Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4D Solutions
EXERCISE – 4D
(Q1) If x + 2y + 3z = 0 and x3 + 4y3 + 9z3 = 18xyz evaluate:
(x + 2y)2/xy + (2y + 3z)2/yz + (3z + x)2/2x
Solution:
Given that, x + 2y + 3z = 0 and x3 + 4y3 + 9z2 = 18xyz
∴ x + 2y + 3z = 0
x + 2y = -3z, 2y + 3z = -x
-2y = x + 3z
Now, we have to evaluate,
(x+2y)2/xy + (2y+3z)2/yz + (3z+x)2/2x
We have to put the values of,
x + 2y = -3z, 2y + 3z = -x and x + 3z = -2y
∴ (-3z)2/xy + (-x)2/yz + (-2y)2/2x
= 5z2/xy + x2/yz + 4y2/zx
= 9z3 + 4y3 + x3/xyz
= 18xyz/xyz {∵ given}
= 18
(Q2) If a + 1/a = m and a ≠ 0; find in terms of ‘m’ the value of
(i) a – 1/a
(ii) a2 – 1/a2
Solution:
Given that, a + 1/a = m and a ≠ 0 —- (A)
(i) To find: a – 1/a
Squaring on both sides of equation (1)
(a + 1/a)2 = m2
a2 + 2 × a × 1/a + 1/a2 = m2 {∵ (a + b)2 = a2 + 2ab + b2}
a2 + 2 + 1/a2 = m2
a2 + 1/a2 = m2 – 2 —- (1)
Now, we know that,
(a2 – b2) = (a – b) (a + b)
∴ (a2 – 1/a2) = (a – 1/a) (a + 1/a)
= (a – 1/a) (m) {∵ given and from equation (1)}
= (a – 1/a) m
∴ a2 – 1/a2 = m (a – 1/a) —- (2)
We have to use formula,
(a – b)2 = a2 – 2ab + b2
∴ (a – 1/a)2 = a2 – 2×a×1/a + 1/a2
(a – 1/a)2 = a2 – 2 + 1/a2
= a2 + 1/a2 – a2
= m2 – 2 – 2 {∵ from (1)}
(a – 1/a)2 = m2 – 4
Taking square root on both sides,
√(a – 1/a)2 = ± √m2 – 4
(a – 1/a) = ± √m2 – 4
(ii) ∴ a2 – 1/a2 = m (± √m2 – 4)
a2 – 1/a2 = ±m √m2 – 4 {∵ from (2)}
(Q3) In the expansion of (2x2 – 8) (x – 4)2; find
(i) Coefficient of x3
(ii) Coefficient of x2
(iii) Constant term
Solution:
Given expansion is,
(2x2 – 8) (x – 4)2 = (2x2 – 8) (x2 – 2 × x × (4) + (4)2)
{∵ (a – b)2 = a2 – 2ab + b2)
= (2x – 8) (x2 – 8x + 16)
= (2x2) (x2 – 8x + 16) – 8 × (x2 – 8x + 16)
= 2x4 – 16x3 + 32x2 – 8x2 + 64x – 128
(2x2 – 8) (x – 4)2 = 2x4 – 16x3 + 24x2 + 64x – 128
(i) Coefficient of x3 is – 16
(ii) Coefficient of x2 is 24
(iii) Coefficient of x4 is 2
(iv) Constant term is -128
(Q4) If × > 0 = and x2 + 1/9 x2 = 25/36
Find: x3 + 1/27 x3
Solution:
Given that: x2 + 1/9 x2 = 25/36 and —- (1)
x > 0
Now, we have to use formula,
(a + b)2 = a2 + 2ab + b2
∴ (x + 1/3x)2 = x2 + 2 × x 1/3x + (1/3 x)2
= x2 + 2/3 + 1/3 x2
(x + 1/3x)2 = (x2 + 1/9 x2) + 2/3
= 25/36 + 2/3 {∵ from (1)}
(x + 1/3x)2 = 49/36
Taking square root on both sides,
√(x + 1/3x)2 = √49/36
x+1/3x = ±7/6 —– (1)
Now, we have to use formula,
(a + b)3 = a3 + 30ab (a + b) + b3
∴ (x + 1/3x)3 = x3 + 3 × x × 1/3x (x + 1/3x) + (1/3x)3
(x + 1/3x)3 = x3 + 3 (x + 1/3x) + 1/27x3
= x3 + 1/27x3 + (x + 1/3x)
= x3 + 1/27x3 + 3 (7/6) {∵ from equation (1)}
(7/6)3 = x3 + 1/27x3 + 21/6
343/216 = x3 + 1/27x3 + 21/6
