Get the simplified Extra Questions for Class 9 Maths and Ganita Manjari Class 9 Maths Chapter 4 Exploring Algebraic Identities Extra Questions with complete explanation.
Class 9 Exploring Algebraic Identities Extra Questions
Extra Questions on Exploring Algebraic Identities Class 9
Class 9 Ganita Manjari Chapter 4 Extra Questions
Question 1.
Find the product of the following.
(i) (x + 2y)(x + 3y)
(ii) (x + 5y)(x – 8y)
Solution:
(i) (x + 2y)(x + 3y) = (x)2 + (2y + 3y)x + (2y)(3y) = x2 + 5xy + 6y2
(ii) (x + 5y)(x – 8y) = (x)2 + (5y – 8y)x + (5y)(-8y) = x2 – 3xy – 40y2
Question 2.
Expand
(i) (9x + 2y + z)2
(ii) (2x + 3y – 5z)2
Solution:
(i) Use the identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(9x + 2y + z)2 = 81x2 + 4y2 + z2 + 36xy + 4yz + 18xz
(ii) Use the identity, (a + b – c)2 = a2 + b2 + c2 + 2ab – 2bc – 2ca
(2x + 3y – 5z)2 = 4x2 + 9y2 + 25z2 + 12xy – 30yz – 20xz
Exploring Algebraic Identities Class 9 Very Short Question Answer
Question 1.
Using a suitable identity, find the value of (97)3.
Solution:
Use the identity, (a – b)3 = a3 – b3 – 3ab(a – b)
(97)3 = (100 – 3)3 = 912673
Question 2.
By using the formula A2 – B2 = (A + B)(A – B), factorise the following questions.
(a) 4x2 – 25
(b) x2 – 21
(c) \(a^2+\frac{1}{a^2}-18\)
(d) a4 – 7a2 + 1
(e) \(x^2+\frac{1}{x^2}-11\)
Solution:
(a) 4x2 – 25 = (2x)2 – (5)2 = (2x + 5)(2x – 5)
(b) x2 – 21 = x2 – (√21)2 = (x – √21)(x + √21)
(c) \(a^2+\frac{1}{a^2}-18\) = \(\left(a-\frac{1}{a}\right)^2-16\)
= \(\left(a-\frac{1}{a}\right)^2-4^2\)
= \(\left(a-\frac{1}{a}+4\right)\left(a-\frac{1}{a}-4\right)\)
(d) a4 – 7a2 + 1 = (a2 + 1)2 – (3a)2 = (a2 + 1 + 3a)(a2 + 1 – 3a)
(e) \(x^2+\frac{1}{x^2}-11\) = \(x^2+\frac{1}{x^2}-2-9\)
= \(x^2+\frac{1}{x^2}-2 \cdot x \cdot \frac{1}{x}-9\)
= \(\left(x-\frac{1}{x}\right)^2-(3)^2\)
= \(\left(x-\frac{1}{x}+3\right)\left(x-\frac{1}{x}-3\right)\)
Question 3.
Find the factors of the following.
