Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 ਬੀਜਗਣਿਤਕ ਵਿਅੰਜਕ ਅਤੇ ਤਤਸਮਕ Ex 9.5 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 8 Maths Chapter 9 ਬੀਜਗਣਿਤਕ ਵਿਅੰਜਕ ਅਤੇ ਤਤਸਮਕ Exercise 9.5
1. ਹੇਠਾਂ ਲਿਖੇ ਗੁਣਨਫਲਾਂ ਵਿਚੋਂ ਹਰੇਕ ਨੂੰ ਪ੍ਰਾਪਤ ਕਰਨ ਦੇ ਲਈ ਢੁੱਕਵੇਂ ਤਤਸਮਕ ਦਾ ਉਪਯੋਗ ਕਰੋ :
ਪ੍ਰਸ਼ਨ (i).
(x + 3) (x + 3)
ਹੱਲ:
(x + 3) (x + 3) = (x + 3)2
(a + b)2 = a2 + 2ab + b2
a = x, b = 3
∴ (x + 3)2 = (x)2 + 2(x) (3) + (3)2
= x2 + 6x + 9.
ਪ੍ਰਸ਼ਨ (ii).
(2y + 5) (2y + 5)
ਹੱਲ:
(2y + 5) (2y + 5) = (2y + 5)2
(a + b)2 = a2 + 2ab + b2
a = 2y, b = 5
∴ (2y + 5)2 = (2y)2 + 2 (2y) (5) + (5)2
= 4y2 + 20y + 25
ਪ੍ਰਸ਼ਨ (iii).
(2a – 7) (2a – 7)
ਹੱਲ:
(2a – 7) (2a – 7) = (2a – 7)2
(a – b)2 = a2 – 2ab + b2
a = 2a, b = 7
∴ (2a – 7)2 = (2a)2 – 2(2a) (7) + (7)2
= 4a2 – 28a + 49.
ਪ੍ਰਸ਼ਨ (iv).
(3a – \(\frac{1}{2}\))(3a – \(\frac{1}{2}\))
ਹੱਲ:
(3a – \(\frac{1}{2}\))(3a – \(\frac{1}{2}\)) = (3a – \(\frac{1}{2}\))2
(a – b)2 = a2 – 2ab + b2
a = 3a, b = \(\frac{1}{2}\)
∴ (3a – \(\frac{1}{2}\))2 = (3a)2 – 2(3a)(\(\frac{1}{2}\)) + (\(\frac{1}{2}\))2
= 9a2 – 3a + \(\frac{1}{4}\)
ਪ੍ਰਸ਼ਨ (v).
(1.1m – 0.4) (1.1m + 0.4)
ਹੱਲ:
(1.1m – 0.4) (1.1m + 0.4)
(a + b) (a – b) = a2 – b2
a = 1.1m, b = 0.4
∴ (1.1m – 0.4) (1.1m + 0.4) = (1.1m)2 – (0.4)2
= 1.21m2 – 0.16.
ਪ੍ਰਸ਼ਨ (vi).
(a2 + b2) (-a2 + b2)
ਹੱਲ:
(a2 + b2) (-a2 + b2) ⇒ (b2 + a2) (b2 – a2)
(a + b) (a – b) = a2 – b2
a = b2, b = a2
∴ (b2 + a2) (b2 – a2) = (b2)2 – (a2)2
= b4 – a4.
ਪ੍ਰਸ਼ਨ (vii).
(6x – 7) (6x + 7)
ਹੱਲ:
(6x – 7) (6x + 7)
(a – b) (a + b) = a2 – b2
a = 6x, b = 7
(6x – 7) (6x + 7) = (6x)2 – (7)2
= 36x2 – 49
ਪ੍ਰਸ਼ਨ (viii).
(-a + c) (-a + c)
ਹੱਲ:
(-a + c) (-a + c)
(c – a) (c – a) = (c – a)2
(a – b)2 = a2 – 2ab + b2
a = c, b = a
∴ (c – a)2 = (c)2 – 2 (c) (a) + (a)2.
= c2 – 2ac + a2.
