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RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3

Other Exercises

• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.2
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.4
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.6
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.7
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.9

Question 1.
Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication :
(i) 25
(ii) 37
(iii) 54
(iv) 71
(v) 96
Solution:
(i) (25)²

Question 2.
Find the squares of the following numbers using diagonal method :
(i) 98
(ii) 273
(iii) 348
(iv) 295
(v) 171
Solution:

Question 3.
Find the squares of the following numbers :
(i) 127
(ii) 503
(iii) 451
(iv) 862
(v) 265
Solution:
(i) (127)² = (120 + 7)²
{(a + b)² = a² + lab + b²}
= (120)² + 2 x 120 x 7 + (7)²
= 14400+ 1680 + 49 = 16129

(ii) (503)² = (500 + 3)²
{(a + b)² = a² + lab + b¹}
= (500)² + 2 x 500 x 3 + (3)²
= 250000 + 3000 + 9 = 353009

(iii) (451)² = (400 + 51)²
{(a + b)² = a² + lab + b²}
= (400)² + 2 x 400 x 51 + (5l)²
= 160000 + 40800 + 2601 = 203401

(iv) (451)² = (800 + 62)²
{(a + b)² = a² + lab + b²}
= (800)² + 2 x 800 x 62 + (62)²
= 640000 + 99200 + 3844 = 743044

(v) (265)²
{(a + b)² = a² + 2ab + b²}
(200 + 65)² = (200)² + 2 x 200 x 65 + (65)²
= 40000 + 26000 + 4225 = 70225

Question 4.
Find the squares of the following numbers
(i) 425
(ii) 575
(iii) 405
(iv) 205
(v) 95
(vi) 745
(vii) 512
(viii) 995
Solution:
(i) (425)²
Here n = 42
∴ n (n + 1) = 42 (42 + 1) = 42 x 43 = 1806
∴ (425)² = 180625

(ii) (575)²
Here n = 57
∴ n (n + 1) = 57 (57 + 1) = 57 x 58 = 3306
∴ (575)² = 330625

(iii) (405)²
Here n = 40
∴ n (n + 1) = 40 (40 + 1) -40 x 41 = 1640
∴ (405)² = 164025

(iv) (205)²
Here n = 20
∴ n (n + 1) = 20 (20 + 1) = 20 x 21 = 420
∴ (205)² = 42025

(v) (95)²
Here n = 9
∴ n (n + 1) = 9 (9 + 1) = 9 x 10 = 90
∴ (95)² = 9025

(vi) (745)²
Here n = 74
∴ n (n + 1) = 74 (74 + 1) = 74 x 75 = 5550
∴ (745)² = 555025

(vii) (512)²
Here a = 1, b = 2
∴ (5ab)² = (250 + ab) x 1000 + (ab)²
∴ (512)² = (250 + 12) x 1000 + (12)²
= 262 x 1000 + 144
= 262000 + 144 = 262144

(viii) (995)²
Here n = 99
∴ n (n + 1) = 99 (99 + 1) = 99 x 100 = 9900
∴ (995)² = 990025

Question 5.
Find the squares of the following numbers using the identity (a + b)1 = a2 + lab + b1
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605
Solution:
a + b)² = a² + lab + b²

(i) (405)² = (400 + 5)²
= (400)² + 2 x 400 x 5 + (5)²
= 160000 + 4000 + 25 = 164025

(ii) (510)² = (500 + 10)²
= (500)² + 2 x 500 x 10 x (10)²
= 250000 + 10000 + 100
= 260100

(iii) (1001)² = (1000+1)²
= (1000)² + 2 X 1000 x 1 + (1)
= 1000000 + 2000 + 1
=1002001

(iv) (209)² = (200 + 9)²
= (200)² + 2 x 200 x 9 x (9)²
= 40000 + 3600 +81
= 43681

(v) (605)² = (600 + 5)²
= (600)² + 2 x 600 x 5 +(5)²
= 360000 + 6000       25
=366025

Question 6.
Find the squares of the following numbers using the identity (a – b)² = a² – 2ab + b² :
(i) 395
(ii) 995
(iii) 495
(iv) 498
(v) 99
(vi) 999
(vii) 599
Solution:
a – b)² = a² – lab + b²

(i) (395)² = (400 – 5)²
= (400)² – 2 x 400 x 5 + (5)²
= 160000-4000 + 25
= 160025-4000
= 156025

(ii) (995)² = (1000 – 5)²
= (1000)² – 2 x 1000 x 5 + (5)²
= 1000000- 10000 + 25
= 1000025- 10000
= 990025

(iii) (495)² = (500 – 5)²
= (500)² – 2 x 500 x 5 + (5)²
= 250000 – 5000 + 25
= 250025 – 5000
= 245025

(iv) (498)² = (500 – 2)²
= (500)² – 2 x 500 x 2 + (2)²
= 250000 – 2000 + 4
= 250004 – 2000
= 248004

