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Algebra Class 9th Mathematics Term 2 Tamilnadu Board Solution

Class 9th Mathematics Term 2 Tamilnadu Board Solution
Exercise 1.1
  1. (5x + 2y + 3z)^2 Expand the following
  2. (2a + 3b - c)^2 Expand the following
  3. (x - 2y - 4z)^2 Expand the following
  4. (P - 2q + r)^2 Expand the following
  5. (x + 1)(x + 4)(x + 7) Find the expansion of
  6. (p + 2)(p - 4)(p + 6) Find the expansion of
  7. (x + 5)(x - 3)(x - 1) Find the expansion of
  8. (x - a)(x - 2a)(x - 4a) Find the expansion of
  9. (3x + 1)(3x + 2)(3x + 5) Find the expansion of
  10. (2x + 3)(2x - 5)(2x - 7) Find the expansion of
  11. (x + 7)(x + 3)(x + 9) Using algebraic identities find the coefficients of x^2…
  12. (x - 5)(x - 4)(x + 2) Using algebraic identities find the coefficients of x^2…
  13. (2x + 3)(2x + 5)(2x + 7) Using algebraic identities find the coefficients of…
  14. (5x + 2)(1 - 5x)(5x + 3) Using algebraic identities find the coefficients of…
  15. If (x + a)(x + b)(x + c) ≡ x^3 - 10x^2 + 45x - 15 find a + b + c, 1/a + 1/b +…
  16. Expand : (i) (3a + 5b)^3 (ii) (4x - 3y)^3 (iii) (2y - 3/y)^3
  17. Evaluate : (i) 99^3 (ii) 101^3 (iii) 98^3 (iv) (102)^3 (v) (1002)^3…
  18. Find 8x^3 + 27y^3 if 2x + 3y = 13 and xy = 6.
  19. If x - y = - 6 and xy = 4, find the value of x^3 - y^3
  20. If x + 1/x = 4 , find the value of x^3 + 1/x^3 .
  21. If, x - 1/x = 4 find the value of x^3 - 1/x^3 .
  22. Simplify: (i) (2x + y + 4z)(4x^2 + y^2 + 16z^2 - 2xy - 4yz - 8zx) (ii) (x - 3y…
  23. Evaluate using identities : (i) 6^3 - 9^3 + 3^3 (ii) 16^3 - 6^3 - 10^3…
Exercise 1.2
  1. 2a^3 - 3a^2 b + 2a^2 c Factorize the following expressions:
  2. 16x + 64x^2 y Factorize the following expressions:
  3. 10x^3 - 25x^4 y Factorize the following expressions:
  4. xy - xz + ay - az Factorize the following expressions:
  5. p^2 + pq + pr + qr Factorize the following expressions:
  6. x^2 + 2x + 1 Factorize the following expressions:
  7. 9x^2 - 24xy + 16y^2 Factorize the following expressions:
  8. b^2 - 4 Factorize the following expressions:
  9. 1 - 36x^2 Factorize the following expressions:
  10. p^2 + q^2 + r^2 + 2pq + 2qr + 2rp Factorize the following expressions:…
  11. a^2 + 4b^2 + 36 - 4ab + 24b - 12a Factorize the following expressions:…
  12. 9x^2 + y^2 + 1 - 6xy + 6x - 2y Factorize the following expressions:…
  13. 4a^2 + b^2 + 9c^2 - 4ab - 6bc + 12ca Factorize the following expressions:…
  14. 25x^2 + 4y^2 + 9z^2 - 20xy + 12yz - 30zx Factorize the following expressions:…
  15. 27x^3 + 64y^3 Factorize the following expressions:
  16. m^3 + 8 Factorize the following expressions:
  17. a^3 + 125 Factorize the following expressions:
  18. 8x^3 - 27y^3 Factorize the following expressions:
  19. x^3 - 8y^3 Factorize the following expressions:
Exercise 1.4
  1. x + 3y = 10; 2x + y = 5 Solve the following equations by substitution method.…
  2. 2x + y = 1; 3x - 4y = 18 Solve the following equations by substitution method.…
  3. 5x + 3y = 21; 2x - y = 4 Solve the following equations by substitution method.…
  4. 1/x + 2/y = 9 2/x + 1/y = 12 (x not equal 0 , y not equal 0) Solve the…
  5. 3/x + 1/y = 7 5/x - 4/y = 6 (x not equal 0 , y not equal 0) Solve the following…
  6. Find two numbers whose sum is 24 and difference is 8.
  7. A number consists of two digits whose sum is 9. The number formed by reversing…
  8. Kavi and Kural each had a number of apples. Kavi said to Kural “If you give me 4…
  9. Solve the following inequations. (i) 2x + 715 (ii) 2(x - 2) 3 (iii) 2(x + 7) ≤ 9…
Exercise 1.5
  1. The expansion of (x + 2)(x - 1) isA. x^2 - x - 2 B. x^2 + x + 2 C. x^2 + x - 2…
  2. The expansion of (x + 1)(x - 2)(x + 3) isA. x^3 + 2x^2 - 5x - 6 B. x^3 - 2x^2 +…
  3. (x - y)(x^2 + xy + y^2) is equal toA. x^3 + y^3 B. x^2 + y^2 C. x^2 - y^2 D. x^3…
  4. Factorization of x^2 + 2x - 8 isA. (x + 4)(x - 2) B. (x - 4)(x + 2) C. (x + 4)(x…
  5. If one of the factors of x^2 - 6x - 16 is (x + 2) then other factor isA. x + 5…
  6. If (2x + 1) and (x - 3) are the factors of ax^2 - 5x + c, then the values of a…
  7. If x + y = 10 and x - y = 2, then value of x isA. 4 B. - 6 C. - 4 D. 6…
  8. The solution of 2 - x 5 isA. x - 3 B. x - 3 C. x 3 D. x 3

Exercise 1.1
Question 1.