ਪ੍ਰਸ਼ਨ (ix).
(\(\frac{x}{2}\) + \(\frac{3y}{4}\))(\(\frac{x}{2}\) = \(\frac{3y}{4}\))
ਹੱਲ:
(\(\frac{x}{2}\) + \(\frac{3y}{4}\))(\(\frac{x}{2}\) + \(\frac{3y}{4}\)) = (\(\frac{x}{2}\) + \(\frac{3y}{4}\))2
(a + b)2 = a2 + 2ab + b2
a = \(\frac{x}{2}\), b = \(\frac{3y}{2}\)
∴ (\(\frac{x}{2}\) + \(\frac{3y}{4}\))2 = (\(\frac{x}{2}\))2 + 2(\(\frac{x}{2}\))(\(\frac{3y}{2}\)) + (\(\frac{3y}{2}\))2
= \(\frac{x^{2}}{4}\) + \(\frac{3xy}{2}\) + \(\frac{9y^{2}}{4}\)
ਪ੍ਰਸ਼ਨ (x).
(7a – 9b) (7a – 9b).
ਹੱਲ:
(7a – 9b) (7a – 9b) = (7a – 9b)2
(a – b)2 = a2 – 2ab + b2
a = 7a, b = 9b
(7a – 9b)2 = (7a)2 – 2 (7a) (9b) + (9b)2
= 49a2 – 126ab + 81b2
2. ਹੇਠਾਂ ਲਿਖੇ ਗੁਣਨਫਲਾਂ ਨੂੰ ਪਤਾ ਕਰਨ ਦੇ ਲਈ, ਤਤਸਮਕ (x + a) (x + b) = x2 + (a + b)x + ab ਦਾ ਉਪਯੋਗ ਕਰੋ ।
ਪ੍ਰਸ਼ਨ (i).
(x + 3) (x + 7).
ਹੱਲ:
(x + 3) (x + 7)
(x + a) (x + b) = x2 + (a + b)x + ab
x = x, a = 3, b = 7
∴ (x + 3) (x + 7) = x2 + (3 + 7) + (3) (7)
= x2 + 10x + 21.
ਪ੍ਰਸ਼ਨ (ii).
(4x + 5) (4x + 1)
ਹੱਲ:
(4x + 5) (4x + 1)
(x + a) (x + b) = x2 + (a + b)x + ab
x = 4x, a = 5, b = 1
∴ (4x + 5) (4x + 1) = (4x)2 + (5 + 1) 4x + (5) (1)
= 16x2 + ( 6) (4x) + 5
= 16x2 + 24x + 5.
ਪ੍ਰਸ਼ਨ (iii).
(4x – 5) (4x – 1)
ਹੱਲ:
(4x – 5) (4x – 1)
(x + a) (x + b) = x2 + (a + b)x + ab.
x = 4x, a = – 5, b = – 1
∴ (4x – 5) (4x – 1) = (4x)2 + (-5 – 1)(4x) + (-5) (-1)
= 16x2 + (-6) (4x) + 5
= 16x2 – 24x + 5
ਪ੍ਰਸ਼ਨ (iv).
(4x + 5) (4x – 1)
ਹੱਲ:
(4x + 5) (4x – 1)
(x + a) (x + b) = x2 + (a + b)x + ab.
x = 4x, a = 5, b = – 1
∴ (4x + 5) (4x – 1) = (4x)2 + (5 – 1) 4x + (5) (-1)
= 16x2 + (4) (4x) – 5
= 16x2 + 16 – 5.
ਪ੍ਰਸ਼ਨ (v).
(2x + 5y) (2x + 3y)
ਹੱਲ:
(2x + 3y) (2x + 3y)
(x + a) (x + b) = x2 + (a + b)x + ab
x = 2x, a = 5y, b = 3y
∴ (2x + 5y) (2x + 3y) = (2x)2 + (5y + 3y)(2x) + (5y) (3y)
= 4x2 + (8y)(2x) + 15y2
= 4x2 + 6xy + 15y2.
ਪ੍ਰਸ਼ਨ (vi).