(a) (i) 8ab2 + 12a2b
(ii) 4(x + y)2 – 3(x + y)
(iii) 2(2x – 5y)(3x + 4y) – 6(2x – 5y)(x – y)
(b) a – 1 – (a – 1)2 + ax – x
(c) 5√5x2 + 20x + 3√5
(d) px2 + (4p2 – 3q)x – 12pq
(e) 4(2a – 3)2 – 3(2a – 3)(a – 1) – 7(a – 1)2
(f) (3p – 1)2 – (3p – 2)(3p – 1)
(g) \(\frac{a}{b} x^2+\left(\frac{a}{b}+\frac{c}{d}\right) x+\frac{c}{d}\), b ≠ 0 and d ≠ 0
(h) 25a(x – y) + 30b(y – x)
(i) px – (px + qy)2 + p2x + pqy + qy
(j) 8x3 – 6x2 + 10x
(k) 18p2q2 – 24pq2 + 30p2q
(l) 6xy2 + 9x2y – 21xy
(m) a(a + b – c) – bc
(n) a2x2 + (ax2 + 1)x + a
(o) 4(x – y)2 – 12(x2 – y2) + 9(x + y)2
(p) a2b + ab2 – abc – b2c + axy + bxy
Solution:
(a) (i) 8ab2 + 12a2b = \(4 a b\left[\frac{8 a b^2}{4 a b}+\frac{12 a^2 b}{4 a b}\right]\) = 4ab(2b + 3a)
(ii) 4(x + y)2 – 3(x + y)
Taking (x + y) common
(x + y)(4x + 4y – 3)
(iii) 2(2x – 5y)(3x + 4y) – 6(2x – 5y)(x – y)
Taking (2x – 5y) common
14y(2x – 5y)
(b) a – 1 – (a – 1)2 + ax – x = (a – 1) – (a – 1)2 + x(a – 1)
Taking (a -1) common
(a – 1)(x – a + 2)
(c) 5√5x2 + 20x + 3√5 = 5√5x2 + 15x + 5x + 3√5 = (5x + √5)(√5x + 3)
(d) px2 + (4p2 – 3q)x – 12pq = px(x + 4p) – 3q(x + 4p) = (x + 4p)(px – 3q)
(e) 4(2a – 3)2 – 3(2a – 3)(a – 1) – 7(a – 1)2
Let x = (2a – 3) and y = (a – 1)
So, the given expression will be
4x2 – 3xy – 7y2 = (4x – 7y)(x + y)
= [4(2a – 3) – 7(a – 1)][(2a – 3) + (a – 1)]
= (a – 5)(3a – 4)
(f) (3p – 1)2 – (3p – 2)(3p – 1) = (3p – 1)[3p – 1 – 3p + 2] = (3p – 1)
(g) \(\frac{a}{b} x^2+\left(\frac{a}{b}+\frac{c}{d}\right) x+\frac{c}{d}\)
= \(\left(\frac{a}{b} x^2+\frac{a}{b} x\right)+\left(\frac{c}{d} x+\frac{c}{d}\right)\)
= \(\frac{a}{b} x(x+1)+\frac{c}{d}(x+1)\)
= \((x+1)\left(\frac{a x}{b}+\frac{c}{d}\right)\)
(h) 25a(x – y) + 30b(y – x) = (x – y)(25a – 30b) = 5(x – y)(5a – 6b)
(i) px – (px + qy)2 + p2x + pqy + qy = (px + qy) – (px + qy)2 + p(px + qy) = (px + qy) (1 – px – qy + p)
(j) 8x3 – 6x2 + 10x
Taking 2x common
2x(4x2 – 3x + 5)
(k) 18p2q2 – 24pq2 + 30p2q
Taking 6pq common
6pq(3pq – 4q + 5p)
(l) 6xy2 + 9x2y – 21xy
Taking 3xy common
3xy(2y + 3x – 7)
(m) a(a + b – c) – bc = a2 + ab – ac – bc
= a(a + b) – c(a + b)
= (a + b)(a – c)
(n) a2x2 + (ax2 + 1)x + a = a2x2 + ax3 + x + a
= ax2(x + a) + 1(x + a)
= (x + a)(ax2 + 1)
(o) 4(x – y)2 – 12(x2 – y2) + 9(x + y)2 = (-x – 5y)2 = (x + 5y)2
(p) a2b + ab2 – abc – b2c + axy + bxy = (a2b + ab2) – (abc + b2c) + (axy + bxy)
= ab(a + b) – bc(a + b) + xy(a + b)
= (a + b)(ab – bc + xy)
Exploring Algebraic Identities Class 9 Short Question Answer
Question 1.
If a2 + b2 + c2 – ab – bc – ca = 0, prove that a = b = c.
