Students often refer to Ganita Manjari Class 9 Solutions and Part 1 Class 9 Maths Chapter 4 Exploring Algebraic Identities Solutions to verify their answers.
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:
Do it yourself.
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:
We know that (a + b)2 = a2 + 2ab + b2
Also, (a + b)2 < a2 + b2
a2 + 2ab + b2 < a2 + b2
2ab < 0
Hence, ab < 0
Therefore, a and b must have opposite signs.
Question 2.
What can you say about a and b if (a + b)2 > a2 + b2 ?
Solution:
We know that (a + b)2 = a2 + 2ab + b2
Also, (a + b)2 > a2 + b2
a2 + 2ab + b2 > a2 + b2
2ab > 0
Hence, ab> 0
Therefore, a and b must have the same sign.
Question 3.
When will (a + b)2 be equal to a2 + b2 ?
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:
We know that, (a + b)2 = a2 + 2ab + b2
Also, (a + b)2 = a2 + b2
∴ a2 + 2ab + b2 = a2 + b2
2ab = 0
Hence, ab = 0
Therefore, either a = 0 or b = 0 or both are 0.
Think and Reflect (NCERT Textbook Page No. 73)
What if we replace b by –b in (a + b)2 = a2 + 2ab + b2?
Solution:
Do it yourself.
Think and Reflect (NCERT Textbook Page No. 76)
Label the squares and rectangles in the given figure, so that it represents the identity
(a + b + c) = a2 + b2 + c2 + 2ab + 2bc + 2ca.
A geometrical model representing the identity
(a + b + c) – a2 + b2 + c2 + 2ab + 2bc + 2ca
Solution:
We know that (a + b + c)2 = (a + b + c) (a + b + c)
Now, consider a square of side (a + b + c) and divide it into parts of lengths a, b and c along both the length and the breadth.
Then, the top-left square of side a represents area a2.
Also, the middle square of side b represents area b2 and the bottom-right squared of side c represents area c2.
Now, the rectangles formed between a and b are of area ab each, and there are two such rectangles.
Hence, total area = 2ab.
Also, the rectangles formed between b and c.
Hence, the total area = 2bc
and the rectangles formed between c and a.
Hence, the total area = 2ca.
On adding all these areas, we get
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Hence, the given figure represents the identity.
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) We have, 352 =(30 + 5)2
Using the identity (a + b)2 = a2 + 2ab + b2, we get
= 302 + 2 × 30 × 5 + 52
= 900 + 300 + 25
= 1225
(ii) We have, 652 = (60 + 5)2
Using the identity (a + b)2 = a2 + 2 ab + b2, we get = 602 + 2 × 60 × 5 +52
= 3600 + 600 + 25
= 4225
(iii) We have, 852 =(80 + 5)2
Using the identity (a + b)2 = a2 + 2ab + b2 , we get
= 802 + 2 × 80 × 5 + 52
= 6400 + 800 + 25
= 7225
(iv) We have, 1052 =(100 + 5)2
Using the identity (a + b)2 = a2 + 2ab + b2 ,we get
= 1002 + 2 × 100 × 5 + 52
= 10000 + 1000 + 25
= 11025
Now, we observe that each number ends in 5, and also each square ends in 25.
For any number ending in 5, let the number be written as ti5.
Now, multiply the leading part n by its successor (n + 1), and write this product.
Then, place 25 at the end of the product.
Hence, the square of a number ending in 5 is obtained.
Question 2.
Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.
Solution:
Given, the figure represents an algebraic identity.
In the first row, we have four squares of sides
(a + b + c),(a + b – c),(a – b + c) and (a – b – c).
Also, in the second row, the same parts are rearranged to form three squares of sides 2a, 2b and 2c.
Now, area of figures in first row
= (a + b + c) + (a + b – c) + (a – b + c) + (a – b – c)
and area of figures in second row = (2a)2 + (2b)2 + (2c)2
Since, both rows are formed by the same parts, their areas are equal.
Hence,(a + b + c)2 + (a + b – c)2 + (a – b + c)2 + (a – b – c)2 = (2a)2 + (2b)2 + (2c)2
= 4a2 + 4b2 + 4c2
Therefore, the required identity is
(a + b + c) + (a + b – c) + (a – b + c)2 + (a – b – c)2 =4(a2 + b2 + c2).
Think and Reflect (NCERT Textbook Page No. 79)
Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.
Solution:
Do it Yourself.
Think and Reflect (NCERT Textbook Page No. 79)
Algebra tiles can be used to represent products and find factors.
Question 1.
Figure out the product of x + 2 and x + 3 using algebra tiles.
Solution:
Consider a rectangle, whose length is x + 2 and breadth is x + 3
Now, divide it into smaller rectangles using algebra tiles.
Then, the area formed by x × x gives x2 and also x × 3 gives 3x.
Now, 2 × x gives 2x and also 2 × 3 gives 6.
Hence, adding all parts,
x2 + 3x + 2x + 6 = x2 + 5x + 6
Hence, the product of x + 2 and x + 3 is x + 5x + 6.
Question 2.
Lay out algebra tiles for x2 + 11x + 30 in such a way that you will see its factors.
Sol Given expression is x2 + 11x + 30.
Now, arrange algebra tiles as one x2 tile, eleven x-tiles and thirty unit tiles.
Arrange in rectangular form,
x2 – 11x + 30 = x2 + 5x + 6x + 30
= x(x + 5) + 6(x + 5)
= (x + 5) (x + 6)
Hence, 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:
Given, (x + 3) (x + 4) = x2 + 7x + 12 and
(x + 6) (x + 7) = x2 + 13x + 42
Now, we observe that
In each case, the coefficient of x is the sum of the numbers and the constant term is their product.
Therefore, consider (x + a) (x + b)
(x + a) (x + b) = x (x + b) + a(x + b)
= x2 +bx + ax + ab = x2 + (a + b)x + ab
Hence, (x + a) (x + b) – x2 + (a + b)x + ab
Question 2.
Now consider the case where we have a rectangle of sidelengths 2x + 3 and 3x + 1 as shown in given figure. What can you say about its area (2x + 3) (3x + 1) ?
Using algebra tiles to represent
(2x + 3) (3x +1)
Solution:
Given dimensions are (2x + 3) and (3x + 1).