(2a2 + 9) (2a2 + 5)
ਹੱਲ:
(2a2 + 9) (2a2 + 5)
(x + a) (x + b) = x2 + (a + b) x + ab.
x = 2a2, a = 9, b = 5
∴ (2a2 + 9) (2a2 + 5) = (2a2)2 + (9 + 5)(2a2) + (9) (5)
= 4a2 + (14) (2a2) + 45
= 4a4 + 28a2 + 45.
ਪ੍ਰਸ਼ਨ (vii).
(xyz – 4) (xyz – 2).
ਹੱਲ:
(xyz – 4) (xyz – 2)
(x + a) (x + b) = x2 + (a + b) x + ab.
x = xyz, a = – 4, b = -2
∴ (xyz – 4) (xyz – 2) = (xyz)2 +(-4 – 2)xyz + (-4)(-2)
= x2y2z2 – 6xyz + 8.
3. ਤਤਸਮਕਾਂ ਦਾ ਉਪਯੋਗ ਕਰਦੇ ਹੋਏ ਹੇਠਾਂ ਲਿਖੇ , ਵਰਗਾਂ ਨੂੰ ਪਤਾ ਕਰੋ :
ਪ੍ਰਸ਼ਨ (i).
(b – 7)2
ਹੱਲ:
(b – 7)2
(a – b)2 = a2 – 2ab + b2
a = b, b = 7
∴ (b – 7)2 = (b)2 – 2 (b) (7) + (7)2
= b2 – 14b + 49
ਪੜਤਾਲ : (b – 7)2 = (b – 7) (b – 7)
= b (b – 7) – 7 (b – 7)
= b2 – 7b – 7b + 49
= b2 – 14b + 49.
ਇਸ ਲਈ, ਇਹ ਸੱਚ ਹੈ ।
ਪ੍ਰਸ਼ਨ (ii).
(xy + 3z)2
ਹੱਲ:
(xy + 3z)2
(a + b)2 = a2 + 2ab + b2
a = xy, b = 3z
∴ (xy + 3z)2 = (xy)2 + 2 (xy) (3z) + (3z)2
= x2y2 + 6xyz + 9z2.
ਪੜਤਾਲ : (xy + 3z)2 = (xy + 3z) (xy + 3z)
= xy(xy + 3z) + 3z (xy + 3z)
= x2y2 + 3xyz + 3xyz + 9z2
= x2y2 + 6xyz + 9z2
ਇਸ ਲਈ, ਇਹ ਸੱਚ ਹੈ ।
ਪ੍ਰਸ਼ਨ (iii).
(6x2 – 5y)2
ਹੱਲ:
(6x2 – 5y)2
(a – b)2 = a2 – 2ab + b2
a = 6x2, b = 5y
∴ (6x2 – 5y)2 – 2(6x)2 – 2(6x2) (5y) + (5y)2
= 36x4 – 60x2y + 25y
ਪੜਤਾਲ :
(6x2 – 5y)2 = (6x2 – 5y) . (6x2 – 5y)
= 6x2(6x2 – 5y) – 5y (6x2 – 5y)
= 36x4 – 30x2y – 30x2y + 25y2
= 36x4 – 60x2y + 25y2
ਇਸ ਲਈ, ਇਹ ਸੱਚ ਹੈ ।
ਪ੍ਰਸ਼ਨ (iv).
(\(\frac{2}{3}\)m + \(\frac{3}{2}\)n)
ਹੱਲ:
(\(\frac{2}{3}\)m + \(\frac{3}{2}\)m)2
(a + b)2 = a2 + 2ab + b2
a = \(\frac{2}{3}\)m, b = \(\frac{3}{2}\)n
(\(\frac{2}{3}\)m + \(\frac{3}{2}\)m)2 = (\(\frac{2}{3}\)m)2 + 2(\(\frac{2}{3}\)m)(\(\frac{3}{2}\)n) + (\(\frac{3}{2}\)n)2
= \(\frac{4}{9}\)m2 +2mn + \(\frac{9}{4}\)n2
ਪਤਾਲ :
(\(\frac{2}{3}\)m + \(\frac{3}{2}\)n)2 = (\(\frac{2}{3}\)m + \(\frac{3}{2}\)n)(\(\frac{2}{3}\)m + \(\frac{3}{2}\)n)
= \(\frac{2}{3}\)m(\(\frac{2}{3}\)m + \(\frac{3}{2}\)n) + \(\frac{3}{2}\)n(\(\frac{2}{3}\)m + \(\frac{3}{2}\)n)
= \(\frac{4}{9}\)m2 + mn + mn + \(\frac{9}{4}\)n2
= \(\frac{4}{9}\)m2 + 2mn + \(\frac{9}{4}\)n2
ਇਸ ਲਈ, ਇਹ ਸੱਚ ਹੈ ।
ਪ੍ਰਸ਼ਨ (v).