Expand the following

(5x + 2y + 3z)2


Answer:

(5x + 2y + 3z)2


The identity is


(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


∴ (5x + 2y + 3z)2


= (5x)2 + (2y)2 + (3z)2 + 2(5x)(2y) + 2(2y)(3z) + 2(3z)(5x)


= 25x2 + 4y2 + 9z2 + 20xy + 12yz + 30zx



Question 2.

Expand the following

(2a + 3b – c)2


Answer:

(2a + 3b – c)2


Here the identity used is


(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


(2a + 3b – c)2 = (2a)2 + (3b)2 + (– c)2 + 2(2a)(3b) + 2(3b)(– c) + 2(– c)(2a)


= 4a2 + 9b2 + c2 + 12ab – 6bc – 4ca



Question 3.

Expand the following

(x – 2y – 4z)2


Answer:

(x – 2y – 4z)2


Here the identity used is


(x + y + z) 2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


∴ (x – 2y – 4z) 2


= x2 + (– 2y)2 + (– 4z)2 + 2(x)(– 2y) + 2(– 2y)(– 4z) + 2 (– 4z)(x)


= x2 + 4y2 + 16z2 – 4xy + 16yz – 8yz



Question 4.

Expand the following

(P – 2q + r)2


Answer:

(P – 2q + r)2


Here we use the identity of


(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


∴ (p – 2q + r)2


= (p)2 + (– 2q)2 + (r)2 + 2(p)(– 2q) + 2(– 2q)(r) + 2(r)(p)


= p2 + 4q2 + r2 – pq – 4qr + 2rp



Question 5.

Find the expansion of

(x + 1)(x + 4)(x + 7)


Answer:

(x + 1) (x + 4) (x + 7)


we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (x + 1)(x + 4)(x + 7)


= x3 + (1 + 4 + 7) x2 + (1 × 4 + 4 × 7 + 7 × 1) x + 1 × 4 × 7


= x3 + 12x2 + (4 + 28 + 7) x + 28


= x3 + 12x2 + 39x + 28



Question 6.

Find the expansion of

(p + 2)(p – 4)(p + 6)


Answer:

(p + 2)(p – 4)(p + 6)


Here the identity used is


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (p + 2)(p – 4)(p + 6)


= p3 + (2 – 4 + 6) p2 + (2(– 4) + (– 4)6 + 6(2))p + (2)(– 4)(6)


= p3 + 4p2 + (– 8 – 24 + 12)p – 48


= p3 + 4p2 – 20p – 48



Question 7.

Find the expansion of

(x + 5)(x – 3)(x – 1)


Answer:

(x + 5)(x – 3)(x – 1)


Here the identity used is


(x + a)(x + b)(x + c) = x3 + (a + b + c) x2 + (ab + bc + ca)x + abc


∴ (x + 5)(x – 3)(x – 1)


= x3 + (5 – 3 – 1) x2 + (5(– 3) + (– 3)(– 1)(– 1)5)x + 5(– 3)(– 1)


= x3 + x2 + (– 15 + 3 – 5) x + 15


= x3 + x2 – 17x + 15



Question 8.

Find the expansion of

(x – a)(x – 2a)(x – 4a)


Answer:

(x – a)(x – 2a)(x – 4a)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (x – a)(x – 2a)(x – 4a)


= x3 + (– a – 2a – 3a)x2 + [(– a)(– 2a) + (– 2a)(– 3a) + (– 3a)(– a)]x + (– a)(– 2a)(– 3a)


= x3 – 6ax2 + (2a2 + 6a2 + 3a2)x – 6a3


= x3 – 6ax2 + 11a2x – 6a3



Question 9.

Find the expansion of

(3x + 1)(3x + 2)(3x + 5)


Answer:

(3x + 1)(3x + 2)((3x + 5)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (3x + 1)(3x + 2)((3x + 5)


= (3x)3 + (1 + 2 + 5)(3x)2 + (1 × 2 + 2 × 5 + 5 × 1)(3x) + 1 × 2 × 5


= 27 x3 + (8)9x2 + (2 + 10 + 5) (3x) + 1 × 2 × 5


= 27 x3 + 72x2 + 51x + 10



Question 10.

Find the expansion of

(2x + 3)(2x – 5)(2x – 7)


Answer:

(2x + 3)(2x – 5)(2x – 7)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (2x + 3)(2x – 5)(2x – 7)


= (2x)3 + (3 – 5 – 7)(2x)2 + [(3)(– 5) + (– 5)(– 7) + (– 7)(3)](2x) + (3)(– 5)(– 7)


= 8x3 + (– 9)(4x2) + (– 15 + 35 – 21)(2x) + 105


= 8x3 – 36 x2 – 2x + 105



Question 11.

Using algebraic identities find the coefficients of x2 term, x term and constant term.

(x + 7)(x + 3)(x + 9)


Answer:

Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (x + 7)(x + 3)(x + 9)


= x3 + (7 + 3 + 9)x2 + (7 × 3 + 3 × 9 + 9 × 7)x + 7 × 3 × 9


= x3 + 19x2 + (21 + 27 + 63)x + 189


= x3 + 19x2 + 111x + 189


Coefficient of


x2 = 19


x = 111


Constant term = 189



Question 12.

Using algebraic identities find the coefficients of x2 term, x term and constant term.

(x – 5)(x – 4)(x + 2)


Answer:

(x – 5)(x – 4)(x + 2)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (x – 5)(x – 4)(x + 2)


= x3 + (– 5 – 4 + 2)x2 + {(– 5) × (– 4) + (– 4) × 2 + 2 × (– 5)}x + (– 5)(– 4)2


= x3 – 7x2 + (20 – 8 – 10)x + 40


= x3 – 7x2 + 2x + 40


Coefficient of


x2 = – 7


x = 2 and


Constant term = 40



Question 13.