We know that the area of a rectangle is the product of its length and breadth.
Now, by applying the distributive law, we get
(2x + 3) (3x +1) = 2x(3x + 1) + 3(3x + 1)
= 6x + 2x + 9x + 3
= 6x2 + 11x + 3
Hence, the area of the given rectangle is 6x2 + 11x + 3.
Question 3.
Fill in the blanks with the appropriate expressions to make the equation true.
(px + a) (qx + b) = (………..)x2 + (………..)x + ….
Also, verify your answer using the distributive property.
Solution:
Given, (px + a) (qx + b)
Now, using the distributive property,
(px + a) (qx + b) = px(qx + b) + a(qx + b)
= pqx2 + pbx + aqx + ab = pqx2 + (pb + aq)x + ab
Hence, the required expression is
(px + a) (qx + b) = (pq)x2 + (pb + aq)x + ab
Thus, the blanks are (pq), (pb + aq) and 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:
James and Reshma both are correct.
James used the identity
(a – b)2 = a2 – 2ab + b2
Therefore,
(a – b)2(a + b) = (a2 – 2ab + b2) (a + b)
= a3 + a2b – 2a2b – 2ab2 + ab2 + b3 = a3 – a2b – ab2 + b3
Also, Reshma used the identity
(a – b)(a + b) = a2 – b2
Therefore,
(a – b)2 (a + b) = (a – b) [(a – b) (a + b)]
= (a – b)(a2 – b2)
= a3 – ab2 – a2b + b3 = a3 – a2b – ab2 + b3
Hence, both James and Reshma get the same result, so both are correct.
More such combinations (new results)
- (a + b)2(a – b) = (a2 + 2ab + b2)(a – b)
- (a + b)(a2 – b2) = (a + b)2(a – b)
- (a – b)3 = (a – b)(a2 – 2ab + b2)
- (a2 – b2) (a2 + b2) = a4 – b4
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:
Yes, x – y is also a factor of x4 – y4 .
We have, x4 – y4 = (x2 )2 – (y2 )2
Using the identity a2 – b2 =(a – b) (a + b), we get
x4 – y4 = (x2 – y2)(x2 + y2)
Now, x2 – y2 = (x – y)(x + y)
So, x4 – y4 = (x – y) (x + y) (x2 + y2)
Hence, x – y is also a factor of x4 – y4.
Also, for x5 – y5.
x5 – y5 = (x – y) (x4 + x3y + x2y2 + xy3 + y4)
Hence, x – y is also a factor of x5 – y5.
Therefore, in general, xn – yn is divisible by x – y, where n is a positive integer.
Think and Reflect (NCERT Textbook Page No. 87)
Question 1.
Try to simplify the following rational expression
\(\frac{36 s^2-12 s t+t^2}{t^2+2 t s-48 s^2}=\frac{(6 s-t)^2}{(\ldots+\ldots)(\ldots+\ldots)} .\)
(Hint Factor t2 + 2ts – 48s2 and simplify the rational expressions assuming that t2 + 2ts – 48s2 ≠ 0).
Solution:
Given, \(\frac{36 s^2-12 s t+t^2}{t^2+2 t s-48 s^2}\)
Now, factorise numerator and denominator.
36s2 – 12st + 12 = (6s – 1)2
Also, t2 + 2ts – 48s2 = (t + 8s) (t – 6s)
So, the expression becomes
\(\frac{(6 s-t)^2}{(t+8 s)(t-6 s)}\)
Now, (6s – t) = -(t – 6s)
On squaring both sides, we get
(6s – t)2 = (t – 6s)2
Therefore, \(\frac{(t-6 s)^2}{(t+8 x)(t-6 s)}=\frac{t-6 s}{t+8 s}\)
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
Solution:
We have, (7x + 4y)2
On comparing (7x + 4y)2 with (a + b)2, we get a = 7x and b = 4y.
Using the identity (a + b)2 = a2 + 2ab + b2,
we get (7x + 4y)2 = (7x)2 + 2(7x) (4y) + (4y)2
= 49x2 + 56xy + 16y2
(ii) (\(\frac{7}{5}\)x + \(\frac{3}{2}\)y)2
Solution:
We have, (\(\frac{7}{5}\)x + \(\frac{3}{2}\)y)2
On comparing (\(\frac{7}{5}\)x + \(\frac{3}{2}\)y)2 with (a + b)2, we get
a = \(\frac{7}{5}\)x and b = \(\frac{3}{2}\)y.
Using the identity (a + b)2 = a2 + 2ab + b2, we get
(\(\frac{7}{5}\)x + \(\frac{3}{2}\)y)2 = \(\left(\frac{7}{5} x\right)^2+2\left(\frac{7}{5} x\right)\left(\frac{3}{2} y\right)+\left(\frac{3}{2} y\right)^2\)
= \(\frac{49}{25}\)x2 + \(\frac{21}{5}\)xy + \(\frac{9}{4}\)y2
(iii) (25p + 15q)2
Solution:
We have, (25p + 15q)2
On comparing (2.5 p + 15q)2 with (a + b)2, we get a = 2.5p and b = 15q.
Using the identity (a + b)2 = a2 + 2ab + b2, we get
(25p + 15q)2 =(25p)2 + 2(25p) (15q) + (15q)2
= 625p2 + 75pq + 225q2
(iv) (\(\frac{3}{4}\)s + 8t)2
Solution:
We have, (\(\frac{3}{4}\)s + 8t)2 with (a + b)2, we get a = \(\frac{3}{4}\)s and b = 8t
Using the identity (a + b)2 = a2 + 2ab + b2, we get
(\(\frac{3}{4}\)s + 8t)2 = (\(\frac{3}{4}\)s)2 + 2(\(\frac{3}{4}\)s)(8t) + (8t)2
= \(\frac{9}{16}\)s2 + 12st + 64t2
(v) (x + \(\frac{1}{2y}\))2
Solution:
We have, (x + \(\frac{1}{2y}\))2
On comparing (x + \(\frac{1}{2y}\))2 with (a + b)2, we get a = x and b = \(\frac{1}{2y}\)
Using the identity (a + b)2 = a2 + 2ab + b2, we get
(x + \(\frac{1}{2y}\))2 = x2 + 2(x)(\(\frac{1}{2y}\)) + (\(\frac{1}{2y}\))2
= x2 + \(\\frac{x}{y}+\frac{1}{4 y^2}\)
(v) \(\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
Solution:
We have, \(\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
On comparing \(\left(\frac{1}{x}+\frac{1}{y}\right)^2\) with (a + b)2, we get a = \(\frac{1}{x}\) and b = \(\frac{1}{y}\).
