By using Ganita Manjari Class 9 Solutions and Part 1 Class 9 Maths Chapter 4 Exploring Algebraic Identities NCERT Solutions, students can improve their problem-solving skills.
Exploring Algebraic Identities Class 9 Solutions
Class 9 Ganita Manjari Chapter 4 Solutions
Class 9 Maths Ganita Manjari Chapter 4 Solutions Exploring Algebraic Identities
Think and Reflect (NCERT Textbook Page No. 69)
Question 1.
Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.
Solution:
Using four consecutive squares,
i.e., n2, (n + 1)2, (n + 2)2 and (n + 3)2, we find a pattern.
Here, n2 + (n + 3)2 = (n + 1)2 + (n + 2)2 + 4
For example, when n = 1
⇒ 1 + 16 = 4 + 9 + 4
⇒ 17 = 17
Think and Reflect (NCERT Textbook Page No. 71)
Question 1.
What can you say about a and b if (a + b)2 < a2 + b2?
Solution:
Given (a + b)2 < a2 + b2
We know (a + b)2 = a2 + 2ab + b2
∴ a2 + 2ab + b2 < a2 + b2
⇒ a2 + 2ab + b2 – a2 – b2 < a2 + b2 – a2 – b2 [Subtracting (a2 + b2) on both sides]
⇒ 2ab < 0, which is not possible when a and b have opposite signs.
Hence, a and b must have opposite signs.
Question 2.
What can you say about a and b if (a + b)2 > a2 + b2?
Solution:
Here (a + b)2 > a2 + b2
We know (a + b)2 = a2 + 2ab + b2
∴ a2 + 2ab + b2 > a2 + b2
⇒ a2 + 2ab + b2 – a2 – b2 > a2 + b2 – a2 – b2 [Subtracting (a2 + b2) on both sides]
⇒ 2ab > 0
It is possible only when both a and b have the same sign.
Hence, a and b must have the same sign for (a + b)2 > a2 + b2.
Question 3.
When will (a + b)2 be equal to a2 + b2?
Solution:
Given (a + b)2 = a2 + b2
⇒ a2 + 2ab + b2 = a2 + b2
⇒ a2 + b2 + 2ab – (a2 + b2) = a2 + b2 – (a2 + b2) [Subtracting (a2 + b2) on both sides]
⇒ a2 + b2 + 2ab – a2 – b2 = a2 + b2 – a2 – b2
⇒ 2ab = 0
Hence, either a = 0 or b = 0 for (a + b)2 = a2 + b2
Question 4.
Did you observe that (a + b)2 and a2 + b2 are both positive? What term will decide which is larger? Use the expansion of (a + b)2 to decide.
Solution:
Yes, we observe that both (a + b)2 and (a2 + b2) are indeed positive, as they are squares of numbers.
Clearly the term lab will determine which of the two expressions is larger.
If a and b have opposite signs, then 2ab < 0.
Hence (a + b)2 < a2 + b2
If a and b have the same sign, then 2ab > 0.
Hence (a + b)2 > a2 + b2
Think and Reflect (NCERT Textbook Page No. 73)
Question 1.
What if we replace b by -b in (a + b)2 = a2 + 2ab + b2?
Solution:
We have (a + b)2 = a2 + 2ab + b2
Put b = -b
then (a + (-b))2 = (a – b)2 = a2 – 2ab + b2
So, (a – b)2 = a2 – 2ab + b2
It is an identity. Also, both are not similar.
Think and Reflect (NCERT Textbook Page No. 76)
Question 1.
Label the squares and rectangles in Fig. (below) so that it represents the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
Solution:
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Think and Reflect (NCERT Textbook Page No. 78)
Question 1.
Try to evaluate the following using a suitable identity:
(i) 352
(ii) 652
(iii) 852
(iv) 1052
Do you observe any interesting pattern?
Solution:
(i) Here 352 = (35 + 5)(35 – 5) + 52 [∵ a2 = (a + b)(a – b) + b2]
= 40 × 30 + 25
= 1225
(ii) Here 652 = (65 + 5)(65 – 5) + 52 [∵ a2 = (a + b)(a – b) + b2]
= 70 × 60 + 25
= 4225
(iii) Here 1052 = (105 + 5)(105 – 5) + 52 [∵ a2 = (a + b)(a – b) + b2]
= 110 × 100 + 25
= 11025
(iv) Here 852 = (85 + 5)(85 – 5) + 52 [∵ a2 = (a + b)(a – b) + b2]
= 90 × 80 + 25
= 7225
∴ We observe that the square of a number ending in 5, same as multiplying the leading part by its successor and writing 25 at the end.
Question 2.
Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.
Solution:
The top row figure shows squares whose sides are (a + b + c) (a + b – c), (a – b + c), (a – b – c)
Hence total area of above squares whose sides are given = (a + b + c)2 + (a + b – c)2 + (a – b + c)2 + (a – b – c)2
Similarly, the total area of squares in the bottom row whose sides are 2a, 2b, and 2c = (2a)2 + (2b)2 + (2c)2
= 4a2 + 4b2 + 4c2
= 4(a2 + b2 + c2)
Hence, the algebraic identity represents = (a + b + c)2 + (a + b – c)2 + (a – b + c)2 + (a – b – c)2 = 4(a2 + b2 + c2)
Think and Reflect (NCERT Textbook Page No. 79)
Question 1.
Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.
Solution:
Do yourself.
Think and Reflect (NCERT Textbook Page No. 79)
Question 1.
Algebra tiles can be used to represent products and find factors. Figure out the product of x + 2 and x + 3 using algebra tiles.
Solution:
Here we have to find (x + 2)(x + 3)
Represent (x + 2)(x + 3) with algebra tiles
Step: Add the areas of all the tiles
x2 + x + x + x + x + x + 1 + 1 + 1 + 1 + 1 + 1 = x2 + 5x + 6
∴ (x + 2)(x + 3) = x2 + 5x + 6.
Question 2.
Lay out algebra tiles for x2 + 11x + 30 in such a way that you will see its factors.
Solution:
Layout algebra tiles for x2 + 11x + 30
Use this:
Arrange all tiles in a rectangle
Hence, x2 + 11x + 30 = (x + 5)(x + 6).
So the factors are (x + 5) and (x + 6).
Think and Reflect (NCERT Textbook Page No. 80)
Question 1.
We have seen that (x + 3)(x + 4) = x2 + 7x + 12. Also (x + 6)(x + 7) = x2 + 13x + 42. Generalise the pattern to get an expression for (x + a)(x + b).
Solution:
We have (x + 3)(x + 4) = x(x + 4) + 3(x + 4)
= x2 + 4x + 3x + 12
= x2 + (3 + 4)x + 12
= x2 + 7x + 12
and (x + 6)(x + 7) = x(x + 7) + 6(x + 7)
= x2 + 7x + 6x + 42
= x2 + (7 + 6)x + 42
= x2 + 13x + 42
Hence, the general pattern for (x + a)(x + b) is x2 + (a + b)x + ab.
Think and Reflect (NCERT Textbook Page No. 82)
Question 1.
James and Reshma were talking about algebraic identities they learnt in school.
James: (a – b)2 (a + b) = (a2 – 2ab + b2) (a + b)
Reshma: I have a different idea: (a – b)2 (a + b) = (a – b) [(a – b) (a + b)] = (a – b)(a2 – b2)
I will find this product to get the answer.
According to you, who is correct and why?
Try to combine more such identities and find new results.
Solution:
Both are correct.
(a – b)2 (a + b) = (a2 – 2ab + b2) (a + b)
= a(a2 – 2ab + b2) + b(a2 – 2ab + b2)
= a3 – 2a2b + ab2 + a2b – 2ab2 + b3
= a3 – a2b – ab2 + b3
and (a – b)2 (a + b) = (a – b) (a – b) (a + b)
= (a – b) (a2 – b2)
= a(a2 – b2) – b(a2 – b2)
= a3 – ab2 – a2b + b3
∴ Both methods lead to the same final polynomial (a3 – ab2 – a2b + b3).
∴ James and Reshma are simply applying different valid algebraic identities to simplify the same expression.
Think and Reflect (NCERT Textbook Page No. 85)
Question 1.
We already know that x2 – y2 = (x – y) (x + y).
Further, we have verified that x3 – y3 = (x – y) (x2 + xy + y2).
Observe that x – y is a common factor of x2 – y2 and x3 – y3.
Do you think x – y is also a factor of x4 – y4?
Note that x4 – y4 = (x2)2 – (y2)2 = (x2 – y2) (x2 + y2).
Can you see how x – y is a factor of x4 – y4?
How about x5 – y5? Does this also have x – y as a factor?
Solution:
Here x4 – y4 = (x2)2 – (y2)2 = (x2 – y2) (x2 + y2)
We know that x2 – y2 = (x – y) (x + y)
So, x4 – y4 = (x – y) (x + y) (x2 + y2)
∴ Yes, (x – y) is also a factor of x4 – y4.
Now x3 – y3 = (x – y) (x2 + xy + y2)
x4 – y4 = (x – y) (x + y) (x2 + y2)
= (x – y) x(x2 + y2) + y(x2 + y2)
– (x – y) (x3 + xy2 + x2y + y3)
Now, x5 – y5 = (x – y) (x4 + x3y + xy3 + x2y2 + y4)
∴ Yes, x – y is also a factor of x5 – y5.
Think and Reflect (NCERT Textbook Page No. 87)
Question 1.
Try to simplify the following rational expression:
\(\frac{36 s^2-12 s t+t^2}{t^2+2 t s-48 s^2}=\frac{(6 s-t)^2}{(-+-)(-+-)}\)
(Hint: Factor t2 + 2ts – 48s2 and simplify the rational expressions assuming that t2 + 2ts – 48s2 ≠ 0).
Solution:
Here \(\frac{36 s^2-12 s t+t^2}{t^2+2 t s-48 s^2}=\frac{(6 s)^2-2 \times 6 s \times t+t^2}{t^2+8 t s-6 t s-48 s^2}=\frac{(6 s-t)^2}{t(t+8 s)-6 s(t+8 s)}\) = \(\frac{(6 s-t)^2}{(t+8 s)(t-6 s)}\)
Thus \(\frac{36 s^2-12 s t+t^2}{t^2+2 t s-48 s^2}=\frac{(6 s-t)^2}{(t+8 s)(t-6 s)}\)
Ex 4.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.1 Solutions
Exercise 4.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.1 Solutions
Question 1.