(0.4p – 0.5q)2
ਹੱਲ:
(0.4p – 0.5q)2
(a – b)2 = a2 – 2ab + b2
a = 0.4p, b = 0.5 q
∴ (0.4p – 0.5q)2 = (0.4p)2 – 2 (0.4p)(0.5q) + (0.5q)2
= 0.16p2 – 0.04pq + 025q2
ਪੜਤਾਲ :
(0.4p – 0.5q)2
= (0.4p – 0.5q) (0.4p – 0.5q)
= 0.4p (0.4p – 0.5q) – 0.5q (0.4p – 0.5q)
= 0.16p2 – 0.02pq – 0.02pq + 0.25q2
= 0.16p2 – 0.04pq + 0.25q2
ਇਸ ਲਈ, ਇਹ ਸੱਚ ਹੈ ।
ਪ੍ਰਸ਼ਨ (vi).
(2xy + 5y)2.
ਹੱਲ:
(2xy +5y)2
(a + b)2 = a2 + 2ab + b2
a = 2xy, b = 5y
∴ (2xy + 5y)2 = (2xy)2 + 2 (2xy) (5y) + (5y)2
= 4x2y2 + 20xy2 + 25y2
ਪੜਤਾਲ :
(2xy + 5y)2 = (2xy + 5y) (2ry + 5y)
= 2xy (2xy + 5y) + 5y (2xy + 5y)
= 4x2y2 + 10xy2 + 10xy2 + 25y2
= 4x2y2 + 20xy2 + 25y2
ਇਸ ਲਈ, ਇਹ ਸੱਚ ਹੈ ।
4. ਸਰਲ ਕਰੋ :
ਪ੍ਰਸ਼ਨ (i).
(a2 – b2)2
ਹੱਲ:
(a2 – b2)2
(a – b)2 = a2 – 2ab + b2
a = a2. b = b2
∴ (a2 – b2)2 = (a2)2 – 2(a2) (b2) + (b2)2
= a4 – 2a2b2 + b4.
ਪ੍ਰਸ਼ਨ (ii).
(2x + 5)2 – (2x – 5)2
ਹੱਲ:
(2x + 5)2 – (2x – 5)2
(a2 – b2) = (a + b) (a – b)
a = (2x + 5), b = (2x – 5)
∴ (2x + 5)2 – (2x – 5)2 = (2x + 5 + 2x – 5)
(2x + 5 – 2x + 5)
= (4x) (10)
= 40x.
ਪ੍ਰਸ਼ਨ (iii).
(7m – 8n)2 + (7m + 8n)2
ਹੱਲ:
(7m – 8n)2 + (7m + 8n)2
(a – b)2 = a2 – 2ab + b2
(a + b)2 = a2 + 2ab + b2
∴ (7m – 8n)2 + (7m + 8n)2 = (7m)2 – 2 (7m) (8n) + (8n)2 + (7m)2 + 2 (7m) (8n) + (8n)2
= 49m2 – 112mn + 64n2 + 49m2 +112mn + 64n2
= 98m2 + 128n2.
ਪ੍ਰਸ਼ਨ (iv).
(4m + 5n)2 + (5m + 4n)2
ਹੱਲ:
(4m + 5n)2 + (5m + 4n)2
(a + b)2 = a2 + 2ab + b2
∴ (4m + 5n)2 + (5m + 4n)2 = (4m)2 + 2(4m) (5n)+ (5n)2 + (5m)2 + 2 (5m) (4n) + (4n)2
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2.