Using the identity (a + b)2 = a2 + 2ab + b2, we get
\(\left(\frac{1}{x}+\frac{1}{y}\right)^2\) = \(=\left(\frac{1}{x}\right)^2+2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)+\left(\frac{1}{y}\right)^2\)
= \(\frac{1}{x^2}+\frac{2}{x y}+\frac{1}{y^2}\)
Question 2.
Using the same identity, find the values of the following.
(i) (64)2
(ii) (105)2
(iii) (205)2
Solution:
(i) We have, (64)2 = (60 + 4)2
= (60)2 + (4)2 + 2(60) (4)
[∵ (a + b)2 = a2 + b2 + 2 ab]
= 3600 + 16 + 480 = 4096
(ii) We have, (105)2 = (100 + 5)2
= (100)2 + (5)2 + 2(100)(5)
[∵ (a + b)2 = a2 + b2 + 2ab]
= 10000 + 25 + 1000
= 11025
(iii) We have, (205)2 = (200 + 5)2
= (200)2 + (5)2 + 2(200) (5)
[∵ (a + b)2 = a2 + b2 + 2ab]
= 40000 + 25 + 2000
= 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
Solution:
We have, 9x2 + 24xy + 16y2
= (3x)2 + 2(3x) (4y) + (4y)2
Now, take 3x = a and 4y = b also using the identity
a2 + 2ab + b2 = (a + b)2
So, the expression becomes (3x + 4y)2.
Hence, the required factors are (3x + 4y) and (3x + 4y).
(ii) 4s2 + 20st + 25t2
Solution:
We have, 4s2 + 20st + 25t2 = (2s)2 + 2(2s)(5t) + (5t)2
Now, take 2s = a and 5t = b, also using the identity
a2 + 2ab + b2 = (a + b)2
So, the expression becomes (2s + 5t)2.
Hence, the required factors are (2s + 5t) and (2s + 5t).
(iii) 49x2 + 28xy + 4y2
Solution:
We have, 49x2 + 28xy + 4y2
= (7x)2 + 2(7x)(2y) + (2y)2
Now, take 7x = a and 2y = b, also using the identity
a2 + 2ab + b2 = (a + b)2
So, the expression becomes (7x + 2y)2.
Hence, the required factors are (7x + 2y) and (7x + 2y).
(iv) 64p2 + \(\frac{32}{2}\)pq + \(\frac{4}{9}\)q
Solution:
We have, 64p2 – \(\frac{32}{2}\)pq + \(\frac{4}{9}\)q2
= (8p)2 + 2(8 p)(\(\frac{2}{3}\)q) + (\(\frac{2}{3}\)q)2
Now, take 8p = a and \(\frac{2}{3}\)q = b, also using the identity
a2 + 2ab + b2 = (a + b)2
So, the expression becomes (8p + \(\frac{2}{3}\)q)2
Hence, the required factors are (8p + \(\frac{2}{3}\)q) and (8p + \(\frac{2}{3}\)q)
(v) 3a2 + 4ab + \(\frac{4}{3}\)b2
Solution:
3a2 + 4ab + \(\frac{4}{3}\)b2
= – [(3a)2 + 2(3a) (2b) + (2b)2]
Now, take 3d =x and 2b=y, also using the identity
x2 + 2xy + y2 = (x + y)2.
So, the expression becomes
\(\frac{1}{3}\)[9d2 + 12db + 4b2] = \(\frac{1}{3}\)(3d + 2b)2.
Hence, the required factors are \(\frac{1}{3}\), (3a + 2b) and (3a + 2b).
(vi) \(\frac{9}{5}\)s2 +6sv + 5v2
Solution:
We have,
\(\frac{9}{5}\)s2 + 6sv + 5v2 = \(\frac{1}{5}\) [9s2 + 30sv + 25v2]
= \(\frac{1}{5}\) [(3s)2 + 2 (3s) (5v) + (5v)2]
Now, take 3s = a and 5v = b, also using the identity
a2 + 2ab + b2 = (a + b)2.
So, the expression becomes
\(\frac{1}{5}\) [9s2 + 30sv + 25v2] = – (3s + 5v)2
Hence, the required factors are \(\frac{1}{5}\), (3s + 5v) and (3s + 5v).
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) We have, (79)2 = (80 – 1)2
= (80)2 + (1)2 – 2(80)(1)
[∵ (a – b)2 = a2 + b2 – 2ab]
= 6400 + 1 – 160
= 6401 – 160
= 6241
(ii) We have, (193)2 =(200 – 7)2
= (200)2 + (7)2 – 2(200) (7)
[∵ (a -b)2 = a2 + b2 – 2ab]
= 40000 + 49 – 2800
= 40049 – 2800
= 37249
(iii) We have, (299)2 =(300 – 1)2
= (300)2 + (1)2 – 2(300) (1)
[∵ (a – b)2 = a2 + b2 – 2ab]
= 90000 + 1 – 600
= 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 identities.
(i) (a + b)2 = a2 + 2ab + b2
(ii) (a – b)2 = a2 – 2ab + b2
(iii) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Determine which of these identities will make these calculations easier.
(i) 1172
Solution:
We have,
1172 =(120 – 3)2
Using the identity, (a – b)2 = a2 – 2ab + b2
= 1202 – 2(120)(3) + 32
= 14400 – 720 + 9
= 13689
(ii) 782
Solution:
We have, 782 = (80 – 2)2
Using the identity, (a – b)2 = a2 – 2ab + b2
= 802 – 2(80)(2) + 22
= 6400 – 320 + 4
= 6084
(iii) 1982
Solution:
We have,
1982 =(200 -2)2
Using the identity, (a – b)2 = a2 – 2ab + b2
= 2002 – 2(200)(2) + 22
= 40000-800 + 4
= 39204
(iv) 2142
Solution:
We have,
2142 = (200 + 10 + 4)2
Using the identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
= 2002 + 102 + 42 + 2(200)(10) + 2(10) (4) + 2(4) (200)
= 40000 + 100 + 16 + 4000 + 80 + 1600
= 45796
(v) 11042
Solution:
We have,
11042 =(1000 + 100 + 4)2
Using the identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
= 10002 + 1002 + 42 + 2 (1000) (100) + 2(100) (4) + 2(4) (1000)
= 1000000 + 10000 + 16 + 200000 + 800 + 8000
= 1218816
(vi) 11202
Solution:
We have,
11202 = (1000 + 100 + 20)2
Using the identity,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
= 10002 + 1002 + 202 + 2(1000) (100) + 2(100)(20) + 2(20)(1000)
= 1000000 + 10000 + 400 + 200000 + 4000 + 40000
= 1254400
Question 2.
