividing 1575 by 7, the quotient is a perfect square
and square root of the quotient will be = 3 x 5 = 15

Question 9.
Find the greatest number of two digits which is a perfect square.
Solution:
The greatest two digit number = 99 We know, 92 = 81 and 102 = 100 But 99 is in between 81 and 100
∴ 81 is the greatest two digit number which is a perfect square.

Question 10.
Find the least number of three digits which is perfect square.
Solution:
The smallest three digit number =100
We know that 92 = 81, 102 = 100, ll2 = 121
We see that 100 is the least three digit number which is a perfect square.

Question 11.
Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.
Solution:
By factorization:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 36
Grouping the factors in pairs, we see that 11 is left unpaired
∴ The least number is 11 by which multiplying 4851, we get a perfect square.

Question 12.
Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.
Solution:
By factorization,
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 37
Grouping the factors in pairs, we see that 13 is left unpaired
∴ Dividing 28812 by 3, the quotient will be a perfect square.

Question 13.
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number.
Solution:
By factorization,
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 38
Grouping the factors in pairs, we see that one 2 is left unpaired.
∴ Dividing 1152 by 2, we get the perfect square and square root of the resulting number 576, will be 2 x 2 x 2 x 3 = 24

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RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2

January 18, 2023

RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2

Question 1.
Rationalise the denominators of each of the following(i – vii):
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q1.1 >
Solution:

Question 2.
Find the value to three places of decimals of each of the following. It is given that
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q2.1
Solution:

Question 3.
Express each one of the following with rational denominator:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q3.1
Solution:

Question 4.
Rationales the denominator and simplify:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q4.1
Solution:

Question 5.
Simplify:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q5.1
Solution:

Question 6.
In each of the following determine rational numbers a and b:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q6.1
Solution:

Question 7.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q7.1
Solution:

Question 8.
Find the values of each of the following correct to three places of decimals, it being given that \(\sqrt { 2 } \) = 1.4142, \(\sqrt { 3 } \) = 1-732, \(\sqrt { 5 } \) = 2.2360, \(\sqrt { 6 } \) = 2.4495 and \(\sqrt { 10 } \) = 3.162.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q8.1
Solution:

Question 9.
Simplify:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q9.1
Solution:

Question 10.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q10.1
Solution:

Question 11.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q11.1
Solution:

Question 12.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q12.1
Solution:

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RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2

January 18, 2023

RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2

Other Exercises

Question 1.
Write each of the following in exponential form :
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 1
Solution:

Question 2.
Evaluate :
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 4
Solution:

Question 3.
Express each of the following as a rational number in the form \(\frac { p }{ q } :\)
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 7
Solution:

Question 4.
Simplify :
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 9
Solution:

Question 5.
Express each of the following rational numbers with a negative exponent :
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 11
Solution:

Question 6.
Express each of the following rational numbers with a positive exponent :
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 13
Solution:

Question 7.
Simplify :
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 15
Solution:

Question 8.
By what number should 5^-1 be multiplied so that the product may be equal to (-7)^-1 ?
Solution:
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 18

Question 9.
By what number should \({ \left( \frac { 1 }{ 2 } \right) }^{ -1 }\) be multiplied so that the product may be equal to \({ \left( \frac { -4 }{ 7 } \right) }^{ -1 }\) ?
Solution:
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 19

Question 10.
By what number should (-15)^-1 be divided so that the quotient may be equal to (-5)^-1 ?
Solution:
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 20

Question 11.
By what number should \({ \left( \frac { 5 }{ 3 } \right) }^{ -2 }\) be multiplied so that the product may be \({ \left( \frac { 7 }{ 3 } \right) }^{ -1 }\) ?
Solution:
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 21

Question 12.
Find x, if
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 22
Solution:

Question 13.
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 27
Solution:

Question 14.
Find the value of x for which 5^2x + 5^-3 = 5⁵.
Solution:
RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2 29