Using the identity (a + b)2 = a2 + 2ab + b2, expand the following:
(i) (7x + 4y)2
(ii) \(\left(\frac{7}{5} x+\frac{3}{2} y\right)^2\)
(iii) (2.5p + 1.5q)2
(iv) \(\left(\frac{3}{4} s+8 t\right)^2\)
(v) \(\left(x+\frac{1}{2 y}\right)^2\)
(vi) \(\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
Solution:
(i) We have (7x + 4y)2
Now using (a + b)2 = a2 + 2ab + b2
Here, a = 7x and b = 4y
Then, (7x + 4y)2 = (7x)2 + 2 × 7x × 4y + (4y)2
= 72 × x2 + 56xy + 42 × y2 [∵ (ab)2 = a2 × b2]
= 49x2 + 56xy + 16y2
(ii) We have \(\left(\frac{7}{5} x+\frac{3}{2} y\right)^2\)
Now using (a + b)2 = a2 + 2ab + b2
Here, a = \(\frac {7}{5}\)x and b = \(\frac {3}{2}\)y
(iii) We have (2.5p + 1.5q)2
Now using (a + b)2 = a2 + 2ab + b2
Here, a = 2.5p and b = 1.5q
Then, (2.5p + 1.5q)2 = (2.5p)2 + 2 × 2.5p × 1.5q + (1.5q)2
= (2.5)2 × p2 + 2 × 2.5 × 1.5 × pq + (1.5)2 × q2 [∵ (ab)2 = a2 × b2]
= 6.25p2 + 7.5pq + 2.25q2
(iv) We have \(\left(\frac{3}{4} s+8 t\right)^2\)
Now using (a + b)2 = a2 + 2ab + b2
Here, a = \(\frac {3}{4}\)s and b = 8t
(v) We have \(\left(x+\frac{1}{2 y}\right)^2\)
Now using (a + b)2 = a2 + 2ab + b2
Here, a = x and b = \(\frac {1}{2y}\)
Then, \(\left(x+\frac{1}{2 y}\right)^2\)
= \(x^2+2 \times x \times \frac{1}{2 y}+\left(\frac{1}{2 y}\right)^2\)
= \(x^2+\frac{x}{y}+\frac{1}{4 y^2}\)
(vi) We have \(\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
Now using (a + b)2 = a2 + 2ab + b2
Here, a = \(\frac {1}{x}\) and b = \(\frac {1}{y}\)
Then, \(\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
= \(\left(\frac{1}{x}\right)^2+2 \times \frac{1}{x} \times \frac{1}{y}+\left(\frac{1}{y}\right)^2\)
= \(\frac{1}{x^2}+\frac{2}{x y}+\frac{1}{y^2}\)
Question 2.
Using the same identity, find the values of the following:
(i) (64)2
(ii) (105)2
(iii) (205)2
Solution:
(i) Here, (64)2
We can write (64)2 as (60 + 4)2
Now using (a + b)2 = a2 + 2ab + b2
Here a = 60 and b = 4
∴ (60 + 4)2 = 602 + 2 × 60 × 4 + 42
= 3600 + 480 + 16
= 4096
(ii) Here (105)2
We can write (105)2 as (100 + 5)2
Now using (a + b)2 = a2 + 2ab + b2
Here a = 100 and b = 5
∴ (100 + 5)2 = 1002 + 2 × 100 × 5 + 52
= 10000 + 1000 + 25
= 11025
(iii) Here (205)2
We can write (205)2 as (200 + 5)2
Now using (a + b)2 = a2 + 2ab + b2
Here a = 200 and b = 5
∴ (200 + 5)2 = 2002 + 2 × 200 × 5 + 52
= 40,000 + 2000 + 25
= 42025
Ex 4.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.2 Solutions
Exercise 4.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.2 Solutions
Question 1.
Factor completely:
(i) 9x2 + 24xy + 16y2
(ii) 4s2 + 20st + 25t2
(iii) 49x2 + 28xy + 4y2
(iv) \(64 p^2+\frac{32}{3} p q+\frac{4}{9} q^2\)
(v) \(3 a^2+4 a b+\frac{4}{3} b^2\)
(vi) \(\frac{9}{5} s^2+6 s v+5 v^2\)
[Hint: 2 was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?]