Factor using suitable identities.
(i) 16y2 – 24y + 9
Solution:
We have, 16y2 – 24y + 9
= (4y)2 – 2(4y) (3) + 32
Now, take 4y = a and 3 = b also using the identity a2 – 2ab + b2 = (a – b)2.
So, the expression becomes (4y – 3)2.
Hence, the required factors are (4y – 3) and (4y – 3).
(ii) \(\frac{9}{4}\)s2 + 6sf + 4t2
Solution:
We have,
\(\frac{9}{4}\)s2 + 6sf + 4t2
= (\(\frac{3}{2}\)s)2 + 2(\(\frac{3}{2}\)s)(2t) + (2t)2
Now, take \(\frac{3}{2}\) = a, 2t = b, also using the identity
a2 + 2ab + b2 = (a + b)2
So, the expression becomes (\(\frac{3}{2}\)s + 2t)2
Hence, the required factors are (\(\frac{3}{2}\)s + 2t) and (\(\frac{3}{2}\)s + 2t).
(iii) \(\frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}\) + 3nk + 2mn + 9n2
Solution:
We have,
\(\frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}\) + 3nk + 2mn + 9n2
= \(\left(\frac{m}{3}\right)^2+\left(\frac{k}{2}\right)^2\) + (3n) + 2\(\left(\frac{m}{3}\right)\left(\frac{k}{2}\right)\) + 2\(\left(\frac{k}{2}\right)\)(3n) + 2(3n)\(\left(\frac{m}{3}\right)\)
Now, take \(\frac{m}{3}\) = a, \(\frac{k}{2}\) = b and 3n = c, also using the identity
a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2
So, the expression becomes \(\left(\frac{m}{3}+\frac{k}{2}+3 n\right)^2\)
Hence, the required factors are \( and [latex]\left(\frac{m}{3}+\frac{k}{2}+3 n\right)\)
(iv) \(\frac{p^2}{16}\) – 2 + \(\frac{16}{p^2}\)
Solution:
We have,
\(\frac{p^2}{16}\) – 2 + \(\frac{16}{p^2}\)
= \(\left(\frac{p}{4}\right)^2-2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right)+\left(\frac{4}{p}\right)^2\)
Now, take \(\frac{p}{4}\) = a and \(\frac{4}{p}\) = b, also using the identity a2 – 2ab + b2 = (a – b)2
So, the expression becomes \(\left(\frac{p}{4}-\frac{4}{p}\right)^2\)
Hence, the required factors are \(\left(\frac{p}{4}-\frac{4}{p}\right)\) and \(\left(\frac{p}{4}-\frac{4}{p}\right)\)
(v) 9a2 + 4b2 + c2 – 12ab + 6ac – 4bc
Solution:
We have,
9a2 + 4b2 + c2 – 12ab + 6ac – 4bc
= (3a)2 + (-2b)2 + (c)2 + 2(3a) (-2b) + 2(-2b) x (c) + 2 (c) (3a)
Now, take 3a = x , (-2b) = y and c = z, also using the identities
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx.
So, the expression becomes (3a – 2b + c)2.
Hence, the required factors are (3a – 2b + c) and (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
Solution:
We have, (p + 3q + 7r)2
On comparing with (a + b + c)2 , we get a = p,b =3q andc = 7r We know that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2 ca
On putting the values of a, b and c in the above identity, we get
(p + 3q + 7r)2 = p2 +(3q)2 + (7r)2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)
= p2 + 9q2 + 49 r2 + 6pq + 42qr + 14rp
(ii) (3x – 2y + 4z)2
Solution:
We have, (3x – 2y + 4z)2
On comparing with (a + b + c)2 , we get a = 3x, b = -2y and c = 4z
We know that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2 ca
On putting the values of a, b and c in the above identity, we get
(3x – 2y + 4z)2 = (3x)2 + (-2y)2 + (4z)2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x)
= 9x2 + 4y2 + 16z2 – 12xy – 16yz + 24zx
Question 4.
Is this an identity?
(a + b – c)2 + (a – b + c)2 + (a – b – c)2 = 2a2 + 2b2 + 2c2
Solution:
LHS = (a + b – c)2 +(a – b + c)2 + (a – b – c)2
Now, using identity
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx,
we get = (a2 + b2 + c2 + 2ab – 2bc – 2ca) + (a2 + b2 + c2 – 2ab – 2bc + 2 ca) + (a + b +c – 2ab + 2bc – 2 ca)
= 3a2 + 3b2 + 3c2 – 2ab – 2bc – 2ca
But, RHS = 2a2 + 2b2 + 2c2
Clearly, LHS ≠ RHS
Hence, the given statement 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) We have,
s2 – 11s + 24
Now, splitting the middle term, we get
= s2 -3 s—8s + 24 = s(s – 3) – 8(s – 3)
= (s – 3)(s – 8)
Hence, s2 – 11s + 24 = (s – 3)(s – 8)
(ii) Let the missing term be (ax + b).
Now,(ax + b)(x + li) = 3x2 – 4x – 7
⇒ ax2 + ax + bx + b = 3x2 – 4x – 7
⇒ ax2 + (a + b)x + b = 3x2 – 4x – 7
On comparing coefficients, we get
a = 3, a + b = -4 and b = – 7
Hence, required term is (3x – 7).