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RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS

January 18, 2023

RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS

Question 1.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS Q1.1
Solution:

Question 2.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS Q2.1
Solution:

Question 3.
If a + b = 7 and ab = 12, find the value of a² + b².
Solution:
a + b = 7, ab = 12
Squaring both sides,
(a + b)² = (7)²⇒ a² + b² + 2ab = 49
⇒ a² + b² + 2 x 12 = 49
⇒ a² + b² + 24 = 49
⇒ a² + b² = 49 – 24 = 25
∴ a² + b² = 25

Question 4.
If a – b = 5 and ab = 12, find the value of a² + b² .
Solution:
a – b = 5, ab = 12
Squaring both sides,
⇒ (a – b)² = (5)²⇒ a² + b² – 2ab = 25
⇒ a² + b² – 2 x 12 = 25
⇒ a² + b² – 24 = 25
⇒ a² + b² = 25 + 24 = 49
∴ a² + b² = 49

Question 5.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS Q5.1
Solution:

Question 6.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS Q6.1
Solution:

Question 7.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS Q7.1
Solution:

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RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself

January 18, 2023

RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself

Other Exercises

MCQ
Question 1.
Solution:
(b) Polynomial is x² – 2x – 3
=> x² – 3x + x – 3
= x(x – 3) + 1(x – 3)
= (x – 3) (x + 1)
Either x – 3 = 0, then x = 3
or x + 1 = 0, then x = -1
Zeros are 3, -1

Question 2.
Solution:
(a) α, β, γ are the zeros of Polynomial is x³ – 6x² – x + 30
Here, a = 1, b = -6, c = -1, d = 30
αβ + βγ + γα = \(\frac { c }{ a }\) = \(\frac { -1 }{ 1 }\) = -1

Question 3.
Solution:
(c)
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 1

Question 4.
Solution:
(c) Let α and β be the zeros of the polynomial 4x² – 8kx + 9
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 3

Short-Answer Questions
Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 4
Either x + 15 = 0, then x = -15
or x – 13 = 0, then x = 13
Zeros are 13, -15

Question 6.
Solution:
The polynomial is (a² + 9) x² + 13x + 6a
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 5

Question 7.
Solution:
Zeros are 2 and -5
Sum of zeros = 2 + (-5) = 2 – 5 = -3
and product of zeros = 2 x (-5) = -10
Now polynomial will be
x² – (Sum of zeros) x + Product of zeros
= x² – (-3)x + (-10)
= x² + 3x – 10

Question 8.
Solution:
(a – b), a, (a + b) are the zeros of the polynomial x³ – 3x² + x + 1
Here, a = 1, b = -3, c = 1, d = 1
Now, sum of zeros = \(\frac { -b }{ a }\) = \(\frac { -(-3) }{ 1 }\) = 3
a – b + a + a + b = 3
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 6

Question 9.
Solution:
Let f(x) = x³ + 4x² – 3x – 18
If 2 is its zero, then it will satisfy it
Now, (x – 2) is a factor Dividing by (x – 2)
Hence, x = 2 is a zero of f(x)
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 7

Question 10.
Solution:
Sum of zeros = -5
and product of zeros = 6
Quadratic polynomial will be
x² – (Sum of zeros) x + Product of zeros
= x² – (-5) x + 6
= x² + 5x + 6

Question 11.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 8

Question 12.
Solution:
p(x) = x³ + 3x² – 5x + 4
g(x) = x – 2
Let x – 2 = 0, then x = 2
Remainder = p(2) = (2)³ + 3(2)² – 5(2) + 4 = 8 + 12 – 10 + 4 = 14

Question 13.
Solution:
f(x) = x³ + 4x² + x – 6
and g(x) = x + 2
Let x + 2 = 0, then x = -2
f(-2) = (-2)³ + 4(-2)² + (-2) – 6 = -8 + 16 – 2 – 6 = 0
Remainder is zero, x + 2 is a factor of f(x)

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 9

Question 15.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 11

Question 16.
Solution:
f(x) = x⁴ + 4x² + 6
=> (x²)² + 4(x²) + 6 = y² + 4y + 6 (Let x² = y)
Let α, β be the zeros of y² + 4y + 6
Sum of zeros = -4
and product of zeros = 6
But there are no factors of 6 whose sum is -4 {Factors of 6 = 1 x 6 and 2 x 3}
Hence, f(x) Has no zero (real).