ਪ੍ਰਸ਼ਨ (v).
(2.5p + 1.5q)2 – (1.5p – 2.5q)2
ਹੱਲ:
(2.5p – 1.5q)2 – (1.5p – 2.5q)2
(a2 – b2) = (a – b)(a + b)
a = (2.5p + 1.5q), b = (1.5p – 2.5q)
∴ (2.5p – 1.5q)2 – (1.5p – 2.5q)2 = [(2.5p – 1.5q) – (1.5p – 2.5q)][2.5p – 1.5q + 1.5p – 2.5q]
= (2.5p – 1.59 – 1.5p + 2.5q)(4p – 4q)
= (1p + 1q) (4p – 4q)
= p (4p – 4q) + q (4p – 4q)
= 4p2 – 4pq + 4pq – 4q2
= 4p2 – 4q2.
ਪ੍ਰਸ਼ਨ (vi).
(ab + bc) – 2ab2c
ਹੱਲ:
(ab + bc)2 – 2ab2c
= (ab)2 + 2(ab) (bc) + (bc)2 – 2ab2c
= a2b2 + 2ab2c + b2c2 – 2ab2c
= a2b2 + b2c2.
ਪ੍ਰਸ਼ਨ (vii).
(m2 – n2m)2 + 2m3n2.
ਹੱਲ:
(m2 – n2m)2 + 2m3n2
= (m2)2 – 2 (m2) (n2m) + (n2m)2 + 2m3n2
= m4 – 2m3n2 + n4m2 + 2m3n2
= m4 + n4m2
5. ਦਰਸਾਉ ਕਿ :
ਪ੍ਰਸ਼ਨ (i).
(3x + 7)2 – 84x = (3x – 7)2
ਹੱਲ:
(3x + 7)2 – 84x = (3x – 7)2
L.H.S. = (3x + 7)2 – 84x
[∵ (a + b)2 = a2 + 2ab + b2]
= (3x)2 + 2(3x) (7) + (7)2 – 84x
= 9x2 + 42x + 49 – 84x
= 9x2 – 42x + 49
= (3x)2 – 2 (3x) (7) + (7)2
= (3x – 7)2
[∵ (a – b)2 = a2 – 2ab + b2]
= R.H.S.
ਪ੍ਰਸ਼ਨ (ii).
(9p – 5q)2 + 180pq = (9p + 5q)2
ਹੱਲ:
(9p – 5q)2 + 180pq = (9p + 5q)2
L.H.S. = (9p – 5q)2 + 180pq
[∵ (a – b)2 = a2 – 2ab + b2
= (9p)2 – 2(9p) (5q) + (5q)2 + 180pq
= 81p2 – 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
= (9p)2 + 2 (9p) (5q) + (5q)2.
[∵ (a + b)2 = a2 + 2ab + b2]
= (9p + 5q)2
= R.H.S.
ਪ੍ਰਸ਼ਨ (iii).
(\(\frac{4}{3}\)m – \(\frac{3}{4}\)n)2 + 2mn = \(\frac{16}{9}\)m2 + \(\frac{9}{16}\)n2
ਹੱਲ:
(\(\frac{4}{3}\)m – \(\frac{3}{4}\)n)2 + 2mn = \(\frac{16}{9}\)m2 + \(\frac{9}{16}\)n2
L.H.S. = (\(\frac{4}{3}\)m – \(\frac{3}{4}\)m)2 + 2mm
[∵ (a – b)2 = a2 – 2ab + b2]
= (\(\frac{4}{3}\)m)2 – 2(\(\frac{4}{3}\)m)(\(\frac{3}{4}\)n) + (\(\frac{3}{4}\)n)2 + 2mn
= \(\frac{16}{9}\)m2 – 2mn + \(\frac{9}{16}\)n2 + 2mn
= \(\frac{16}{9}\)m2 + \(\frac{9}{16}\)n2 = R.H.S.
ਪ੍ਰਸ਼ਨ (iv).