Solution:
a2 + b2 + c2 – ab – bc – ca = 0
Multiply by 2 on both sides, we get
(a – b)2 + (b – c)2 + (c – a)2 = 0
(a – b)2 = 0, (b – c)2 = 0, (c – a)2 = 0
Question 2.
Find the expansion of (x + y – 1)3.
Solution:
(x + y – 1)3 = [(x + y) – 1]3
= (x + y)3 – (1)3 – 3(x + y)(1)[(x + y) – 1]
= (x + y)3 – 1 – 3(x + y)[(x + y) – 1]
= x3 + y3 + 3x2y + 3xy2 – 3x2 – 6xy – 3y2 + 3x + 3y – 1
Question 3.
If 3(x2 + 1) = 10x, where x > 1 then find
(i) \(x-\frac{1}{x}\)
(ii) \(x^3-\frac{1}{x^3}\)
Solution:
(i) \(x+\frac{1}{x}=\frac{10}{3}\)
Now, \(x-\frac{1}{x}=\sqrt{\left(x+\frac{1}{x}\right)^2-4}\) = \(\pm \frac{8}{3}\)
(ii) \(x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)\) = \(\frac {728}{27}\)
Question 4.
If x > 0 and \(x^2+\frac{1}{9 x^2}=\frac{25}{36}\) then find \(x^3+\frac{1}{27 x^3}\).
Solution:
\(\left(x+\frac{1}{3 x}\right)^2=x^2+\frac{1}{9 x^2}+\frac{2}{3}\)
Now, \(x^3+\frac{1}{27 x^3}=x^3+\left(\frac{1}{3 x}\right)^3\)
= \(\left(x+\frac{1}{3 x}\right)^3-\left(x+\frac{1}{3 x}\right)\)
= \(\pm \frac{91}{216}\)
Question 5.
The sum of two numbers is 9, and their product is 20. Find the sum of their
(i) Squares
(ii) Cubes
Solution:
Let x + y = 9 and xy = 20
(i) x2 + y2 = (x + y)2 – 2xy
= (9)2 – 2(20)
= 81 – 40
= 41
(ii) x3 + y3 = (x + y)3 – 3xy(x + y)
= (9)3 – 3(20)(9)
= 729 – 540
= 189
Question 6.
If the sum of two numbers is 7 and the sum of their cubes is 133, then find the sum of their squares.
Solution:
Let x + y = 7 and x3 + y3 = 133
Then, use (x + y)3 = x3 + y3 + 3xy(x + y)
⇒ (7)3 = 133 + 3xy(7)
⇒ 343 = 133 + 21xy
⇒ 210 = 21xy
⇒ xy = 10
Finally, use the identity,
a2 + b2 = (a + b)2 – 2ab
= (7)2 – 2(10)
= 49 – 20
= 29
Question 7.
Factorise
(i) 32a2x3 – 8b2x3 – 4a2y3 + b2y3
(ii) x3p2 – 8y3p2 – 4x3q2 + 32y3q2
(iii) \(x^2-\left(\frac{a+b}{a}+\frac{a}{a+b}\right) x+1\)
(iv) x3 + xy(2 – 3x) – 6y2
Solution:
(i) (32a2x3 – 8b2x3) – (4a2y3 – b2y3) = (2a + b)(2a – b)(2x – y)(4x2 + 2xy + y2)
(ii) x3p2 – 8y3p2 – 4x3q2 + 32y3q2 = p2(x3 – 8y3) – 4q2(x3 – 8y3)
= (x3 – 8y3)(p2 – 4q2)
= (x – 2y)(x2 + 4y2 + 2xy)(p + 2q)(p – 2q)
(iii)\(x^2-\left(\frac{a+b}{a}+\frac{a}{a+b}\right) x+1\)
= \(x^2-\frac{a+b}{a} x-\frac{a}{a+b} x+1\)
= \(x\left(x-\frac{a+b}{a}\right)-\frac{a}{a+b}\left(x-\frac{a+b}{a}\right)\)
= \(\left(x-\frac{a+b}{a}\right)\left(x-\frac{a}{a+b}\right)\)
(iv) x3 + xy(2 – 3x) – 6y2 = x3 + 2xy – 3x2y – 6y2
= (x3 – 3x2y) + (2xy – 6y2)
= x2(x – 3y) + 2y(x – 3y)
= (x – 3y)(x2 + 2y)
Question 8.