Using algebraic identities find the coefficients of x2 term, x term and constant term.

(2x + 3)(2x + 5)(2x + 7)


Answer:

(2x + 3)(2x + 5)(2x + 7)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)(x)2 + (ab + bc + ca)x + abc


∴ (2x + 3) (2x + 5)(2x + 7)


= (2x)3 + 4(3 + 5 + 7)x2 + (15 + 35 + 21)2x + 105


= 8x3 + 60x2 + 142x + 105


Coefficient of


x2 = 60


x = 142


Constant term = 105



Question 14.

Using algebraic identities find the coefficients of x2 term, x term and constant term.

(5x + 2)(1 – 5x)(5x + 3)


Answer:

(5x + 2)(1 – 5x)(5x + 3)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ – (5x + 2)(5x – 1)(5x + 3)


here x = 5x, a = 2, b = – 1 and c = 3


= – (5x)3 – (2 – 1 + 3)(5x)2 – (– 2 – 3 + 6)5x + 6


= – 125x3 – 100x2 – 5x + 6


Here coefficient of


x2 = – 100


x = – 5 and


constant term = 6



Question 15.

If (x + a)(x + b)(x + c) ≡ x3 – 10x2 + 45x – 15 find a + b + c,  and a2 + b2 + c2.


Answer:

Here, we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


Comparing this with


x3 – 10x2 + 45x – 15 = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


⇒ – 10 = a + b + c


Now, 



And (ab + bc + ca) = 45


Now (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)


⇒ (– 10)2 – 2 × 45 = a2 + b2 + c2


⇒ 100 – 90 = a2 + b2 + c2


⇒ a2 + b2 + c2 = 10



Question 16.

Expand :

(i) (3a + 5b)3

(ii) (4x – 3y)3

(iii) 


Answer:

(i) (3a + 5b)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(3a + 5b)3


= (3a)3 + 3 (3a)2(5b) + 3(3a)(5b)2 + (5b)3


= 27a3 + 135a2b + 225ab2 + 125b3


(ii) (4x – 3y)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(4x – 3y)3


= (4x)3 + 3(4x)2(– 3y) + 3 × 4x (– 3y)2 + (– 3y)3


= 64x3 – 192 x2 + 108y2 – 27y3


(iii)


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


=




Question 17.

Evaluate :

(i) 993

(ii) 1013

(iii) 983

(iv) (102)3

(v) (1002)3


Answer:

(i) (100 – 1)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(100 – 1)3


= (100)3 + 3(100)2 (– 1) + 3(100) (– 1)2 + (– 1)3


= 1000000 – 30000 + 300 – 1


= 970299


(ii) 1013


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(101)3 = (100 + 1)3


= 1003 + 3 × 1002 × 1 + 3 × 100 × 12 + 13


= 1000000 + 30000 + 300 + 1


= 1030301


(iii) 983


= (100 – 2)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(100 – 2)3


= 1003 + (– 2)3 + 3(100)2(– 2) + 3(100)(– 2)2


= 1000000 – 8 – 60000 + 1200


= 941192


(iv) (102)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(100 + 2)3


= 1003 + 3(100)2 (2) + 3(2)2(100) + (2)3


= 1000000 + 60000 + 1200 + 8


= 1061208


(vi) (1002)3


= (1000 + 2)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(1000 + 2)3


= 10003 + 3 × 10002(2) + 3 × 1000 × (2)2 + 23


= 1000000000 + 6000000 + 12000 + 8


= 1006012008



Question 18.

Find 8x3 + 27y3 if 2x + 3y = 13 and xy = 6.


Answer:

Given 2x + 3y = 13


⇒ (2x + 3y)3 = 133


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(2x + 3y)3 = (2x)3 + 3(2x)23y + 3(2x)(3y)2 + (3y)3


⇒ 8x3 + 36x2y + 54xy2 + 27y3 = 2197


⇒ 8x3 + 27y3 + 18xy(2x + 3y) = 2197


⇒ 8x3 + 27y3 = 2197 – 108 (2x + 3y) (∵ xy = 6 given)


⇒ 8x3 + 27y3 = 2197 – 108 × 13


⇒ 8x3 + 27y3 = 2197 – 1404


⇒ 8x3 + 27y3 = 793



Question 19.

If x – y = – 6 and xy = 4, find the value of x3 – y3


Answer:

The identity used here is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


∴ (x – y)3 = x3 + 3x2(– y) + 3x(– y)2 + (– y)3


⇒ x3 – 3x2y + 3xy2 – y3


⇒ (x – y)3 = x3 – 3x2y + 3xy2 – y3


⇒ (– 6)3 = x3 – y3 + 3xy(– x + y)


⇒ – 216 + 3 × 4 (– 6) = x3 – y3


⇒ – 216 – 72 = x3 – y3


⇒ – 288 = x3 – y3



Question 20.

If , find the value of .


Answer:

Here we use the identity of


(a + b)3 = a3 + 3a2b + 3ab2 + b3


∴ 


⇒ 


⇒ 


⇒ 



Question 21.

If,  find the value of .


Answer:

Here the identity used


(a + b)3 = a3 + 3a2b + 3ab2 + b3


∴ 


⇒ 


⇒ 


⇒ 



Question 22.