Solution:
(i) Given expression is 9x2 + 24xy + 16y2
Writing this in the form a2 + 2ab + b2
9x2 + 24xy + 16y2 = (3x)2 + 2 × 3x × 4y + (4y)2
where, a = 3x and b = 4y
Therefore, 9x2 + 24xy + 16y2 = (3x + 4y)2
(ii) Given expression is 4s2 + 20st + 25t2
Writing this in the form a2 + 2ab + b2
4s2 + 20st + 25t2 = (2s)2 + 2 × 2s × 5t + (5t)2
where, a = 2s and b = 5t
Therefore, 4s2 + 20st + 25t2 = (2s + 5t)2
(iii) Given expression is 49x2 + 28xy + 4y2
Writing this in the form a2 + 2ab + b2
49x2 + 28xy + 4y2 = (7x)2 + 2 × 7x × 2y + (2y)2
where, a = 7x and b = 2y
Therefore, 49x2 + 28xy + 4y2 = (7x + 2y)2
(iv) Given expression is \(64 p^2+\frac{32}{3} p q+\frac{4}{9} q^2\)
Writing this in the form a2 + 2ab + b2
\(64 p^2+\frac{32}{3} p q+\frac{4}{9} q^2\) = \((8 p)^2+2 \times 8 p \times \frac{2}{3} q+\left(\frac{2}{3} q\right)^2\)
where, a = 8p and b = \(\frac {2}{3}\)q
Therefore, \(64 p^2+\frac{32}{3} p q+\frac{4}{9} q^2\) = \(\left(8 p+\frac{2}{3} q\right)^2\)
(v) Given expression is \(3 a^2+4 a b+\frac{4}{3} b^2\)
Writing this in the form a2 + 2ab + b2
\(3 a^2+4 a b+\frac{4}{3} b^2\)
Question 2.
Find the values of the following using the identity (a – b)2 = a2 – 2ab + b2.
(i) (79)2
(ii) (193)2
(iii) (299)2
Solution:
(i) (79)2
We can write (79)2 as (80 – 1)2
∴ (80 – 1)2 = 802 – 2 × 80 × 1 + 12 [∵ (a – b)2 = a2 – 2ab + b2]
= 6400 – 160 + 1
= 6401 – 160
= 6241
(ii) (193)2
We can write (193)2 as (200 – 7)2
∴ (200 – 7)2 = 2002 – 2 × 200 × 7 + 72 [∵ (a – b)2 = a2 – 2ab + b2]
= 40000 – 2800 + 49
= 40049 – 2800
= 37249
(iii) (299)2
We can write (299)2 as (300 – 1)2
∴ (300 – 1)2 = 3002 – 2 × 300 × 1 + 12 [∵ (a – b)2 = a2 – 2ab + b2]
= 90000 – 600 + 1
= 90001 – 600
= 89401
Ex 4.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.3 Solutions
Exercise 4.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.3 Solutions
Question 1.
Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) 1172
(ii) 782
(iii) 1982
(iv) 2142
(v) 11042
(vi) 11202
Solution:
(i) (117)2
We can write it as (110 + 7)2 = 1102 + 2 × 110 × 7 + 72 [Using (a + b)2 = a2 + 2ab + b2]
= 12100 + 1540 + 49
= 13689
(ii) (78)2
We can write it as (80 – 2)2 = 802 – 2 × 80 × 2 + 22 [Using (a – b)2 = a2 – 2ab + b2]
= 6400 – 320 + 4
= 6404 – 320
= 6084
(iii) (198)2
We can write it as (200 – 2)2 = 2002 – 2 × 200 × 2 + 22 [Using (a – b)2 = a2 – 2ab + b2]
= 40,000 – 800 + 4
= 40004 – 800
= 39204
(iv) (214)2
We can write it as (210 + 4)2 = 2102 + 2 × 210 × 4 + 42 [Using (a + b)2 = a2 + 2ab + b2]
= 44100 + 1680 + 16
= 45796
(v) (1104)2
We can write it as (1100 + 2 + 2)2 = 11002 + 22 + 22 + 2 × 1100 × 2 + 2 × 2 × 2 + 2 × 2 × 1100
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 1210000 + 4 + 4 + 4400 + 8 + 4400
= 1218816
(vi) (1120)2
We can write it as (1100 + 10 + 10)2 = 11002 + 102 + 102 + 2 × 1100 × 10 + 2 × 10 × 10 + 2 × 10 × 1100
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 1210000 + 100 + 100 + 22000 + 200 + 22000
= 1254400
Question 2.
Factor using suitable identities:
(i) 16y2 – 24y + 9
(ii) \(\frac{9}{4} s^2+6 s t+4 t^2\)
(iii) \(\frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}+3 n k+2 m n+9 n^2\)
(iv) \(\frac{p^2}{16}-2+\frac{16}{p^2}\)
(v) 9a2 + 4b2 + c2 – 12ab + 6ac – 4bc
Solution:
(i) Here, 16y2 – 24y + 9 = (4y)2 – 2 × 4y × 3 + 32
[Using (a – b)2 = a2 – 2ab + b2]
Here, a = 4y and b = 3
= (4y – 3)2
= (4y – 3)(4y – 3)
(v) Here, 9a2 + 4b2 + c2 – 12ab + 6ac – 4bc
= (-3a)2 + (2b)2 + (-c)2 – 12ab – 4bc + 6ca
= (-3a)2 + (2b)2 + (-c)2 + 2 × (-3a) × 2b + 2 × 2b × (-c) + 2 × (-c) × (-3a)
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
Here, a = (-3a), b = 2b and c = (-c)
= [(-3a) + 2b + (-c)]2
= (-3a + 2b – c)2
= (-3a + 2b – c) (-3a + 2b – c)
Question 3.