Factorise
(i) a4 + b4 – 11a2b2
(ii) x4 + 5x2 + 9
(iii) \(\left(2 x-\frac{1}{2 x}\right)^2-4 x^2+\frac{1}{4 x^2}\)
(iv) x8 – y8
Solution:
(i) a4 + b4 – 11a2b2 = (a2)2 + (b2)2 – 2a2b2 – 9a2b2
= (a2 – b2)2 – (3ab)2
= (a2 – b2 + 3ab)(a2 – b2 – 3ab)
(ii) x4 + 5x2 + 9 = x4 + 6x2 + 9 – x2 = (x2 + 3 + x)(x2 + 3 – x)
(iii) \(\left(2 x-\frac{1}{2 x}\right)^2-4 x^2+\frac{1}{4 x^2}\)
= \(\left(2 x-\frac{1}{2 x}\right)^2-\left[(2 x)^2-\frac{1}{(2 x)^2}\right]\)
= \(\left(2 x-\frac{1}{2 x}\right)^2-\left(2 x-\frac{1}{2 x}\right)\left(2 x+\frac{1}{2 x}\right)\)
= \(\left(2 x-\frac{1}{2 x}\right)\left(2 x-\frac{1}{2 x}-2 x-\frac{1}{2 x}\right)\)
= \(-\frac{1}{x}\left(2 x-\frac{1}{2 x}\right)\)
(iv) x8 – y8 = (x4 + y4)(x4 – y4) = (x4 + y4)(x – y)(x + y)(x2 + y2)
Question 9.
The park is in the shape of a square with a side of 50 m. A walking strip of width x m is made all around the park. Find an expression for the area of the strip in terms of x.
Solution:
We have, side of square park = 50 m
Nutv, a walking strip of width x m, is made around it.
So, outer side of the new square = 50 + 2x
Now, area of outer square = (50 + 2x)2
Also, area of park = 502 = 2500
Now, area of strip = Outer area – Inner area = (50 + 2x)2 – 2500
Using identity (a + b)2 = a2 + 2ab + b2, we get
= (2500 + 200x + 4x2) – 2500
= 200x + 4x2
= 4x2 + 200x
Hence, the required expression for the area of the strip is 4x2 + 200x m2.
Question 10.
If a number added to its reciprocal gives \(\frac {5}{2}\), then find the number.
Solution:
Let the number be x.
According to the statement,
\(x+\frac{1}{x}=\frac{5}{2}\)
On multiplying both sides by 2x, we get
⇒ 2x2 + 2 = 5x
⇒ 2x2 – 5x + 2 = 0
Now, splitting the middle term, we get
⇒ 2x2 – 4x – x + 2 = 0
⇒ 2x(x – 2) – 1(x – 2) = 0
⇒ (2x – 1)(x – 2) = 0
⇒ 2x – 1 = 0 or x – 2 = 0
⇒ x = \(\frac {1}{2}\) or x = 2
Hence, the required numbers are \(\frac {1}{2}\) and 2.
Question 11.
The area of a rectangular field is 3y2 + 11y + 10 square units. If its width is y + 2 units, then find its length.
Solution:
We have,
Area = 3y2 + 11y + 10 sq units
and width = y + 2 units
Now, length = \(\frac{\text { Area }}{\text { Width }}\) = \(\frac{3 y^2+11 y+10}{y+2}\)
Now, factorising the numerator, we get
3y2 + 11y + 10 = 3y2 + 6y + 5y + 10
= 3y(y + 2) + 5(y + 2)
= (3y + 5)(y + 2)
So, length = \(\frac{(3 y+5)(y+2)}{y+2}\) = 3y + 5 units
Hence, the length of the field = 3y + 5 units.