Long-Answer Questions
Question 17.
Solution:
3 is one zero of p(x) = x³ – 6x² + 11x – 6
(x – 3) is a factor of p(x)
Dividing, we get
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 12

Question 18.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 13

Question 19.
Solution:
p(x) = 3x⁴ + 5x³ – 7x² + 2x + 2
Dividing by x² + 3x + 1,
we get,
Quotient = 3x² – 4x + 2
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 15

Question 20.
Solution:
Let p(x) = x³ + 2x² + kx + 3
g(x) = x – 3
and r(x) = 21
Dividing p(x) by g(x), we get
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself 16
But remainder = 21
3 + 3k + 45 = 21
3k = 21 – 45 – 3
=> 3k = 21 – 48 = -27
k = -9
Second method:
x – 3 is a factor of p(x) : x = 3
Substituting the value of x in p(x)
p(3) = 3³ + 2 x 3² + k x 3 + 3
= 27 + 18 + 3k + 3
48 + 3k = 21
=> 3k = -48 + 21 = -27
k = -9
Hence, k = -9

Hope given RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Test Yourself are helpful to complete your math homework.

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Value Based Questions in Science for Class 9 Chapter 1 Matter in Our Surroundings

January 18, 2023

Value Based Questions in Science for Class 9 Chapter 1 Matter in Our Surroundings

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 1 Matter in Our Surroundings.

Question 1.
Mohan was asked by his teacher to determine the boiling point of a liquid. He set up the apparatus as shown in the figure. He recorded the boiling point of the liquid as 140°C and reported the reading to his teacher. The teacher asked him to repeat the experiment after dipping the bulb of the thermometer in the liquid and also placing the beaker on a tripod stand covered with a wire gauze. After reading this narration, answer the following questions:
Value Based Questions in Science for Class 9 Chapter 1 Matter in Our Surroundings image - 1

Was the reading taken by the student correct ?
What was the necessity of wire gauze ?
How did teacher help the student ?
What is the value based information associated with this ?

Answer:

No, it was wrong. The bulb of the thermometer must dip in the liquid.
If the beaker is in direct contact with the flame, it might crack.
Teacher guided the student properly. By direct heating, an accident can occur and can cause serious injuries to the student.
In the laboratory or in the Kitchen, we should never heat glass beakers directly on the flame. These are likely to crack. However, Borocil beakers can be heated on direct flame.

More Resources

Question 2.
A student was determining the melting point of ice in the laboratory. The teacher was taking the round. He found that the student was not using any stirrer. Teacher asked him to use a glass stirrer and then determine the melting point of ice.

Is the use of a stirrer necessary during heating ?
What is its purpose ?

Answer:

Yes, it is necessary.
Stirring with a glass stirrer or glass rod provides uniform heating otherwise the results may not be correct.

Question 3.
Ashutosh was getting late for the office. He tried to sip boiling hot coffee from a cup. He felt very uncomfortable. His wife immediately brought a plate and asked him to sip the coffee from the plate. Ashutosh followed her advice and did not face any problem.

Why was Ashutosh feeling uncomfortable ?
How did sipping coffee from a plate was more comfortable ?
How did wife help Ashutosh ?
What is the value associated with it ?