Expand the following using the identity
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(i) (p + 3q + 7r)2
(ii) (3x – 2y + 4z)2
Solution:
(i) (p + 3q + 7r)2
Here, a = p, b = 3q and c = 7r
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
∴ (p + 3q + 7r)2 = p2 + (3q)2 + (7r)2 + 2 × p × 3q + 2 × 3q × 7r + 2 × 7r × p = p2 + 9q2 + 49r2 + 6pq + 42qr + 14rp
(ii) (3x – 2y + 4z)2
Here, a = 3x, b = -2y and c = 4z
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
∴ (3x – 2y + 4z)2 = (3x)2 + (-2y)2 + (4z)2 + 2 × 3x × (-2y) + 2 × (-2y) × 4z + 2 × 4z × 3x = 9x2 + 4y2 + 16z2 – 12yx – 16yz + 24zx
Question 4.
Is this an identity?
(a + b – c)2 + (a – b + c)2 + (a – b – c)2 = 2a2 + 2b2 + 2c2
Solution:
Here (a + b – c)2 = a2 + b2 + c2 + 2ab – 2bc – 2ac …..(i)
and (a – b + c)2 = a2 + b2 + c2 – 2ab – 2bc + 2ac ……(ii)
and (a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ac ……(iii)
Adding (i), (ii), and (iii)
LHS = a2 + b2 + c2 + 2ab – 2bc – 2ac + a2 + b2 + c2 – 2ab – 2bc + 2ac + a2 + b2 + c2 – 2ab + 2bc – 2ac
= 3a2 + 3b2 + 3c2 – 2ab – 2bc – 2ac
This is not equal to 2a2 + 2b2 + 2c2
Thus, the given statement is not true for all values of a, b, and c.
Hence, it is not an identity.
Ex 4.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.4 Solutions
Exercise 4.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.4 Solutions
Question 1.
Fill in the blanks to complete the following identities:
(i) s2 – 11s + 24 = (_________) (_________)
(ii) (_________) (x + 1) = (3x2 – 4x – 7)
(iii) 10x2 – 11x – 6 = (2x – _________) (_________ + 2)
(iv) 6x2 + 7x + 2 = (_________) (_________)
Solution:
(i) Comparing s2 – 11s + 24 with s2 + (a + b)s + ab, we get
a + b = -11 and ab = 24
Above equations are satisfied together When a = -8 and b = -3
∴ s2 – 11s + 24 = s2 + [(-8) + (-3)]s + (-8) × (-3)
= s2 – 8s – 3s + 24
= s(s – 8) – 3(s – 8)
= (s – 8)(s – 3)
(ii) RHS = 3x2 – 4x – 7
Here, the coefficient of x2 = 3
So, multiplying the constant term by +3
3 × (-7) = -21
Now we need two numbers whose sum = -4 and product = -21
These numbers are -7 and 3.
Now 3x2 – 4x – 7 = 3×2 – (7 – 3)x – 7
= 3×2 – 7x + 3x – 7
= x(3x – 7) + 1(3x – 7)
= (3x – 7)(x + 1)
(iii) Here LHS = 10x2 – 11x – 6
Coefficient of x2 = 10
So, multiplying the constant term by 10
10 × (-6) = -60
Now we need two numbers whose sum = -11 and product = -60
These numbers are -15 and 4.
10x2 – 11x – 6 = 10x2 – 15x + 4x – 6
= 5x(2x – 3) + 2(2x – 3)
= (2x – 3)(5x + 2)
(iv) Here 6x2 + 7x + 2
∴ Coefficient of x2 = 6
So, multiplying the constant term by 6
6 × 2 = 12
Now we need two numbers whose sum = 7 and product = 12
These numbers are 3 and 4
Now 6x2 + 7x + 2 = 6x2 + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (2x + 1)(3x + 2)
Question 2.
Select and use the identity that will help you to find the following products without multiplying directly:
(i) (41)2
(ii) (27)2
(iii) (23 × 17)
(iv) (135)2
(v) (97)2
(vi) (18 × 29)
(vii) (34 × 43)
(viii) (205)2
Solution:
(i) We have (41)2 = (40 + 1)2
[∵ (a + b)2 = a2 + 2ab + b2]
Here, a = 40 and b = 1
= 402 + 2 × 40 × 1 + 12
= 1600 + 80 + 1
= 1681
(ii) We have (27)2 = (25 + 2)2
[∵ (a + b)2 = a2 + 2ab + b2]
Here, a = 25 and b = 2
= 252 + 2 × 25 × 2 + 22
= 625 + 100 + 4
= 729
(iii) We have (23 × 17) = (20 + 3) (20 – 3)
Here, a = 20 and b = 3
= 202 – 32 [∵ (a + b) (a – b) = a2 – b2]
= 400 – 9
= 391
(iv) We have (135)2 = (130 + 5)2
Here, a = 130 and 6 = 5
[∵ (a + b)2 = a2 + 2ab + b2]
= 1302+ 2 × 130 × 5 + 52
= 16900 + 1300 + 25
= 18225
(v) We have (97)2 = (100 – 3)2
Here, a = 100 and b = 3
[∵ (a – b)2 = a2 – 2ab + b2]
= 1002 – 2 × 100 × 3 + 32
= 10000 – 600 + 9
= 9409
(vi) We have (18 × 29) = (20 – 2) (20 + 9)
= {20 + (- 2)} (20 + 9)
[∵ (x + a) (x + b) = x2 + (a + b)x + ab]
= 202 + (-2 + 9) × 20 + (-2 × 9)
= 400 + 7 × 20 – 18
= 400 + 140 – 18
= 540 – 18
= 522
(vii) We have (34 × 43)
= (40 – 6) (40 + 3)
= {40 + (-6)} (40 + 3)
= 402 + (-6 + 3) × 40 + (-6 × 3)
[∵ (x + a) (x + b) = x2 + (a + b)x + ab]
= 1600 × (-3) × 40 – 18
= 1600 – 120 – 18
= 1600 – 138
= 1462
(viii) We have (205)2 = (200 + 5)2
[∵ (a + b)22 = a2 + 2ab + b2]
Here, a = 200 and b = 5
= 2002 + 2 × 200 × 5 + 52
= 40000 + 2000 + 25
= 42025
Question 3.