(iii) We have, 10x2 – 11x – 6
Now, splitting the middle term, we get
= 10x2 – 15x + 4x-6 = 5x (2x – 3) + 2 (2x – 3)
= (2x – 3) (5x + 2)
Hence, 10x2 – 11x – 6 = (2x – 3) (5x + 2)
(iv) We have, 6x2 + 7x + 2
Now, splitting the middle term, we get
= 6x2 + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (3x + 2) (2x + 1)
Hence, 6x2 + 7x + 2 = (3x + 2) (2x + 1)
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
Using the identity (a + b)2 = a2 – 2ab + b2, we get
(41)2 = (40)2 + 2(40) (1) + (1)2
= 1600 + 80 + 1
= 1681
(ii) We have, (27)2 = (30 – 3)2
Using the identity (a – b)2 = a2 – 2ab + b2,
we get (27)2 =(30)2 -2(30) (3) + (3)2
= 900 – 180 + 9
= 729
(iii) We have, (23 × 17) = (20 + 3) (20 – 3)
Using the identity (a + b)(a – b) = a2 – b2, we get
23 × 17 =(20)2 – (3)2
= 400 – 9
= 391
(iv) We have, (135)2 = (100 + 35)2
Using the identity (a + b)2 = a2 + 2ab + b2, we get
(135)2 = (100)2 + 2(100)(35) + (35)2
= 10000 + 7000 + 1225
= 18225
(v) We have, (97)2 = (100 – 3)2
Using the identity (a – b)2 = a2 – 2ab + b2, we get (97)2 =(100)2 – 2(100) (3) + (3)2
= 10000 – 600 + 9
= 9409
(vi) We have, 18 × 29 = (23 – 5) (23 + 6)
Using the identity, (a —b) (a + c) = a2 +a(c – b) – bc, we get
(18 × 29) =(23)2 + 23(6 – 5) – (5 × 6)
= 529 + 23 – 30
= 522
(vii) We have, 34 × 43 = (38 – 4) (38 + 5)
Using identity (x – a)(x + b) = x2 + (b – a)x – ab, we get
34 × 43 = (38)2 + (5 – 4) (38) – 20
= 1444 + 38 – 20
= 1462
(viii) We have, (205)2 = (200 + 5)2
Using the identity (a + b)2 = a2 + 2 ab + b2, we get
(205)2 = (200)2 + 2(200) (5) + (5)2
= 40000 + 2000 + 25
= 42025
Question 3.
Factor the following.
(i) 9a2 + b2 + 4c2 – 6ab + 12ac – 4bc
Solution:
We have, 9a2 + b2 + 4c2 – 6ab + 12ac – 4bc
Using identity a2 + b2 + c2 + 2ab + 2bc + 2 ca = (a + b + c)2, we get
= (3a)2 + (- b)2 + (2c)2 + 2(3a)(- b) + 2(-b)(2c) + 2(2c)(3a)
= [(3a) +(-b) + (2c)]2
= (3a – b + 2c)2
= (3a – b + 2c) (3a – b+ 2c),
which are the required factorisation of the given expression.
(ii) 16s2 + 25t2 – 40st
Solution:
We have, 16s2 + 25t2 – 40st
= (4s)2 + (5t)2 – 2(4s)(5t)
Using identity a2 + b2 – 2ab = (a – b)2, we get
= (4s – 5t)2 = (4s – 5t)(4s – 5t)
(iii) r2 – r – 42
Solution:
We have, r2 – r – 42
Now, splitting the middle term, we get
= r2 – 7r + 6r – 42
= r(r – 7) + 6(r – 7)
= (r + 6)(r – 7)
(iv) 49g2 + 14gh + h2
Solution:
We have, 49g2+ 14gh + h2 = (7g)2 + (h)2 + 2(7g)(h)
Using identity a2 + b2 + 2ab = (a + b)2, we get
= (7g + h)2
= (7g + h)(7g + h)
(v) 64u2 + 121v2 + 4w2 – 176wv – 32uw + 44vw
Solution:
We have,
64u2 + 121v2 + 4w2 – 176wv – 32uw + 44vw
= (8u)2 + (-11v)2 + (-2w)2 + 2(8u)(-11v) + 2(-11v) (-2w) + 2(-2w)(8u)
Using identity a2 + b2 + c2 + 2 ab + 2 bc + 2 ca = (a + b + c)2,we get
= [(8w) + (- 11v) + (-2w)]2
= (8u – 11v – 2w)2
= (8u – 11v – 2w) (8u – 11v – 2w)
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) \(\frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}\)
Solution:
We have, \(\frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}\)
= \(\frac{3\left(p^2-p q-6 q^2\right)}{p^2+5 p q-2 p q-10 q^2}\)
= \(\frac{3(p-3 q)(p+2 q)}{(p+5 q)(p-2 q)}\)
(ii) \(\frac{n^3-3 n^3 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2}\)
Solution:
We have,
\(\frac{n^3-3 n^3 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2}\)
Using the identity a3 – 3a2b + 3ab2 – b3 = (a – b)3, we get
= \(\frac{(n-m)^3}{5\left(m^2-2 m n+n^2\right)}\)
Also, using identity a2 – 2ab + b2 = (a – b)2
= \(\frac{(n-m)^3}{5(n-m)^2}\)
= \(\frac{n-m}{5}\)
(iii) \(\frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x}\)
Solution:
We have,
\(\frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x}\)
Using the identity a3 + b3 + c3 – 3abc = (a + b + c)
(a2 + b2 + c2 -ab – bc – ca),
take a = w,b = -v, c = x, then
w3 – v3 + x3 + 3wvx
= (w – v + x) (w2 + v2 + x2 + wv + vx – wx)
Also, denominator
w2 + v2 + x2 – 2wv – 2vx + 2wx = (w – v + x)2
Hence,
= \(\frac{(w-v+x)\left(w^2+v^2+x^2+w v+v x-w x\right.}{(w-v+x)^2}\)
= \(\frac{w^2+v^2+x^2+w v+v x-w x}{w-v+x}\)
(iv) \(\frac{4 y^2-20 y z+25 z^2}{\left(25 z^2-4 y^2\right)}\)
Solution:
We have,