Answer:

Since coffee was very hot, it was not possible for Ashutosh to sip it hurridely. It could burn his tongue.
A plate has more surface area than the cup. Evaporation becomes faster in this case. Since cooling is always caused during evaporation, the temperature got lowered. It became comfortable to sip coffee from a plate,
With the timely help by his wife, Ashutosh could sip coffee quite comfortably and could reach office in time.
We should never drink milk, tea or coffee when these are boiling hot. These are injurious to our throat. It is always better to allow them to cool. In case we are in a hurry, we must place them in a container with large surface area.

Question 4.
During vacation, Neha got an opportunity to visit a village. She found ladies cooking food in open pans. She sat with them for sometime and advised them to use pressure cookers instead of open containers in order to save time as well as fuel.

How does cooking in a pressure cooker help in saving time ?
How does cooking in a pressure cooker help in saving fuel ?
What is the value associated with this gesture of Neha ?

Answer:

Inside the pressure cooker, the temperature gets raised since the pressure is higher than the atmospheric pressure. The cooking becomes fast and this saves time.
Since cooking takes place in lesser time, fuel is always saved.
Neha was quite concerned about the welfare and safety of the villagers. She therefore, gave a very sincere advise.

Hope given Value Based Questions in Science for Class 9 Chapter 1 Matter in Our Surroundings are helpful to complete your science homework.

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RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6

January 18, 2023

RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6

Other Exercises

Question 1.
Verify the property : x x y = y x x by taking :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6 1
Solution:

Question 2.
Verify the property : x x (y x z) = (x x y) x z by taking :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6 5
Solution:

Question 3.
Verify the property :xx(y + 2) = xxy + x x z by taking :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6 9
Solution:

Question 4.
Use the distributivity of multiplication of rational numbers over their addition to simplify :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6 11
Solution:

Question 5.
Find the multiplicative inverse (reciprocal) of each of the following rational numbers :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6 14

Solution:

Question 6.
Name the property of multiplication of rational numbers illustrated by the following statements :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6 18
Solution:

Question 7.
(i) The product of two positive rational numbers is always______.
(ii) The product of a positive rational number and a negative rational number is always________.
(iii) The product of two negative rational numbers is always________.
(iv) The reciprocal of a positive rational number is________.
(v) The reciprocal of a negative rational number is________.
(vi) Zero has reciprocal. The product of a rational number and its reciprocal is______.
(viii) The numbers and are their own reciprocals______.
(ix) If a is reciprocal of b, then the reciprocal of b is______.
(x) The number 0 is the reciprocal of any number______.
(xi) Reciprocal of \(\frac { 1 }{ a }\), a≠ 0 is______.
(xii) (17 x 12)^-1 = 17^-1 x________ .

Solution:
The product of two positive rational numbers is always positive.
(ii) The product of a positive rational number and a negative rational number is always negative.
(iii) The product of two negative rational numbers is always positive.
(iv) The reciprocal of a positive rational number is positive.
(v) The reciprocal of a negative rational number is negative.
(vi) Zero has no reciprocal.
(vii) The product of a rational number and its reciprocal is 1.
(viii)The numbers 1 and -1 are their own reciprocals.
(ix) If a is reciprocal of b, then the reciprocal of b is a.
(x) The number 0 is not the reciprocal of any number.
(xi) Reciprocal of \(\frac { 1 }{ a }\), a≠ 0 is a.
(xii) (17 x 12)^-1 = 17^-1 x________ .