Factor the following:
(i) 9a2 + b2 + 4c2 – 6ab + 12ac – 4bc
(ii) 16s2 + 25t2 – 40st
(iii) r2 – r – 42
(iv) 49g2 + 14gh + h2
(v) 64u2 + 121v2 + 4w2 – 176uv – 32uw + 44vw
Solution:
(i) We have 9a2 + b2 + 4c2 – 6ab + 12ac – 4bc
= 9a2 + b2 + 4c2 – 6ab – 4bc + 12ac
= (-3a)2 + b2 + (-2c)2 + 2 × (-3a) × b + 2 × b × (-2c) + 2 × (-2c) × (-3a)
= (-3a + b – 2c)2
(ii) We have 16s2 + 25t2 – 40st
= 16s2 – 40st + 25t2
= (4s)2 – 2 × 4s × 5t + (5t)2
= (4s – 5t)2
(iii) We have r2 – r – 42
= r2 + (-7 + 6)r + (-7) (6)
= (r + (-7)) (r + 6) [∵ (x + a) (x + b) = x2 + (a + b)x + ab]
= (r – 7) (r + 6)
(iv) We have 49g2 + 14gh + h2
= (7g)2 + 2 × 7g × h + h2 [∵ (a + b)2 = a2 + b2 + 2ab]
= (7g + h)2
(v) We have 64u2 + 121v2 + 4w2 – 176uv – 32uw + 44vw
= 64u2 + 121v2 + 4w2 – 176uv + 44vw – 32uw
= (-8u)2 + (11v)2 + (2w)2 + 2 × (-8u) × (11v) + 2 × 11v × 2w + 2 × 2w × (-8w)
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]
= (-8u + 11v + 2w)2
Ex 4.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.5 Solutions
Exercise 4.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.5 Solutions
Question 1.
Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
(i) \(\frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}\)
(ii) \(\frac{n^3-3 n^2 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2}\)
(iii) \(\frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x}\)
(iv) \(\frac{4 y^2-20 y z+25 z^2}{\left(25 z^2-4 y^2\right)}\)
(v) \(\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}\)
(vi) \(\frac{p^4-16}{p^2-4 p+4}\)
Solution:
Ganita Manjari Class 9 Maths Chapter 4 End of Chapter Exercise Solutions
Exploring Algebraic Identities End of Chapter Exercise Solutions
Question 1.
Use suitable identities to find the following products:
(i) (-3x + 4)2
(ii) (2s + 7) (2s – 7)
(iii) \(\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)\)
(iv) (2n + 7) (2n – 7)
(v) (s – 2t) (s2 + 2st + 4t2)
(vi) \(\left(\frac{1}{2 r}-4 r\right)^2\)
(vii) (-3m + 4k – l)2
(viii) \(\left(x-\frac{1}{3} y\right)^3\)
(ix) \(\left(\frac{7}{2} k-\frac{2}{3} m\right)^3\)
Solution:
(i) Here (-3x + 4)2
= (-3x)2 + 2 × (-3x) × 4 + 42
[∵ (a + b)2 = a2 + b2 + 2ab]
= 9x2 – 24x + 16
(ii) (2s + 7) (2s – 7) = (2s)2 – 72
[∵ (a + b) (a – b) = a2 – b2]
= 4s2 – 49
(iii) \(\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)\)
= \(\left(p^2\right)^2-\left(\frac{1}{2}\right)^2\)
[∵ (a – b) (a + b) = a2 – b2]
= p2 – \(\frac {1}{4}\)
(iv) (2n + 7) (2n – 7)
= (2n)2 – 72
[∵ (a – b) (a + b) = a2 – b2]
= 4n2 – 49
(v) (s – 2t) (s2 + 2st + 4t2)
Here, a = s and b = 2t
= s3 – (2t)3
[∵ a3 – b3 = (a – b) (a2 + ab + b2)]
= s3 – 8t3
(vii) (-3m + 4k – l)2
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
Here, a = -3m, b = 4k and c = -l
= (-3m)2 + (4k)2 + (-l)2 + 2 × (-3m) × 4k + 2 × 4k × (-l) + 2 × (-l) (-3m)
= 9m2 + 16k2 + l2 – 24mk – 8kl + 6lm
Question 2.