\(\frac{4 y^2-20 y z+25 z^2}{\left(25 z^2-4 y^2\right)}\)
Using the identities a2 – 2ab + b2 = (a – b)2 and a2 – b2 = (a – b) (a + b), we get
(v) \(\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}\)
Solution:
We have,
\(\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}\)
Now, numerator = (x2 + x – 6)(x2 -7x + 12)
= [x2 + 3x – 2x – 6].[x2 – 4x – 3x + 12]
= [x(x + 3) – 2(x + 3)].[x(x – 4) -3(x – 4)]
= [(x + 3)(x – 2)].[(x – 4)(x – 3)]
and denominator = (x2 – 6x + 8)(x2 – 9)
= [x2 – 4x – 2x + 8].(x – 3)(x + 3)
[va2 – b2 = (a + b)(a – b)]
= [x(x – 4) – 2(x – 4)] .(x – 3)(x + 3)
= (x – 4) (x – 2) (x – 3) (x + 3)
Now,
\(\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}\)
= \(\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}\) = 1
(vi) \(\frac{p^4-16}{p^2-4 q+4}\)
Solution:
We have,
\(\frac{p^4-16}{p^2-4 q+4}\)
Using the identities a2 – 2ab + b2 = (a – b)2
and a2 – b2 = (a – b) (a + b), we get
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
Solution:
We have, (-3x + 4)2
Using the identity (a + b)2 = a2 + 2ab + b2, we get
= (-3x)2 + 2(-3x)(4) + (4)2
= 9x2 – 24x + 16
(ii) (2s + 7) (2s – 7)
Solution:
We have, (2s + 7) (2s – 7)
Using the identity (a + b) (a – b) = a2 – b2, we get
= (2s)2 – (7)2
= 4s2 – 49
(iii) \(\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)\)
Solution:
We have, \(\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)\)
Using the identity (a + b) (a – b) = a2 – b2, we get
= (p2)2 – (\(\frac{1}{2}\))2
= p4 – \(\frac{1}{4\)
(iv) (2n + 7) (2n – 7)
Solution:
We have, (2n + 7) (2n – 7)
Using the identity (a + b) (a – b) = a2 – b2, we get
= (2n)2 – (7)2
= 4n2 – 49
(v) (s – 2t) (s2 + 2st + 4t2)
Solution:
We have, (s – 2t) (s2 + 2st + 4t2)
Using the identity (a – b) (a2 + ab + b2) = a3 – b3, we get
= s3 – (2t)3
= s3 – 8t3
(vi) (\(\frac{1}{2r}\) – 4r)2
Solution:
We have, (\(\frac{1}{2r}\) – 4r)2
Using the identity (a – b) (a2 + ab + b2) = a3 – b3, we get
= (\(\frac{1}{2r}\))2 – 2(\(\frac{1}{2r}\))(4r) + (4r)2
= \(\frac{1}{4 r^2}\) – 4 + 16r2
(vii) (-3m + 4k – l)2
Solution:
We have, (-3m + 4k – l)2
Using the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca, we get
= (-3m)2 + (4k)2 + (-l)2 + 2(-3m)(4k) + 2(4lc)(-l) + 2(-l)(-3m)
= 9m2 + 16k2 + l2 – 24mk – 8kl + 6ml
(viii) (x – \(\frac{1}{3}\)y)3
Solution:
We have, (x – \(\frac{1}{3}\)y)3
Using the identity (a – b)3 = a2 – 3a2b + 3ab2 – b3, we get
= x3 – 3x2(\(\frac{1}{3}\)y) + 3x(\(\frac{1}{3}\)y)2 – (\(\frac{1}{3}\)y)3
= x3 – x2y + \(\frac{1}{3}\)xy2 – \(\frac{1}{27}\)y3
(ix) (\(\frac{7}{2}\)k – \(\frac{2}{3}\)m)3
Solution:
We have, (\(\frac{7}{2}\)k – \(\frac{2}{3}\)m)3
Using the identity (a – b)3 = a2 – 3a2b + 3ab2 – b3, we get
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
(viii) (-299)3
Solution:
(i) We have, 17 × 21 = (19 – 2) (19 + 2)
Using the identity (a – b) (a + b) = a2 – b2
= 192 – 22
= 361 – 4
= 357
(ii) We have, 104 × 96 = (100 + 4) (100 – 4)
Using the identity (a + b) (a – b) – a2 – b2, we get
= 1002 – 42
= 10000 – 16
= 9984
(iii) We have, 24 × 16 = (20 + 4) (20 – 4)
Using the identity (a + b) (a – b) = a2 – b2, we get
= 202 – 42
= 400 – 16
= 384
(iv) We have, 1473 =(150 – 3)3
Using the identity
(a – b)3 = a3 – 3a2b + 3ab2 – b3, we get
= 1503 – 3(150)2(3) + 3(150)(3)2 – (3)3
= 3375000 – 202500 + 4050 – 27
= 3176523
(v) We have, 1993 = (200 – 1)3
Using the identity
(a – b)3 = a3 – 3a2b + 3ab2 – b3, we get
= 2003 – 3(200)2 (1) + 3(200) (1)2 – 1
= 8000000 – 120000 + 600 – 1
= 7880599
(vi) We have, 1273 = (100 + 27)3
Using the identity
(a + b)3 = a3 + 3a2b + 3ab2 + b3, we get
= 1003 + 3(100)2 (27) + 3(100) (27)2 + 273
= 1000000 + 810000 + 218700 + 19683
= 2048383
(vii) We have,
(-107)3 = -(107)3 = -(100 + 7)3
Using the identity
(a + b)3 = a3 + 3a2b + 3ab2 + b3, we get
= -[1003 + 3(100)2(7) + 3(100)(7)2 + 73]
= -(1000000 + 210000 + 14700 + 343)
= -1225043
(viii) We have,
(-299)3 = -(299)3 = -(300 – 1)3
Using the identity
(a – b)3 = a3 – 3a2b + 3ab2 – b3, we get
= -[(3003 – 3(300)2 (1) + 3 (300) (1)2 – (1)3]
= -(27000000 – 270000 + 900 – 1)
= -(26730900 – 1)
= -26730899
Question 3.
Factor the following algebraic expressions.