Simplify:

(i) (2x + y + 4z)(4x2 + y2 + 16z2 – 2xy – 4yz – 8zx)

(ii) (x – 3y – 5z)(x2 + 9y2 + 25z2 + 3xy – 15yz + 5zx)


Answer:

(i) (2x + y + 4z) (4x2 + y2 + 16z2 – 2xy – 4yz – 8zx)


⇒ Here the identity used is


a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)


Comparing this with given condition


(2x + y + 4z) (4x2 + y2 + 16z2 – 2xy – 4yz – 8zx) = (2x)3 + y3 + (4z)3 – 3 × 2x × y × 4z


⇒ 8x3 + y3 + 64z3 – 24xyz


(ii) (x – 3y – 5z)(x2 + 9y2 + 25z2 + 3xy – 15yx + 5zx)


⇒ Here the identity used is


a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)


Comparing this with given condition


(x – 3y – 5z)(x2 + 9y2 + 25z2 + 3xy – 15yx + 5zx)


= x3 + (– 3y)3 + (– 5z)3 – 3(x) (– 3y)(– 5z)


= x3 – 27y3 – 125z3 – 45xyz



Question 23.

Evaluate using identities :

(i) 63 – 93 + 33

(ii) 163 – 63 – 103


Answer:

(i) 63 – 93 + 33


Here the identity used will be


a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)


∴ 63 – 93 + 33


⇒ (6 + (– 9) + 3)(62 + (– 9)2 + 32 – 6 × – 9 – (– 9)3 – 6 × 3) + 3 × 6 × – 9 × 3


⇒ 0 (36 + 81 + 9 + 54 + 27 – 18) – 486


⇒ – 486


(ii) 163 – 63 – 103


Here the identity used will be


a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)


∴ 163 – 63 – 103


= (16 – 6 – 10) (162 + (– 6)2 + (– 10)2 – 16 × – 6 – (– 6)(– 10) – (16)(– 10)) + 3(16) (– 6) (– 10)


= 0 (256 + 36 + 100 + 96 – 60 + 160) + 2880


= 2880




Exercise 1.2
Question 1.

Factorize the following expressions:

2a3 – 3a2b + 2a2c


Answer:


= 2a3 – 3a2b + 2a2c
Highest common factor is a2

Taking a2 common in the term – 3a2b + 2a2c


we get,
= 2a3 + a2(– 3b + 2c)
Taking a2 common in both the terms
we get,


= a2(2a + (– 3b + 2c))
= a2(2a – 3a + 2c)
2a3 – 3a2b + 2a2c = a2(2a – 3a + 2c)



Question 2.

Factorize the following expressions:

16x + 64x2y


Answer:


= 16x + 64x2y

Highest common factor is 16x


Taking x common in the term 16x + 64x2y


we get,


= x(16 + 64xy)


= x(16 + 16(4)xy)


Taking 16 common in the term (16 + 64xy)


we get,


= 16x(1 + 4xy)


16x + 64x2y = 16x(1 + 4xy)



Question 3.

Factorize the following expressions:

10x3 – 25x4y


Answer:


= 10x3 – 25x4y
Highest common factor is 5x3

Taking x3common in the term 10x3 – 25x4y


we get,


= x3 (10– 25xy)


= x3 (5(2)– 5(2)xy)
Taking 5 common in both the terms


we get,


= 5x3 (2– 5xy)
10x3 – 25x4y = 5x3 (2– 5xy)



Question 4.

Factorize the following expressions:

xy – xz + ay – az


Answer:


= xy – xz + ay – az
Taking x common in xy – xz

we get,


= x(y – z) + ay – az


Taking ‘a’ common in ay – az


we get,


= x(y – z) + a(y – z)
Taking (y – z) common in both the terms


we get,
= (y – z)(x + a)
xy – xz + ay – az = (y – z)(x + a)



Question 5.

Factorize the following expressions:

p2 + pq + pr + qr


Answer:

= p2 + pq + pr + qr
Taking p common in p2 + pq


we get,


= p(p + q) + pr + qr
Taking r common in pr + qr
we get,
= p(p + q) + r(p + q)


Taking (p + q) common in both the terms


we get,
= (p + q)(p + r)
p2 + pq + pr + qr = (p + q)(p + r)



Question 6.

Factorize the following expressions:

x2 + 2x + 1


Answer:


= x2 + 2x + 1

Method 1


It is of the form a2 + 2ab + b2
Where a = x and b = 1.


Using the identity: (a + b)2 = a2 + 2ab + b2


We get,


⇒ x2 + 2x + 1 = (x + 1)2


Method 2
Here 2x can be elaborated as x + x


we get,


= x2 + x + x + 1


Taking x common in x2 + x


we get,


= x(x + 1) + x + 1
Taking (x + 1) common in both the terms


we get,
= (x + 1)(x + 1)
= (x + 1)2
x2 + 2x + 1 = (x + 1)2



Question 7.

Factorize the following expressions:

9x2 – 24xy + 16y2


Answer:


= 9x2 – 24xy + 16y2
Given equation can be simplified as,

= (3x)2 – 2(3x)(4y) + (4y)2


It is of the form a2 – 2ab + b2
Where a = 3x and b = 4y
∴ using the identity (a – b)2 = a2 – 2ab + b2


We get,
⇒ 9x2 – 24xy + 16y2 = (3x – 4y)2


9x2 – 24xy + 16y2 = (3x – 4y)2



Question 8.

Factorize the following expressions:

b2 – 4


Answer:

= b2 – 4
Given equation can be simplified as,


= (b)2 – (2)2


It is of the form (p)2 – (q)2
where p = b and q = 2
∴ using the identity (p)2 – (q)2 = (p + q)(p – q)


We get,
⇒ b2 – 22 = (b + 2)(b – 2)
b2 – 4 = (b + 2)(b – 2)



Question 9.

Factorize the following expressions:

1 – 36x2


Answer:


= 1 – 36x2
Given equation can be simplified as,

= (1)2 – (6x)2


It is of the form (p)2 – (q)2
Where p = 1 and q = 6x
∴ using the identity: (p)2 – (q)2 = (p + q)(p – q)


We get,
⇒ (1)2 – (6x)2 = (1 + 6x)(1 – 6x)


1 – 36x2 = (1 + 6x)(1 – 6x)



Question 10.