Find the values using suitable identities:
(i) 17 × 21
(ii) 104 × 96
(iii) 24 × 16
(iv) 1473
(v) 1993
(vi) 1273
(vii) (-107)3
(vii) (-299)3
Solution:
(i) Here, 17 × 21 = (19 – 2) (19 + 2)
[∵ (a – b) (a + b) = a2 – b2]
Here, a = 19 and b = 2
= 192 – 22
= 361 – 4
= 357
(ii) Here, 104 × 96 = (100 + 4) (100 – 4)
[∵ a2 – b2 = (a + b) (a – b)]
= 1002 – 42
= 10000 – 16
= 9984
(iii) 24 × 16 = (20 + 4) (20 – 4)
[∵ (a + b) (a – b) = a2 – b2]
Here, a = 20 and b = 4
= 202 – 42
= 400 – 16
= 384
(iv) (147)3 = (140 + 7)3
[∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3]
Here, a = 140 and b = 7
= 1403 + 3 × 1402 × 7 + 3 × 140 × 72 + 73
= 2744000 + 3 × 19600 × 7 + 3 × 140 × 49 + 343
= 2744000 + 411600 + 20580 + 343
= 3176523
(v) 1993 = (200 – 1)3
[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
Here, a = 200 and b = 1
= 2003 – 3 × 2002 × 1 + 3 × 200 × 12 – 13
= 8000000 – 3 × 40000 × 1 + 3 × 200 × 1 – 1
= 8000000 – 120000 + 600 – 1
= 8000600 – 120001
= 7880599
(vi) 1273 = (130 – 3)3
[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
Here, a = 130 and b = 3
= 1303 – 3 × 1302 × 3 + 3 × 130 × 32 – 33
= 2197000 – 3 × 16900 × 3 + 3 × 130 × 9 – 27
= 2197000 – 152100 + 3510 – 27
= 2200510 – 152127
= 2048383
(vii) (-107)3 = -(100 + 7)3
[∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3]
Here, a = 100 and b = 7
= -[(100)3 + 3 × (100)2 × 7 + 3 × (100) × 72 + 73]
= -[1000000 + 3 × 10000 × 7 + 3 × (100) × 49 + 343]
= -[1000000 + 210000 + 14700 + 343]
= -12250443
(viii) (-299)3 = -(300 – 1)3
[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
Here, a = 300 and b = 1
= -[3003 – 3 × 3002 × 1 + 3 × 300 × 12 – 13]
= -[27000000 – 3 × 90000 × 1 + 3 × 300 × 1 – 1]
= -[27000000 – 270000 + 900 – 1]
= -(27000900 – 270001)
= -(26730899)
= -26730899
Question 3.
Factor the following algebraic expressions:
(i) \(4 y^2+1+\frac{1}{16 y^2}\)
(ii) \(9 m^2-\frac{1}{25 n^2}\)
(iii) \(27 b^3-\frac{1}{64 b^3}\)
(iv) \(x^2+\frac{5 x}{6}+\frac{1}{6}\)
(v) \(27 u^3-\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25}\)
(vi) \(64 y^3+\frac{1}{125} z^3\)
(vii) p3 + 27q3 + r3 – 9pqr
(viii) 9m2 – 12m + 4
(ix) \(9 x^3-\frac{8}{3} y^3+\frac{z^3}{3}+6 x y z\)
(x) 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy
(xi) \(27 u^3-\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4}\)
Solution:
Question 4.
Simplify the following:
(i) \(\frac{4 x^2+4 x+1}{4 x^2-1}\)
(ii) \(\frac{9\left(3 a^3-24 b^3\right)}{9 a^2-36 b^2}\)
(iii) \(\frac{s^3+125 t^3}{s^2-2 s t-35 t^2}\)
Note: Assume that the denominators are not equal to 0.
Solution:
Question 5.
Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) 25a2 – 30ab + 9b2
(ii) 36s2 – 49t2
Solution:
(i) Here, 25a2 – 30ab + 9b2
= (5a)2 – 2 × 5a × 3b + (3b)2
= (5a – 3b)2
∴ Possible length = 5a – 3b and breadth = 5a – 3b
(ii) 36s2 – 49t2
= (6s)2 – (7t)2
= (6s + 7t) (6s – 7t)
∴ Hence possible length = (6s + 7t) and breadth = (6s – 7t)
Question 6.
Find possible expressions for the length, breadth, and height of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) 6a2 – 24b2
(ii) 3ps2 – 15ps + 12p
Solution:
(i) 6a2 – 24b2
= 6(a2 – 4b2)
= 6{a2 – (2b)2}
= 6{(a + 2b) (a – 2b)}
∴ Possible dimensions are 6, (a + 2b), (a – 2b).
(ii) 3ps2 – 15ps + 12p
= 3p(s2 – 5s + 4)
= 3p(s2 – s – 4s + 4)
= 3p[s(s – 1) – 4(s – 1)]
= 3p(s – 1) (s – 4)
∴ Possible dimensions are 3p, (s – 1) and (s – 4).
Question 7.