(4pq + 3q)2 – (4pq – 3q)2 = 48pq2
ਹੱਲ:
(4pq + 3q)2 – (4pq – 3q)2 = 48pq2
L.H.S. = (4pq + 3q)2 – (4pq – 3q)2
a2 – b2 = (a + b)(a – b)
a = 4pq + 3q, b = 4pq – 3q
∴(4pq + 3q)2 – (4pq – 3q)2 = [(4pq + 3q) + (4pq – 3q)][(4pq + 3q) – (4pq – 3q)]
= (8pq) (4pq + 34 – 4pq + 3q)
= (8pq) (6q)
= 48pq2 = R.H.S.
ਪ੍ਰਸ਼ਨ (v).
(a – b) (a + b) + (b – c)(b + c) + (c – a) (c + a) = 0
ਹੱਲ:
(a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
L.H.S. = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= a2 – b2 + b2 – c2 + c2 – a2
= 0
= R.H.S.
6. ਤਤਸਮਕਾਂ ਦੇ ਉਪਯੋਗ ਨਾਲ ਹੇਠਾਂ ਲਿਖਿਆਂ ਦਾ ਮੁੱਲ ਪਤਾ ਕਰੋ :
ਪ੍ਰਸ਼ਨ (i).
(71)2
ਹੱਲ:
(71)2
= (70 + 1)2
[(a + b)2 = a2 + 2ab + b2]
= (70)2 + 2 (70) (1) + (1)2
= 4900 + 140 + 1
= 5041.
ਪ੍ਰਸ਼ਨ (ii).
(99)2
ਹੱਲ:
(99)2
= (100 – 1)2
[(a – b)2 = a2 – 2ab + b2]
= (100)2 – 2 (100) (1) + (1)2
= 10,000 – 200 +1
= 9801.
ਪ੍ਰਸ਼ਨ (iii).
(102)2
ਹੱਲ:
(102)2
= (100 + 2)2
[(a + b)2 = a2 + 2ab + b2]
= (100)2 + 2 (100) (2) + (2)2
= 10,000 + 400 + 4
= 10404.
ਪ੍ਰਸ਼ਨ (iv).
(998)2
ਹੱਲ:
(998)2
= (1000 – 2)2
[(a – b)2 = a2 – 2ab + b2]
= (1000)2 – 2 (1000) (2) + (2)2
= 1000000 – 4000 + 4
= 996004.
ਪ੍ਰਸ਼ਨ (v).
(5.2)2
ਹੱਲ:
(5.2)2
= (5 + 0.2)2
(a + b)2 = a2 + 2ab + b2
= (5)2 + 2 (5) + (0.2) + (0.2)2
= 25 + 2 + 0.04
= 27.04.
ਪ੍ਰਸ਼ਨ (vi).
297 × 303
ਹੱਲ:
297 × 303
= (300 – 3) (300 + 3)
(a – b) (a + b) = a2 – b2
a = 300, b = 3
∴ (300 – 3) (300 + 3) = (300)2 – (3)2
= 90000 – 9
= 89991.
ਪ੍ਰਸ਼ਨ (vii).
78 × 82
ਹੱਲ:
78 × 82
= (80 – 2) (80 + 2)
(a – b)(a + b) = a2 – b2
a = 80, b = 2
∴ (80 – 2) (80 + 2) = (80)2 – (2)2
= 6400 – 4
= 6396.
ਪ੍ਰਸ਼ਨ (viii).
(8.9)2
ਹੱਲ:
(8.9)2
= (9 – 0.1)2
(a – b)2 = a2 – 2ab + b2
a = 9, b = 0.1
∴ (9 – 0.1)2 = (9)2 – 2 (9) (0.1) + (0.1)2
= 81 – 1.8 + 0.1
= 79.21.
ਪ੍ਰਸ਼ਨ (ix).
1.05 × 0.95.
ਹੱਲ:
1.05 × 0.95
= (1 + 0.05) (1 – 0.05)
(a – b) (a + b) = a2 – b2
a = 1, b = 0.05
⇒ (1.05) (1 – 0.05) = (1)2 – (0.05)2
= 1 – 0.0025
= 0.9975.