(i) 4y2 + 1 + \(\frac{1}{16 y^2}\)
Solution:
We have,
4y2 + 1 + \(\frac{1}{16 y^2}\)
Using the identity a2 + b2 + 2ab = (a + b)2, we get
(ii) 9m2 – \(\frac{1}{25 n^2}\)
Solution:
We have,
9m2 – \(\frac{1}{25 n^2}\)
Using the identity a2 – b2 = (a – b)(a + b), we get
= (3m)2 – \(\left(\frac{1}{5 n}\right)^2\)
= (3m – \(\frac{1}{5n}\))(3m + \(\frac{1}{5n}\))
(iii) 27b3 – \(\frac{1}{64 b^3}\)
Solution:
We have,
27b3 – \(\frac{1}{64 b^3}\)
Using the identity a3 – b3 = (a – b)(a2 + ab + b2), we get
= (3b)2 – \(\left(\frac{1}{4 b}^3\right)\)
= (3b – \(\frac{1}{4 b}\))(9b2 + \(\frac{3}{4}+\frac{1}{16 b^2}\))
(iv) x2 + \(\frac{5 x}{6}+\frac{1}{6}\)
Solution:
We have, x2 + \(\frac{5 x}{6}+\frac{1}{6}\)
= x2 + \(\frac{x}{2}+\frac{x}{3}+\frac{1}{6}\)
= x\(\left(x+\frac{1}{2}\right)+\frac{1}{3}\left(x+\frac{1}{2}\right)\)
= (x + \(\frac{1}{2}\))(x + \(\frac{1}{3}\))
(v) 27u3 – \(\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25}\)
Solution:
We have,
27u3 – \(\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25}\)
Using the identity
a3 – 3a2b + 3ab2 – b3 = (a – b)3, we get
= (3u)2 – 3(3u)2(\(\frac{1}{5}\)) + 3(3u)\(\left(\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^3\)
= (3u – \(\frac{1}{5}\))3
= \(\left(3 u-\frac{1}{5}\right)\left(3 u-\frac{1}{5}\right)\left(3 u-\frac{1}{5}\right)\)
(vi) 64y3 + \(\frac{1}{125}\)z3
Solution:
We have,
64y3 + \(\frac{1}{125}\)z3
Using the identity
a3 + b3 = (a + b)(a2 – ab + b2), we get
= (4y)3 + \(\left(\frac{z}{5}\right)^3\)
= \(\left(4 y+\frac{z}{5}\right)\left(16 y^2-\frac{4 y z}{5}+\frac{z^2}{25}\right)\)
(vii) p3 +27q3 + r3 – 9pqr
Solution:
We have,
p3 +27q3 + r3 – 9pqr
Using the identity a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 -ab – bc – ca), we get
= p3 + (3q)3 + r3 – 3(p)(3q)(r)
= (p + 3q + r) (p2 + 9q2 + r2 – 3pq – 3pr – pr)
(viii) 9m2 – 12m + 4
Solution:
We have, 9m2 – 12m + 4
Using the identity a2 + b2 – 2ab = (a – b)2, we get
= (3m)2 + (2)2 – 2(3m)(2)
= (3m – 2)2
= (3m – 2) (3m – 2)
(ix) 9x3 – \(\frac{8}{3}\) y3 + \(\frac{z^3}{3}\) + 6xyz
Solution:
We have,
9x3 – \(\frac{8}{3}\) y3 + \(\frac{z^3}{3}\) + 6xyz
Using the identity
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 -ab – bc – ca), we get
= \(\frac{1}{3}\) [(3x)3 + (-2y)3 + z3 – 3(3x) (-2y) (z)]
= \(\frac{1}{3}\)(3x – 2y + z)
(9x2 + 4y2 + z2 + 6xy + 2yz – 3xz)
(x) 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy
Solution:
We have,
4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy
Using the identity a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2, we get
= (2x)2 + (3y)2 + (6z)2 + 2(2x)(3y) + 2(3y)(6z) + 2(2x)(6z)
= (2x + 3y + 6z)2
= (2x + 3y + 6z) (2x + 3y + 6z)
(xi) 27u3 – \(\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4}\)
Solution:
We have, 27u3 – \(\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4}\)
Using the identity
a3 – 3a2b + 3ab2 – b3 = (a – b)3, we get
= (3u)3 – 3(3u)2(\(\frac{1}{6}\)) + 3(3u)\(\left(\frac{1}{6}\right)^2-\left(\frac{1}{6}\right)^3\)
= (3u – \(\frac{1}{6}\))3
= (3u – \(\frac{1}{6}\))(3u – \(\frac{1}{6}\))(3u – \(\frac{1}{6}\))
Question 4.
Simplify the following
(i) \(\frac{4 x^2+4 x+1}{4 x^2-1}\)
Solution:
We have,
\(\frac{4 x^2+4 x+1}{4 x^2-1}\)
= \(\frac{(2 x)^2+2(2 x)(1)+1^2}{(2 x)^2-1^2}\)
Using the identities a2 + 2ab + b2 = (a + b)2 and a2 – b2 =(a – b) (a + b), we get
= \(\frac{(2 x+1)^2}{(2 x-1)(2 x+1)}\)
= \(\frac{2 x+1}{2 x-1}\)
(ii) \(\frac{9\left(3 a^3-24 b^3\right)}{9 a^2-36 b^2}\)
Solution:
We have,
\(\frac{9\left(3 a^3-24 b^3\right)}{9 a^2-36 b^2}\)
= \(\frac{9 \cdot 3\left(a^3-8 b^3\right)}{9\left(a^2-4 b^2\right)}\)
= \(\frac{3\left(a^3-(2 b)^3\right)}{a^2-(2 b)^2}\)
Using the identities a3 – b3 = (a – b) (a2 + ab + b2)
and a2 – b2 = (a – b) (a + b), we get
= \(\frac{3(a-2 b)\left(a^2+2 a b+4 b^2\right)}{(a-2 b)(a+2 b)}\)
= \(\frac{3\left(a^2+2 a b+4 b^2\right)}{a+2 b}\)
(iii) \(\frac{s^3+125 t^3}{s^2-2 s t-35 t^2}\)
Solution:
We have,
\(\frac{s^3+125 t^3}{s^2-2 s t-35 t^2}\)
Using the identity
a3 + b3 =(a + b)(a2 – ab + b2), we get
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) We have, 25a2 – 30ab + 9b2
= (5a)2 – 2(5a) (3b) + (3b)2
Using the identity a2 – 2ab + b2 = (a – b)2, we get
= (5a – 3b)2 = (5a – 3b) (5a – 3b)
Hence, possible length = 5a – 3b
and breadth = 5a – 3b.