The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Solution:
Area of the playground = 40 × 40 = 1600 m2
New length = 40 + 2s (both sides)
New area = (40 + 2s)2
= 402 + 2 × 40 × 2s + (2s)2
= 1600 + 160s + 4s2
∴ Area of the path = 1600 + 160s + 4s2 – 1600
= 160s + 4s2
= 4s2 + 160s, which is the required expression.
Question 8.
If a number plus its reciprocal equals \(\frac {10}{3}\), find the number.
Solution:
Let the number be x.
Then its reciprocal is \(\frac {1}{x}\)
According to the question,
the sum of reciprocal = \(\frac {10}{3}\)
⇒ \(x+\frac{1}{x}=\frac{10}{3}\)
⇒ \(\frac{x^2+1}{x}=\frac{10}{3}\)
⇒ 3(x2 + 1) = 10x
⇒ 3x2 + 3 – 10x = 0
⇒ 3x2 – 10x + 3 = 0
⇒ 3x2 – 9x – x + 3 = 0
⇒ 3x(x – 3) – 1(x – 3) = 0
⇒ (x – 3)(3x – 1) = 0
⇒ x – 3 = 0
⇒ x = 3
OR
3x – 1 = 0
⇒ 3x = 1
⇒ x = \(\frac {1}{3}\)
∴ x = 3 or \(\frac {1}{3}\)
∴ The number is 3 or \(\frac {1}{3}\).
Question 9.
A rectangular pool has area 2x2 + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
Solution:
The area of rectangle pool
Area = Length × Width = 2x2 + 7x + 3
Width = 2x + 1
∴ The length of the pool is (x + 3) hastas.
Question 10.
If both x – 2 and x – \(\frac {1}{2}\) are factors of px2 + 5x + r, show that p = r.
Solution:
Given (x – 2) and (x – \(\frac {1}{2}\)) are factors, then
px2 + 5x + r = k(x – 2) (x – \(\frac {1}{2}\))
= \(k\left(x^2-\frac{1}{2} x-2 x+1\right)\)
= \(k\left(x^2-\frac{5}{2} x+1\right)\)
= \(k x^2-\frac{5}{2} k x+k\)
= px2 + 5x + r (given)
Comparing the coefficients on both sides, we get
Coeff. of x2 ⇒ k = p ……(i)
Coeff. of constant ⇒ k = r ……(ii)
From (i) and (ii), we get k = p = r.
Hence proved.
Question 11.
If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 – 3abc = -25.
Solution:
We know, a3 + b3 + c3 – 3abc = (a + b + c) [(a2 + b2 + c2 + ab + bc + ca)]
and (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Putting values of a + b + c = 5 and ab + bc + ca = 10
⇒ 52 = a2 + b2 + c2 + 2 × 10
⇒ 25 = a2 + b2 + c2 + 20
⇒ a2 + b2 + c2 = 25 – 20
⇒ a2 + b2 + c2 = 5 ……(i)
Hence, a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – (ab + bc + ca))
= 5(5 – 10) [Using eq. (i)]
= 5 × (-5)
= -25
Hence proved.
Question 12.
By factoring the expression, check that n3 – n is always divisible by 6 for all natural numbers n. Give reasons.
Solution:
We have, (n3 – n) = n(n2 – 1)
⇒ (n3 – n) = n(n2 – 12)
⇒ (n3 – n) = n[(n + 1) (n – 1)]
[∵ a2 – b2 = (a – b) (a + b)]
The Reasons for Divisibility
The factored expression represents the product of three consecutive integers.
For any three consecutive integers:
Divisibility by 2: In any two consecutive integers, at least one must be even (a multiple of 2).
Therefore, the product is always divisible by 2.
Divisibility by 3: In any three consecutive integers, exactly one must be a multiple of 3.
Therefore, the product is always divisible by 3.
Conclusion: Since the expression is always divisible by both 2 and 3, and since 2 and 3 are coprime (they have no common factors), the expression must be divisible by their product: 2 × 3 = 6.
Thus, (n3 – n) is always divisible by 6 for all natural numbers (n).
Question 13.
Find the value of:
(i) x3 + y3 – 12xy + 64, when x + y = -4
(ii) x3 – 8y3 – 36xy – 216, when x = 2y + 6
Solution:
(i) Here, x3 + y3 – 12xy + 64 and x + y = -4
⇒ x + y + 4 = 0 …..(i)
Now, x3 + y3 + 64 – 12xy
= x3 + y3 + 43 – 3 × x × y × 4
= (x + y + 4) (x2 + y2 + 42 – x × y – y × 4 – 4 × x)
= (0) (x2 + y2 + 16 – xy – 4y – 4x) [Using eq. (i)]
= 0
∴ The value of the expression x + y = -4 is 0.
(ii) Here we have to find the value of x3 – 8y3 – 36xy – 216 at x = 2y + 6
Now, x – 2y – 6 = 20 ……(i)
= x3 – 8y3 – 36xy – 216
= x3 + (-2y)3 + (-6)3 – 3 × x × 2y × 6
= (x – 2y – 6) (x2 + (-2y)2 + (-6)2 – x × (-2y) – (-2y) × (-6) – (-6 × x))
= (0) (x2 + 4y2 + 36 + 2xy – 12y – 6x) [Using eq. (i)]
= 0