7. a2 – b2 = (a + b) (a – b) ਦਾ ਉਪਯੋਗ ਕਰਦੇ ਹੋਏ ਹੇਠਾਂ ਲਿਖਿਆਂ ਦਾ ਮੁੱਲ ਪਤਾ ਕਰੋ :
ਪ੍ਰਸ਼ਨ (i).
(51)2 – (49)2
ਹੱਲ:
(51)2 – (49)2
a2 – b2 = (a + b) (a – b)
a = 51, b = 49
∴ (51)2 – (49)2 = (51 + 49) (51 – 49)
= (100) (2)
= 200.
ਪ੍ਰਸ਼ਨ (ii).
(1.02)2 – (0.98)2
ਹੱਲ:
(1.02)2 – (0.98)2
a2 – b2 = (a + b) (a – b)
a = 1.02, b = 0.98
∴ (1.02)2 – (0.98)2 = (1.02 + 0.98) (1.02 – 0.98)
= (2) (0.04)
= 0.08.
ਪ੍ਰਸ਼ਨ (iii).
(153)2 – (147)2
ਹੱਲ:
(153)2 – (147)2
a2 – b2 = (a + b) (a – b)
a = 153, b = 147
∴ (153)2 – (147)2 = (153 + 147) (153 – (147)
= (300) (6)
= 1800.
ਪ੍ਰਸ਼ਨ (iv).
(12.1)2 – (7.9)2.
ਹੱਲ:
(12.1)2 – (7.9)2
a2 – b2 = (a + b) (a – b)
a = 12.1, b = 7.9
∴ (12.1)2 – (7.9)2 = (12.1 + 7.9) (12.1 – 7.9)
= (20) (4.2)
= 84.
8. (x + a) (x + b) = x2 + (a + b)x + ab, ਦਾ ਉਪਯੋਗ ਕਰਦੇ ਹੋਏ, ਹੇਠਾਂ ਲਿਖਿਆਂ ਦਾ ਮੁੱਲ ਪਤਾ ਕਰੋ :
ਪ੍ਰਸ਼ਨ (i).
103 × 104
ਹੱਲ:
103 × 104
⇒ (100 + 3) (100 + 4)
(x + a) (x + b) = x2 + (a + b) x + ab
x = 100, a = 3, b = 4
(100 + 3) (100 + 4) = (100)2 + (3 + 4) (100) + (3) (4)
= 10000 + (7) (100) + 12
= 10000 + 700 + 12
= 10712.
ਪ੍ਰਸ਼ਨ (ii).
5.1 × 5.2
ਹੱਲ:
5.1 × 5.2
⇒ (5 + 0.1) (5 + 0.2)
(x + a) (x + b) = x2 + (a + b) x + ab
x = 5, a = 0.1, b = 0.2
= (5 + 0.1) (5 + 0.2)
= (5)2 + (0.1 + 0.2)2 + (0.1) (0.2)
= 25 + (0.3)5 + 0.02
= 25 + 1.5 + 0.02
26.52.
ਪ੍ਰਸ਼ਨ (iii).
103 × 98
ਹੱਲ:
103 × 98
(100 + 3) (100 – 2)
(x + a) (x + b) = x2 + (a + b) x + ab
x = 100, a = 3, b = -2
(100 + 3) (100 – 2) = (100)2 + (3 – 2)(100) + (3) (-2)
= 10000 + (1) (100) – 6
= 10000 + 100 – 6
= 10094.
ਪ੍ਰਸ਼ਨ (iv).
9.7 × 9.8.
ਹੱਲ:
9.7 × 9.8
(10 – 0.3) (10 – 0.2)
(x + a) (x + b) = x2 + (a + b) x + ab
x = 10, a = -0.3, b = -0.2
∴ (10 – 0.3) (10 – 0.2) = (10)2 + (-0.3 – 0.2) (10) + (0.3) (0.2)
= 100 + (-0.5) (10) + 0.06
= 100 – 5 + 0.06
= 95 + 0.06
= 95.06.