Factorize the following expressions:

p2 + q2 + r2 + 2pq + 2qr + 2rp


Answer:


Method 1
= p2 + q2 + r2 + 2pq + 2qr + 2rp
= (p)2 + (q)2 + (r)2 + 2(p)(q) + 2(q)(r) + 2(r)(p)
It is of the from a2 + b2 + c2 + 2ab + 2bc + 2ca

where a = p, b = q, c = r


Using the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca


We get,
⇒ p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2


Method 2


= p2 + q2 + r2 + 2pq + 2qr + 2rp
The above equation can be simplified as


= p2 + 2pq + q2 + r2 + 2qr + 2rp
= (p + q)2 + r2 + 2qr + 2rp (∵ a2 + 2ab + b2 = (a + b)2)


Taking 2r common in the term 2qr + 2rp


we get,


= (p + q)2 + r2 + 2r(q + p)
Rearranging the above equation as
= (p + q)2 + 2r(q + p) + r2


This is of the from a2 + 2ab + b2


Where, a = p + q and b = r
Using the identity: (a + b)2 = a2 + 2ab + b2


We get,


⇒ p2 + q2 + r2 + 2pq + 2qr + 2rp = ((p + q) + r)2


= (p + q + r)2
p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2



Question 11.

Factorize the following expressions:

a2 + 4b2 + 36 – 4ab + 24b – 12a


Answer:

= a2 + 4b2 + 36 – 4ab + 24b – 12a


Method 1


The above equation can be simplified as : 
= (– a)2 + (2b)2 + (6)2 + 2(– a)(2b) + 2(2b)(6) + 2(– a)(6)
This equation is of the form: p2 + q2 + r2 + 2pq + 2qr + 2rp


where p = – a, q = 2b, r = 6
Using the identity: p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2
We get,
⇒ a2 + 4b2 + 36 – 4ab + 24b – 12a = (– a + 2b + 6)2


= (– a + 2b + 6)2


= (– 1)2(a – 2b – 6)2


= (a – 2b – 6)2


Method 2


The above equation can be simplified as


= (– a)2 + (2b)2 + (6)2 + 2(– a)(2b) + 2(2b)(6) + 2(– a)(6)
= {(– a)2 + 2(– a)(2b) + (2b)2} + (6)2 + 2(2b)(6) + 2(– a)(6)
(∵ a2 + 2ab + b2 = (a + b)2)
(– a + 2b)2 + (6)2 + 2(2b)(6) + 2(– a)(6)
Taking 2(6) common in term 2(2b)(6) + 2(– a)(6)
= (– a + 2b)2 + 2(6)(– a + 2b) + (6)2


This of the form: (p + q)2 + 2r(p + q) + r2


Where, p = – a, q = 2b and r =6


Using the identity: (p + q)2 = p2 + 2pq + q2


We get,


⇒ a2 + 4b2 + 36 – 4ab + 24b – 12a = (– a + 2b + 6)2


= (– a + 2b + 6)2


= (– 1)2(a – 2b – 6)2


= (a – 2b – 6)2


a2 + 4b2 + 36 – 4ab + 24b – 12a = (– a + 2b + 6)2




Question 12.

Factorize the following expressions:

9x2 + y2 + 1 – 6xy + 6x – 2y


Answer:

Method 1


The above equation can be simplified as : 
= (3x)2 + (– y)2 + (1)2 + 2(3x)(– y) + 2(3x)(1) + 2(– y)(1)
This equation is of the form: p2 + q2 + r2 + 2pq + 2qr + 2rp


where p =3x, q = – y, r = 1
Using the identity: p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2
We get,
⇒ 9x2 + y2 + 1 – 6xy + 6x – 2y = (3x – y + 1)2


= (3x – y + 1)2


Method 2


The above equation can be simplified as:


= (3x)2 + (– y)2 + (1)2 + 2(3x)(– y) + 2(3x)(1) + 2(– y)(1)
= {(3x)2 + 2(3x)(– y) + (– y)2} + (1)2 + 2(3x)(1) + 2(– y)(1)
(∵ p2 + 2pq + q2 = (p + q)2)
= (3x – y)2 + (1)2 + 2(3x)(1) + 2(– y)(1)
Taking 2(1) common in term 2(3x)(1) + 2(– y)(1)
= (3x – y)2 + (1)2 + 2(1)(3x – y)
= (3x – y)2 + 2(1)(3x – y) + (1)2


This of the form: (p + q)2 + 2r(p + q) + r2


Where, p = 3x, q = – y and r = 1


Using the identity: (a + b)2 = a2 + 2ab + b2


We get,


⇒ 9x2 + y2 + 1 – 6xy + 6x – 2y = (3x – y + 1)2
9x2 + y2 + 1 – 6xy + 6x – 2y = (3x – y + 1)2



Question 13.

Factorize the following expressions:

4a2 + b2 + 9c2 – 4ab – 6bc + 12ca


Answer:


Method 1

The above equation can be simplified as : 
= (2a)2 + (– b)2 + (3c)2 + 2(2a)(– b) + 2(– b)(3c) + 2(3c)(2a)
This equation is of the form: p2 + q2 + r2 + 2pq + 2qr + 2rp


where p =2a, q = – b, r = 3c
Using the identity: p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2
We get,
⇒ 4a2 + b2 + 9c2 – 4ab – 6bc + 12ca = (2a – b + 3c)2


= (2a – b + 3c)2


Method 2


The above equation can be simplified as:


= (2a)2 + (– b)2 + (3c)2 + 2(2a)(– b) + 2(– b)(3c) + 2(3c)(2a)
= {(2a)2 + 2(2a)(– b) + (– b)2} + (3c)2 + 2(– b)(3c) + 2(3c)(2a)
(∵ p2 + 2pq + q2 = (p + q)2)


= (2a – b)2 + (3c)2 + 2(– b)(3c) + 2(3c)(2a)
Taking 2(3c) common in term 2(– b)(3c) + 2(3c)(2a)
= (2a – b)2 + (3c)2 + 2(3c)(– b + 2a)
= (2a – b)2 + 2(3c)(2a – b) + (3c)2


This of the form: (p + q)2 + 2r(p + q) + r2


Where, p = 2a, q = – b and r = 3c


Using the identity: (p + q)2 = p2 + 2pq + q2


We get,


⇒ 4a2 + b2 + 9c2 – 4ab – 6bc + 12ca = (2a – b + 3c)2
4a2 + b2 + 9c2 – 4ab – 6bc + 12ca = (2a – b + 3c)2




Question 14.