(v) (99)² = (100 – l)²
= (100)² – 2 x 100 x 1 + (1)²
= 10000 – 200 + 1
= 10001 – 200
= 9801

(vi) (999)² = (1000- l)²
= (1000)² – 2 x 1000 x 1+ (1)2
= 1000000-2000+1
= 10000001-2000=998001

(vii) (599)² = (600 – 1)²
= (600)² -2 x 600 X 1+ (1)2
= 360000 -1200+1
= 360001 – 1200 = 358801

Question 7.
Find the squares of the following numbers by visual method :
(i) 52
(ii) 95
(iii) 505
(iv) 702
(v) 99
Solution:
(a + b)² = a² – ab + ab + b²
(i) (52)² = (50 + 2)²
= 2500 + 100 + 100 + 4
= 2704

(ii) (95)² = (90 + 5)²
= 8100 + 450 + 450 + 25
= 9025

(iii) (505)² = (500 + 5)²
= 250000 + 2500 + 2500 + 25
= 255025

(iv) (702)² = (700 + 2)²
= 490000 + 1400+ 1400 + 4
= 492804

(v) (99)² = (90 + 9)²
= 8100 + 810 + 810 + 81
= 9801

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 are helpful to complete your math homework.

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

RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

Other Exercises

• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.3
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.4
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials VSAQS
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS

Question 1.
If f(x) = 2x³ – 13x² + 17x + 12, find
(i) f (2)
(ii) f (-3)
(iii) f(0)
Solution:
f(x) = 2x³ – 13x² + 17x + 12
(i) f(2) = 2(2)³ – 13(2)² + 17(2) + 12
= 2 x 8-13 x 4+17 x 2+12
= 16-52 + 34 + 12
= 62 – 52
= 10
(ii) f(-3) = 2(-3)³ – 13(-3)² + 17 x (-3) + 12
= 2 x (-27) – 13 x 9 + 17 x (-3) + 12
= -54 – 117 -51 + 12
= -222 + 12
= -210
(iii) f(0) = 2 x (0)³ – 13(0)² + 17 x 0 + 12
= 0-0 + 0+ 12 = 12

Question 2.
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases: [NCERT]

Solution:

Question 3.
If x = 2 is a root of the polynomial f(x) = 2x²-3x + la, find the value of a.
Solution:
p(x) = 2x² – 3x + 7a
∵ x = 2 is its zero, then
p(0) = 0
∴ p( 2) = 2(2)² – 3×2 + la = 0
⇒2 x 4-3 x2 + 7a = 0
⇒ 8 – 6 + 7o = 0
⇒2 + 7a = 0
⇒ 7a = -2 ⇒ a =\(\frac { -2 }{ 7 }\)
∴ Hence a = \(\frac { -2 }{ 7 }\)

Question 4.
If x = –\(\frac { 1 }{ 2 }\) is a zero of the polynomial p(x) = 8x³ – ax² – x + 2, find the value of a.
Solution:

Question 5.
If x = 0 and x = -1 are the roots of the polynomial f(x) = 2x³ – 3x² + ax + b, find the value of a and b.
Solution:
f(x) = 2x³ – 3x² + ax + b
∵ x = 0 and x = -1 are its zeros
∴ f(0) = 0 and f(-1) = 0
Now, f(0) = 0
⇒  2(0)³ – 3(0)² + a x 0 + b = 0
⇒ 0-0 + 0 + b= 0
∴ b = 0
and f(-1) = 0
⇒ 2(-1)³ – 3(-1)² + a(-1) + b = 0
⇒  2 x (-1) – 3 x 1 + a x (-1) + b = 0
⇒ -2 -3-a + b = 0
⇒ -2-3-a + 0 = 0
⇒ -5- a = 0=>a =-5
Hence a = -5, b = 0

Question 6.
Find the integral roots of the polynomial f(x) = x³ + 6x² + 11x + 6.
Solution:
f(x) = x³ + 6x² + 11x + 6
Construct = 6 = ±1, ±2, +3, ±6
If x = 1, then
f(1) = (1)³ + 6(1)² + 11 x 1 + 6
= 1+ 6+11+ 6 = 24
∵  f(x) ≠ 0, +0
∴ x = 1 is not its zero
Similarly, f(-1) = (-1)³ + 6(-1)² + 11(-1) + 6
= -1 + 6 x 1-11+6
=-1+6-11+6
= 12-12 = 0
∴  x = -1 is its zero
f(-2) = (-2)³ + 6(-2)² + 11 (-2) + 6
= -8 + 24 – 22 + 6
= -30 + 30 = 0
∴ x = -2 is its zero
f(-3) = (-3)³ + 6(-3)² + 11 (-3) + 6
= -27 + 54 – 33 + 6 = 60 – 60 = 0
∴  x = -3 is its zero
x = -1, -2, -3 are zeros of f(x)
Hence roots of f(x) are -1, -2, -3

Question 7.
Find the rational roots of the polynomial f(x) = 2x³ + x² – 7x – 6.
Solution:

Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 are helpful to complete your math homework.

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