Exploring Algebraic Identities Class 9 Long Question Answer
Question 1.
(i) If x = 7 + 4√3 then find the value of \(\sqrt{x}+\frac{1}{\sqrt{x}}\).
(ii) If x = 3 + 2√2 then find the value of \(x^3-\frac{1}{x^3}\)
(iii) If x = 5 – 2√6 then find the value of \(x^3-\frac{1}{x^3}\).
Solution:
x = 7 + 4√3
\(\frac{1}{x}=\frac{1}{7+4 \sqrt{3}}=\frac{7-4 \sqrt{3}}{(7)^2-(4 \sqrt{3})^2}\) = 7 – 4√3
Thus, x + \(\frac {1}{x}\) = 14
⇒ \((\sqrt{x})^2+\left(\frac{1}{\sqrt{x}}\right)^2=14\)
⇒ \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=14+2\)
⇒ \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\) = ±4
(ii) Do the same as Part (i)
(iii) Do the same as Part (i)
Question 2.
Factorise the following questions.
(i) \(\left(x-\frac{1}{x}\right)^2+6\left(x-\frac{1}{x}\right)+9\)
(ii) (a2 – 3a)(a2 – 3a + 7) + 10
(iii) 5(3a + b)2 + 6(3a + b) – 8
(iv) 3(2a – b)2 – 19(2a – b) + 28
(v) (a2 – a)(4a2 – 4a – 5) – 6
(vi) (p2 + 4p)2 + 21(p2 + 4p) + 98
Solution:
(i) \(\left(x-\frac{1}{x}\right)^2+6\left(x-\frac{1}{x}\right)+9\)
Put \(\left(x-\frac{1}{x}\right)\) = y, we get y2 + 6y + 9.
\(\left(x-\frac{1}{x}+3\right)^2\)
(ii) (a2 – 3a)(a2 – 3a + 7) + 10 = (a2 – 3a + 5)(a – 2)(a – 1)
(iii) Let 3a + b = x
Given expression becomes 5x2 + 6x – 8
(3a + b + 2)(15a + 5b – 4)
(iv) 3(2a – b)2 – 19(2a – b) + 28
Let 2a – b = x
Now, factorise 3x2 -19x + 28 = (2a – b – 4)(6a – 3b – 7)
(v) (a2 – a)(4a2 – 4a – 5) – 6
Let a2 – a = x
∴ (a2 – a)[4(a2 – a) – 5] – 6 = x(4x – 5) – 6
= 4x2 – 5x – 6
= 4x2 – 8x + 3x – 6
= 4x(x – 2) + 3(x – 2)
= (4x + 3)(x – 2)
On putting x = a2 – a, we get
[4(a2 – a) + 3](a2 – a – 2) = (4a2 – 4a + 3)(a2 – 2a + a – 2) = (4a2 – 4a + 3)(a – 2) (a + 1)
(vi) (p2 + 4p)2 + 21(p2 + 4p) + 98
Let p2 + 4p = x
Now, factorise x2 + 21x + 98 = (p2 + 4p + 7)(p2 + 4p + 14)
Question 3.
Find suitable algebraic expressions for the dimensions (length, breadth, and height) of each of the following cuboids, whose volumes are given below (in cubic units).
(i) 12a2 – 48b2
(ii) 4x2y + 8xy2
Solution:
(i) We have, 12a2 – 48b2 = 12(a2 – 4b2) = 12(a2 – (2b)2)
Using identity a2 – b2 = (a – b)(a + b), we get
= 12(a – 2b)(a + 2b)
= (12)(a – 2b)(a + 2b)
Hence, possible length = 12, breadth = a – 2b, and height = a + 2b.
(ii) We have, 4x2y + 8xy2 = 4xy(x + 2y) = (4x)(y) (x + 2y)
Hence, possible length = 4x, breadth = y, and height = x + 2y.