Factorize the following expressions:

25x2 + 4y2 + 9z2 – 20xy + 12yz – 30zx


Answer:


Method 1

The above equation can be simplified as : 
= (– 5x)2 + (2y)2 + (3z)2 + 2(– 5x)(2y) + 2(2y)(3z) + 2(3z)(– 5x)
This equation is of the form: p2 + q2 + r2 + 2pq + 2qr + 2rp


where p = – 5x, q = 2y, r = 3z
Using the identity: p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2
We get,
⇒ 25x2 + 4y2 + 9z2 – 20xy + 12yz – 30zx = (– 5x + 2y + 3z)2


= (– 5x + 2y + 3z)2


= (– 1)2(5x – 2y – 3z)2


= (5x – 2y – 3z)2


Method 2


The above equation can be simplified as:


= (– 5x)2 + (2y)2 + (3z)2 + 2(– 5x)(2y) + 2(2y)(3z) + 2(3z)(– 5x)
= {(– 5x)2 + 2(– 5x)(2y) + (2y)2} + (3z)2 + 2(2y)(3z) + 2(3z)(– 5x)
(∵ p2 + 2pq + q2 = (p + q)2)
= (– 5x + 2y)2 + (3z)2 + 2(2y)(3z) + 2(3z)(– 5x)
Taking 2(3z) common in term 2(2y)(3z) + 2(3z)(– 5x)
= (– 5x + 2y)2 + (3z)2 + 2(3z)(2y – 5x)
= (– 5x + 2y)2 + 2(3z)(2y – 5x) + (3z)2


This of the form: (p + q)2 + 2r(p + q) + r2


Where, p = – 5x, q = 2y and r = 3z


Using the identity: (p + q)2 = p2 + 2pq + q2


We get,


⇒ 25x2 + 4y2 + 9z2 – 20xy + 12yz – 30zx = (– 5x + 2y + 3z)2


= (– 5x + 2y + 3z)2


= (– 1)2(5x – 2y – 3z)2


= (5x – 2y – 3z)2


25x2 + 4y2 + 9z2 – 20xy + 12yz – 30zx = (5x – 2y – 3z)2



Question 15.

Factorize the following expressions:
27x3 + 64y3


Answer:

The above equation can be simplified as


= (3x)3 + (4y)3
Both the given terms are perfect cubes, factor using the sum of cubes formula,


(a)3 + (b)3 = (a + b)(a2 – ab + b2)


Here a = 3x and b = 4y


⇒ (3x)3 + (4y)3 = (3x + 4y)((3x)2 – (3x)(4y) + (4y)2)


= (3x + 4y)(9x2 – 12xy + 16y2)
27x3 + 64y3 = (3x + 4y)(9x2 – 12xy + 16y2)



Question 16.

Factorize the following expressions:
m3 + 8


Answer:

The above equation can be simplified as


= (m)3 + (2)3
Both the given terms are perfect cubes, factor using the sum of cubes formula,


(a)3 + (b)3 = (a + b)(a2 – ab + b2)
Here a = m and b = 2


⇒ (m)3 + (2)3 = (m + 2)((m)2 – (m)(2) + (2)2)


= (m + 2)(m2 – 2m + 4)
m3 + 8 = (m + 2)(m2 – 2m + 4)



Question 17.

Factorize the following expressions:
a3 + 125


Answer:

The above equation can be simplified as


= (a)3 + (5)3
Both the given terms are perfect cubes, factor using the sum of cubes formula,


(a)3 + (b)3 = (a + b)(a2 – ab + b2)
Here a = a and b = 5


⇒ (a)3 + (5)3 = (a + 5)((a)2 – (a)(5) + (5)2)


= (a + 5)(a2 – 5a + 25)
a3 + 123 = (a + 5)(a2 – 5a + 25)



Question 18.

Factorize the following expressions:
8x3 – 27y3


Answer:

The above equation can be simplified as


= (2x)3 – (3y)3
Both the given terms are perfect cubes, factor using the subtraction of cubes formula,


(a)3 – (b)3 = (a – b)(a2 + ab + b2)
Here a = 2x and b = 3y


⇒ (2x)3 – (3y)3 = (2x – 3y)((2x)2 + (2x)(3y) + (3y)2)


= (2x – 3y)(4x2 + 6xy + 9y2)
8x3 – 27y3 = (2x – 3y)(4x2 + 6xy + 9y2)



Question 19.

Factorize the following expressions:
x3 – 8y3


Answer:

The above equation can be simplified as


= (x)3 – (2y)3
Both the given terms are perfect cubes, factor using the subtraction of cubes formula,


(a)3 – (b)3 = (a – b)(a2 + ab + b2)
Here a = x and b = 2y


⇒ (x)3 – (2y)3 = (x – 2y)((x)2 + (x)(2y) + (2y)2)


= (x – y)(x2 + 2xy + 4y2)
x3 – 8y3 = (x – y)(x2 + 2xy + 4y2)





Exercise 1.4
Question 1.