(ii) We have, 36s2 – 49t2
= (6s)2 – (7t)2
Using the identity a2 – b2 = (a – b) (a + b), we get
= (6s – 7t)(6s + 7t)
Hence, possible length = 6s – 7t
and breadth = 6s + 7t.
Or length = 6s + 7t and breadth = 6s – 7t.
Question 6.
Find possible expressions for the length, breadth and heights 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) We have, 6a2 – 24b2
= 6(a2 – 4b2)
= 6[a2 – (2b)2]
Using the identity a2 – b2 = (a – b) (a + b), we get = 6(a – 2b) (a + 2b)
Hence, possible length = 6, breadth = a – 2b and height =a+2b.
(ii) We have, 3ps2 – 15ps + 12p = 3p(s2 – 5s + 4)
= 3p(s2 – 4s – s + 4)
= 3p[s(s – 4) -1 (s – 4)]
= 3p(s – 1)(s – 4)
Hence, possible length = 3p,
breadth = s – 1 and height = s – 4.
Question 7.
The village playground is shaped as a square of side 40 m. A path of width s m is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Solution:
We have, side of square playground = 40 m
Now, a path of width s m is created around it.
So, outer side of the new square = (40 + 2s) m
Now, area of outer square = (40 + 2s)2 m2
So, area of playground = 402 = 1600 m2
Now, area of path = Outer area – Inner area = (40 + 2s)2 – 1600
Using the identity (a + b)2 = a2 + 2ab + b2, we get
= (1600 + 160s + 4s2) – 1600
= 160s + 4s2
= 4s2 + 160s
Hence, required expression for area of the path is 4s2 + 160s sq m.
Question 8.
If a number plus its reciprocal equals \(\frac{10}{3}\), find the number.
Solution:
Let the number be x
According to the statement, we have
x + \(\frac{1}{x}=\frac{10}{3}\)
On multiplying both sides by lx, we get
⇒ 3x2 + 3 = 10x
⇒ 3x2 – 10x + 3 = 0
Now, splitting the middle term, we get
⇒ 3x2 – 9x – x + 3 = 0
⇒ 3x(x – 3) – 1(x – 3) = 0
⇒ (3x – 1) (x – 3) = 0
⇒ 3x – 1 = 0 or x – 3 = 0
⇒ x = \(\frac{1}{3}\) or x = 3
Hence, the required numbers are – and 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:
Given,
Area of rectangular poal = 2x2 + 7x + 3 sq hastas
and width = 2x + 1 hastas.
Now, length = \(\frac{\text { Area }}{\text { Width }}=\frac{2 x^2+7 x+3}{2 x+1}\)
Now, factorising the numerator, we get
2x2 + 7x + 3 = 2x2 + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (2x + 1) (x + 3)
∴ Length = \(\frac{(2 x+1)(x+3)}{2 x+1}\) = x + 3
Hence, 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 of px2 + 5x + r.
First, taking x – 2 as a factor, so putting x =2, we get
p(2)2 + 5(2) + r = 0
4p + 10 + r = 0 …(i)
Now, taking [ x – \(\frac{1}{2}\)] as a factor, so putting x = \(\frac{1}{2}\), we get
p(\(\frac{1}{2}\))2 + 5(\(\frac{1}{2}\)) + r = 0
\(\frac{p}{4}+\frac{5}{2}\) + r = 0 ……(ii)
Now, multiplying Eq. (ii) by 4, we get
p + 10 + 4r = 0 … (iii)
Now, subtracting Eq. (iii) from Eq. (i), we get
(4p + 10+r)-(p + 10 + 4r) = 0
⇒ 4p + 10 + r – p – 10 – 4r = 0
⇒ 3p – 3r = 0
⇒ 3(p – r) = 0
∴ p = r
Question 11.
If a + b + c = 5 and ab + bc + ca = 10 then prove that a3 + b3 + c3 – 3abc = -25.
Solution:
We have, a + b + c = 5 and ab + bc + ca = 10
Using the identity a3 + b3 + c3 – 2abc
= (a + b + c) [(a + b + c)2 – 3(ab + bc + ca)]
On substituting the given values, we get a3 + b3 + c3 – 3abc
= 5[52 -3(10)]
= 5(25 – 30)
= 5(—5)
= -25
Hence, a3 + b3 + c3 – 2abc = -25.
Question 12.
By factorising the expression, check that n3 – n is always divisible by 6 for all natural numbers n. Give reason.
Solution:
Given, n3 – n
Now, factorising the expression, we get n3 – n = n(n2 – 1)
Also, using the identity a2 – b2 = (a – b) (a + b), we get
n(n2 – 1) = n(n – 1) (n + 1)
Now, n – 1, n, n + 1 are three consecutive natural numbers and among any three consecutive numbers, one is divisible by 2 and also one is divisible by 3.
Hence, the product (n – 1) (n + 1) is divisible by 2 × 3 = 6.
Therefore, 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) x3 + y3 – 12xy + 64, when x + y = -4
We know that
x3 + y3 = (x + y)3 – 3xy(x + y)
Now, x3 + y3 – 12xy + 64
= (x + y)3 – 3xy(x + y) – 12xy + 64
Also, x + y = -4
= (-4)3 – 3xy(-4) – 12xy + 64
= -64 + 12xy – 12xy + 64 = 0
Hence, the value of x3 + y3 – 12xy + 64 is 0.
(ii) x3 – 8y3 -36xy – 216, when x = 2y + 6
Given, x = 2y + 6
So, x – 2y = 6
Now,
x3 – 8y3 = x3 – (2y)3
Using the identity
a3 – b3 =(a – b)(a2 + ab + b2), we get
x3 – 8y3 =(x – 2y) (x2 + 2xy + 4y2)
Then,
x3 – 8y3 -36xy – 216
= (x – 2y) (x2 + 2xy + 4y2) – 36xy – 216
= 6(x2 + 2xy + 4y2) – 36xy – 216
= 6x2 + 12xy + 24y2 – 36xy – 216
= 6x2 – 24xy + 24y2 – 216
= 6(x2 – 4xy + 4y2) – 216
= 6(x – 2y)2 – 216
= 6 (6)2 – 216
= 216 – 216
= 0
Hence, the value of x3 – 8y3 – 36xy – 216 is 0,