Question 8.
Fill in the blanks :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6 21

Solution:
Fill in the blanks :

Hope given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.6 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS

January 18, 2023

RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS

Mark the correct alternative in each of the following:
Question 1.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q1.1
Solution:

Question 2.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q2.1
Solution:

Question 3.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q3.1
Solution:

Question 4.
The rationalisation factor of 2 + \(\sqrt { 3 } \) is
(a) 2 – \(\sqrt { 3 } \)
(b) \(\sqrt { 2 } \) + 3
(c) \(\sqrt { 2 } \) – 3
(d) \(\sqrt { 3 } \) – 2
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q4.1

Question 5.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q5.1
Solution:

Question 6.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q6.1
Solution:

Question 7.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q7.1
Solution:

Question 8.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q8.1
Solution:

Question 9.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q9.1
Solution:

Question 10.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q10.1
Solution:

Question 11.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q11.1
Solution:

Question 12.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q12.1
Solution:

Question 13.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q13.1
Solution:

Question 14.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q14.1
Solution:

Question 15.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q15.1
Solution:

Question 16.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q16.1
Solution:

Question 17.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q17.1
Solution:

Question 18.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q18.1
Solution:

Question 19.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q19.1
Solution:

Question 20.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q20.1
Solution:

Question 21.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q21.1
Solution:

Question 22.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q22.1
Solution:

Question 23.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q23.1
Solution:

Question 24.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q24.1
Solution:

Question 25.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS Q25.1
Solution:

Hope given RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2

January 18, 2023

RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2

Other Exercises

Factorize each of the following expressions:
Question 1.
p³ + 27
Solution:
We know that a³ + b³ = (a + b) (a² – ab + b²)
a³ – b³ = (a – b) (a² + aft + b²)
p³ + 21 = (p)³ + (3)³= (p + 3) (p²– p x 3 + 3²)
= (p + 3) (p² – 3p + 9)

Question 2.
y³ + 125
Solution:
y³ + 125 = (p)³ + (5)³= (p + 5) (p² – 5y + 5²)
= (P + 5) (p² – 5y + 25)

Question 3.
1 – 21a³Solution:
1 – 21a³ = (1)³ – (3a)³= (1 – 3a) [1² + 1 x 3a + (3a)²]
= (1 – 3a) (1 + 3a + 9a²)

Question 4.
8x³y³ + 27a³
Solution:
8x³y³ + 27a³= (2xy + 3a) [(2xy)² – 2xy x 3a + (3a)²]
= (2xy + 3a) (4x²y – 6xya + 9a²)

Question 5.
64a³ – b³Solution:
64a³ – b³ = (4a)³ – (b)³= (4a – b) [(4a)² + 4a x b + (b)²]
= (4a – b) (16a² + 4ab + b²)

Question 6.
RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2 Q6.1
Solution:

Question 7.
10x⁴– 10xy⁴Solution:
I0x⁴y- 10xy⁴ = 10xy(x³ -y³)
= 10xy(x – y) (x² + xy + y²)

Question 8.
54x⁶y + 2x³y⁴Solution:
54 x⁶y + 2x³y⁴ = 2x³y(27x³ + y³)
= 2x³y[(3x)³ + (y)³]
= 2x³y(3x + y) [(3x)² -3x x y + y²]
= 2x³y(3x + y) (9x² -3xy + y²)

Question 9.
32a³ + 108b³
Solution:
32a³ + 108b³= 4(8a³ + 27b³) = 4 [(2a)³ + (3 b)³]
= 4(2a + 3b) [(2a)² – 2a x 3b + (3b)²]
= 4(2a + 3b) (4a² – 6ab + 9b²)

Question 10.
(a – 2b)³ – 512b³
Solution:
(a – 2b)³ – 512b³= (a – 2b)³ – (8b)³= (a – 2b- 8b) [(a – 2b)² + (a – 2b) x 8b + (8b)²]
= (a – 10b) [a² + 4b² – 4ab + 8ab – 16b² + 64b²]
= (a – 10b) (a² + 4ab + 52b²)

Question 11.
8x²y³ – x⁵Solution:
8x²y³ – x⁵ = x²(8y³ – X³)
= x²(2y)³ – (x)³]
= x²[(2y – x) (2y)² + 2y x x + (x)²]
= x²(2y – x) (4y² + 2xy + x²)

Question 12.
1029 -3x³Solution:
1029 – 3X³ = 3(343 – x³) ‘
= 3 [(7)³ – (x)³]
= 3(7 – x) (49x + 7x + x²)