Solve the following equations by substitution method.

x + 3y = 10; 2x + y = 5


Answer:

[In substitution method, we have to find value of one variable in terms of the other variable from any one of the two given equations . Then, we will substitute this value in the other equation. By doing this, we will get a equation in one variable which can be solved easily]


x + 3y = 10……(1)


2x + y = 5 …..(2)


From eq(1),


X + 3y = 10


⇒ x = 10 – 3y


Substituting the value of x in eq(2)


2x + y = 5


⇒ 2(10 – 3y) + y = 5


⇒ 20 – 6y + y = 5


⇒ 20 – 5y = 5


⇒ 5y = 20 – 5


⇒ 5y = 15




Substituting value of y in eq(1)


x + 3y = 10


⇒ x + 3×3 = 10


⇒ x + 9 = 10


⇒ x = 10 – 9


⇒ x = 1


Hence,


x = 1, y = 3




Question 2.

Solve the following equations by substitution method.

2x + y = 1; 3x – 4y = 18


Answer:

[In substitution method, we have to find value of one variable in terms of the other variable from any one of the two given equations . Then, we will substitute this value in the other equation. By doing this, we will get a equation in one variable which can be solved easily]


2x + y = 1….(1)


3x – 4y = 18….(2)


From eq(1),


2x + y = 1….(1)


⇒ y = 1 – 2x


Substituting the value of y in eq(2)


3x – 4y = 18….(2)


⇒ 3x – 4(1 – 2x) = 18


⇒ 3x – 4 + 8x = 18


⇒ 11x + 4 = 18


⇒ 11x = 18 + 4


⇒ 11x = 22


⟹x = 


⇒ x = 2


Substituting the value of x in eq(1),


⇒ 2x + y = 1


⇒ 2×2 + y = 1


⇒ y = 1 – 4


⇒ y = – 3


Hence,


x = 2,y = – 3



Question 3.

Solve the following equations by substitution method.

5x + 3y = 21; 2x – y = 4


Answer:

[In substitution method, we have to find value of one variable in terms of the other variable from any one of the two given equations . Then, we will substitute this value in the other equation. By doing this, we will get a equation in one variable which can be solved easily]


5x + 3y = 21……(1)


2x – y = 4…….(2)


From eq(2)


2x – y = 4


⟹y = 2x – 4…..(3)


Substituting value of y in eq(1)


5x + 3y = 21


⇒ 5x + 3(2x – 4) = 21


⇒ 5x + 6x – 12 = 21


⇒ 11x = 33


⇒ x = 


⇒ x = 3


From eq(3)


y = 2x – 4


⇒ y = 2×3 – 4


⇒ y = 2


Hence,


x = 3, y = 2



Question 4.

Solve the following equations by substitution method.



Answer:

In such questions, if we start question directly, then it will involve LCM of x and y, and a term of xy, which will be inconvenient for us. So, we will assume 1/x as “u “and 1/y as v”]


Let


 = u


 = v


Then,


 = 9


⇒ u + 2v = 9……(1)


And


 = 12


⇒ 2u + v = 12……(2)


From eq(1)


u + 2v = 9


⇒ u = 9 – 2v……(3)


Substituting value of u in eq(2)


2u + v = 12


⇒ 2(9 – 2v) + v = 12


⟹18 – 4v + v = 12


⇒ 18 – 3v = 12


⇒ 3v = 18 – 12


⇒ 3v = 6


⇒ v = 2


⇒  = 2



Substituting value of v in eq(3)


u = 9 – 2v


⇒ u = 9 – 2×2


⇒ u = 5


⇒  = 5


⇒ 


Hence,


x = ,y = 



Question 5.

Solve the following equations by substitution method.



Answer:

In such questions , if we start question directly, then it will involve LCM of x and y, and a term of xy, which will be inconvenient for us. So, we will assume 1/x as “u “and 1/y as v”]


 +  = 7


⇒ 3u + v = 7……(1)


 = 6


⇒ 5u – 4v = 6 ……..(2)


From eq(1)


v = 7 – 3u…..(3)


Substituting the value of v in eq(2)


5u – 4(7 – 3u) = 6


⇒ 5u – 28 + 12u = 6


⇒ 17u = 6 + 28


⇒ 17u = 34


⇒ u = 34/17


⇒ u = 2


⇒  = 2



Substituting the value of u in eq(3)


v = 7 – 3×2


⇒ v = 1


⇒  = 1


⇒ y = 1


Hence, x = , y = 1



Question 6.

Find two numbers whose sum is 24 and difference is 8.


Answer:

Let the two numbers be x and y.


Since, it is given that their sum is 24,


We can write, x + y = 24


For forming second equation, we will use their difference, which is 8.


So, we can write,


x – y = 8


(we can assume any number to be greater. Here, we assumed the greater number is x)


Here, coefficient of y is equal in magnitude and opposite in sign.


So, if we add the above two equations, it will be eliminated.


Adding equation (1)and(2)


We get


2x = 32


⇒ x = 


⇒ x = 16


Substituting the value of x in eq(1)


We get


x + y = 24


⇒ 16 + y = 24


⇒ y = 24 – 16


⇒ y = 8


Hence, the two required numbers are 16 and 8.



Question 7.

A number consists of two digits whose sum is 9. The number formed by reversing the digits exceeds twice the original number by 18. Find the original number.


Answer:

Let unit digit of the number be x and its ten’s digit be y.


Then the number will be 10×y + x = 10y + x


Sum of its digits is 9.


Therefore, the sum of x and y is 9


x + y = 9……(1)


On reversing the number, ten’s digit will become unit’s digit and unit’s digit will become ten’s digit.