Question 13.
x³y³+ 1
Solution:
x³y³ + 1 = (xy)³ + (1)³= (xy + 1) [(xy)² – xy x 1 + (1)²]
= (xy + 1) (x²y² – xy + 1)
= (xy + 1) (x²y – xy + 1)

Question 14.
x⁴y⁴ – xy
Solution:
x⁴y⁴ – xy = xy(x³y³ – 1)
= xy[(xy³-(1)³]
= xy (xy – 1) [x²y² + 2xy + 1]

Question 15.
a³ + b³ + a + b
Solution:
a³ + b³ + a + b
= (a + b) (a² – ab + b²) + 1 (a + b)
= (a + b) (a² – ab + b² + 1)

Question 16.
Simplify:
RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2 Q16.1
Solution:

Question 17.
(a + b)³ – 8(a – b)³
Solution:
(a + b)³ – 8(a – b)³= (a + b)² – (2a – 2b)³= (a+ b – 2a + 2b) [(a + b)² + (a + b) (2a-2b) + (2a – 2b)²)]
= (3b – a) [a² + b² + 2ab + 2a² – 2ab + 2ab – 2b² + 4a² – 8ab + 4b²]
= (3b – a) [7a² – 6ab + 3b²]

Question 18.
(x + 2)³ + (x- 2)³
Solution:
(x + 2)³ + (x – 2)³= (x + 2 + x – 2) [(x + 2)² – (x + 2) (x – 2) + (x – 2)²]
= 2x [x² + 4x + 4 – (x² + 2x – 2x – 4) + x²4x + 4]
= 2x[x² + 4x + 4- x²-2x + 2x + 4+ x²– 4x + 4]
= 2x[x² + 12]

Question 19.
x⁶ +y⁶Solution:
x⁶ + y⁶= (x²)³ + (y²)³= (x² + y²) [x⁴ – x²y² + y⁴]

Question 20.
a¹² + b¹²
Solution:
a¹² + b¹² = (a⁴)³ + (b⁴)³= (a⁴ + b⁴) [(a⁴)² – a⁴b⁴ + (b⁴)²]
= (a⁴ + b⁴) (a⁸ – a⁴b⁴ + b⁸)

Question 21.
x³ + 6x² + 12x + 16
Solution:
x³ + 6x² + 12x + 16
= (x)³ + 3.x².2 + 3.x.4 + (2)³ + 8 {∵ a³ + 3a²b + 3ab² +b³ = (a + b)³}
= (x + 2)³ + 8 = (x + 2)³ + (2)³= (x + 2 + 2) [(x + 2)² – (x + 2) x 2 + (2)²] {∵ a³ + b² = (a + b) (a² – ab + b²}
= (x + 4) (x² + 4x + 4 – 2x – 4 + 4)
= (x + 4) (x² + 2x + 4)

Question 22.
RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2 Q22.1
Solution:

Question 23.
a³ + 3a²b + 3ab² + b³ – 8
Solution:
a³ + 3a²b + 3ab² + b³ – 8
= (a + b)³ – (2)³= (a + b -2)[(a + b)² + (a +b)x2 + (2)²]
= (a + b-2) (a² + b² + 2ab + 2a + 2b + 4)
= (a + b – 2) (a² + b² + 2ab + 2(a + b) + 4]
= (a + b – 2) [(a + b)² + 2(a + b) + 4}

Question 24.
8a³ – b³ – 4ax + 2bx
Solution:
8a³ – b³ – 4ax + 2bx
(2a)³ – (b)³ – 2x(2a – b)
= (2a-b)[(2a)² + 2a x b + (b)²]- 2x(2a-b)
= (2a – b) [4a² + 2ab + b²] – 2x(2a – b)
= (2a – b) [4a² + 2ab + b² – 2x]

Hope given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.