So, after reversing the number,


Its ten’s digit = x and unit’s digit = y


Therefore, the new number formed = 10x + y


New number formed exceeds the twice the original number by 18


⇒ 10x + y = 2(10y + x) + 18


⇒ 10x + y = 20y + 2x + 18


⇒ 8x – 19y = 18…..(2)


From equation(1),


x + y = 9


⇒ x = 9 – y…..(3)


Substituting the value of x in eq(2)


8(9 – y) – 19y = 18


⇒ 72 – 8y – 19y = 18


⇒ – 27y = 18 – 72


⇒ – 27y = – 54


⇒ y = 2


Substituting the value of y in eq(3)


x = 9 – 2


⇒ x = 7


Hence, the original number is 10y + x = 10×2 + 7 = 27



Question 8.

Kavi and Kural each had a number of apples. Kavi said to Kural “If you give me 4 of your apples, my number will be thrice yours”. Kural replied, “If you give me 26, my number will be twice yours”. How many did each have with them?.


Answer:

Let Kavi has x apples and Kural has y apples.


According to Kavi’s statement,


We can write


x + 4 = 3(y – 4)


x + 4 = 3y – 12


⇒ x = 3y – 16…..(1)


According to Kural’s statement


y + 26 = 2(x – 26)


y + 26 = 2x – 52…….(2)


Substituting the value of x from eq(1) in eq(2)


y + 26 = 2(3y – 16) – 52


⇒ y + 26 = 6y – 32 – 52


⇒ y + 26 = 6y – 84


⇒ 5y = 110


⇒ y = 22


Substituting value of y in eq(1)


x = 3×22 – 16


⇒ x = 50


Hence, Kavi has 50 apples and Kural has 22 apples.



Question 9.

Solve the following inequations.

(i) 2x + 7>15

(ii) 2(x – 2) < 3

(iii) 2(x + 7) ≤ 9

(iv) 3x + 14 ≥ 8


Answer:

(i) 2x + 7>15


2x>15 – 7


⇒ 2x>8


⇒ x>


⇒ x>4


(ii) 2x – 4<3


⇒ 2x>3 + 4


⇒ 2x>7


⇒ x>7/2


(iii) 2(x + 7)≤9


⇒ 2x + 14≤9


⇒ 2x≤9 – 14


⇒ 2x≤ – 5



(iv) 3x + 14≥8


⇒ 3x≥8 – 14


⇒ 3x≥ – 6


⇒ x≥


⇒ x≥ – 2




Exercise 1.5
Question 1.

The expansion of (x + 2)(x – 1) is
A. x2 – x – 2

B. x2 + x + 2

C. x2 + x – 2

D. x2 – x + 2


Answer:

Using identity,


(x + a)(x + b) = x2 + (a + b)x + ab


We have,


(x + 2)(x – 1)


= x2 + (2 – 1) x + (2x – 1) = x2 + x – 2


Question 2.

The expansion of (x + 1)(x – 2)(x + 3) is
A. x3 + 2x2 – 5x – 6

B. x3 – 2x2 + 5x – 6

C. x3 + 2x2 + 5x – 6

D. x3 + 2x2 + 5x + 6


Answer:

Using identity,


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


We have,


(x + 1)(x – 2)(x + 3) = x3 + (1 – 2 + 3)x2 + (– 2 – 6 + 3)x – 6


= x3 + 2x2 – 5x – 6


Question 3.

(x – y)(x2 + xy + y2) is equal to
A. x3 + y3

B. x2 + y2

C. x2 – y2

D. x3 – y3


Answer:

(x – y)(x2 + xy + y2) is simply the expansion of x3 – y3.


we can also obtain this result by multiplying them


i.e. (x – y)(x2 + xy + y2) = x(x2 + xy + y2) – y(x2 + xy + y2)


= x3 + x2y + xy2 – x2y – xy2 – y3


= x3 – y3


Question 4.

Factorization of x2 + 2x – 8 is
A. (x + 4)(x – 2)

B. (x – 4)(x + 2)

C. (x + 4)(x + 2)

D. (x – 4)(x – 2)


Answer:

Suppose (x + p)(x + q) are two factors of x2 + 2x – 8.


Then, x2 + 2x – 8 = (x + p)(x + q)


= x2 + (p + q)x + pq


So, to factorize we have to find p and q, such that pq = – 8 and p + q = 2.



∴ x2 + 2x – 8 = (x + 4)(x – 2)


Question 5.

If one of the factors of x2 – 6x – 16 is (x + 2) then other factor is
A. x + 5

B. x – 5

C. x + 8

D. x – 8


Answer:

Let the other factor be (x + p)


Then, x2 – 6x – 16 = (x + p)(x + 2)


= x2 + (p + 2)x + 2p


On comparing the coefficients on both sides, we get,


p + 2 = – 6


⇒ p = – 8


∴ The other factor is (x – 8).


Question 6.

If (2x + 1) and (x – 3) are the factors of ax2 – 5x + c, then the values of a and c are respectively
A. 2, 3

B. – 2, 3

C. 2, – 3

D. 1, – 3


Answer:

ax2 – 5x + c = (2x + 1)(x – 3)


= 2x2 – 6x + x – 3


= 2x2 – 5x – 3


On comparing the coefficients on both the sides, we get,


a = 2 and c = – 3


Question 7.

If x + y = 10 and x – y = 2, then value of x is
A. 4

B. – 6

C. – 4

D. 6


Answer:

x + y = 10


⇒ y = 10 – x substituting this value in x – y = 2


We get,


x – (10 – x) = 2


⇒ 2x – 10 = 2


⇒ 2x = 12


⇒ x = 6


Question 8.

The solution of 2 – x <5 is
A. x > – 3

B. x < – 3

C. x > 3

D. x < 3


Answer:

2 – x <5


Adding x on both the sides,


2 <5 + x


⇒ – 3 < x


⇒ x > – 3


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