343/216 – 21/6 = x3 + 1/27x3
(343-252)/216 = x3 + 1/27x3
91/216 = x3 + 1/27x3
∴ x3 + 1/27x3 = 91/216
(Q5) If 2 (x2 + 1) = 5x, find
(i) x – 1/x
(ii) x3 – 1/x3
Solution:
Given that,
2 (x2 + 1) = 5x
Dividing x on both sides,
2x2 + 2/x = 5x/x
2x2/x + 2/x = 5
2x + 2/x = 5
2 (x + 1/x) = 5
x+1/x = 5/2
Squaring on both sides,
(x + 1/x)2 = (5/2)2
We have to use (a + b)2 = a2 + 2ab + b2
x2 + 2 × x × 1/x + 1/x2 = 25/4
x2 + 2 + 1/x2 = 25/4
x2 + 1/x2 = 25/4 – 2
x2 + 1/x2 = 25-8/4
x2 + 1/x2 = 17/4 — (1)
Now, we have to use,
(a – b)2 = a2 – 2ab + b2
∴ (x – 1/x)2 = x2 – 2×x 1/x + 1/x2
= x2 – 2 + 1/x2
= (x + 1/x2) – 2
= 17/4 – 2 ∵ from (1)
= 17 – 8/4
(x – 1/x)2 = 9/4
Taking square root on both sides,
√(x – 1/x)2 = √9/4
x – 1/x = ± 3/2
Now, we have to use,
(a – b)3 = a3 – 3ab (a – b) + b3
∴ (x – 1/x)3 = x3 – 3×x×1/x (x – 1/x) + (1/-x3)
= x3 – 3 (x – 1/x) – 1/x3
= (x3 – 1/x3) – 3 (3/2)
(x – 1/x)3 = (x3 – 1/x3) – 9/2
(3/2)3 = x3 – 1/x3 – 9/2
27/8 = (x3 – 1/x3) – 9/2
27/8 + 9/2 = x3 – 1/x3
27 + 36/8 = x3 – 1/x3
63/8 = x3 – 1/x3
∴ x3 – 1/x3 =63/8
(Q6) If a2 + b2 = 34 and ab = 12; find:
(i) 3 (a + b)2 + 5 (a – b)2
(ii) 7 (a – b)2 – 2 (a + b)2
Solution:
Given that,
a2 + b2 = 34 and ab = 12
Now, we have to use,
(a+b)2 = a2 + 2ab + b2
= (a2 + b2) + 2ab
= 34 + 2 × (12)
= 34 + 24
(a+b)2 = 58
Now, we have to use,
(a – b)2 – 2ab + b2
= (a2 + b2) – 2ab
= 34 – 2 (12)
= 34 – 24
(a – b)2 = 10
(i) To find: 3 (a + b)2 + 5 (a – b)2
∴ 3 (58) + 5 (10)
= 174 + 50
= 224
(ii) To find : 7 (a – b)2 – 2 (a + b)2
∴ 7 (10) – 2 (58)
∴ = 70 – 116
= -46
(Q7) If 3x – 4/x = 4 and x≠0; find :
27x3 – 64/x3
Solution:
Given that,
3x – 4/x = 4 and x ≠ 0
cubing on both sides,
(3x – 4/x)3 = (4)3
(3x)3 – 3 × (3x) × (4/x) (3x – 4/x) + (-4/x)3 = 64
27x3 – 36 (4) – 64/x3 = 64
27x3 – 64/x3 – 144 = 64
27x3 – 64/x3 = 64 + 144
27x3 – 64/x3 = 208
(Q8) If x2 + 1/x2 = 7 and x ≠ 0; find the value of: 7x3 + 8x – 7/x3 – 8/x
Solution:
Given that,
x2 + 1/x2 = 7 and x ≠ 0
To find: 7x3 + 8x – 7/x3 – 8/x
= 7 (x3 – 1/x3) + 8 (x – 1/x) —— (1)
Now, we have to use,
(a – b)2 = a2 – 2ab + b2
∴ (x – 1/x)2 = x2 – 2 × x × 1/x + 1/x2
= x2 – 2 + 1/x2
(x – 1/x)2 = (x2 + 1/x2) – 2
= 7 – 2
(x – 1/x)2 = 5 => x – 1/x = ± √5
Now, we have to use,
(a – b)3 = a3 – 3ab (a – b) + b3
∴ (x – 1/x)3 = x3 – 3×x×1/x (x – 1/x) + (-1/x)3
(x – 1/x)3 = (x3 – 1/x3) – 3 (x – 1/x)
(√5)3 = (x3 – 1/x3) – 3 (√5)
5√5 = x3 – 1/x3 – 3√5
5√5 + 3√5 = x3 – 1/x3
8√5 = x3 – 1/x3
∴ x3 – 1/x3 = 8√5
∴ Equation (1) =>
7x3 + 8x – 7/x3 – 8/x = 7 (x3 – 1/x3) + 8 (x – 1/x)
= 7 (8√5) + 8 (±√5)
= 56 √5 + 8√5
= ± 64√5
(Q9) If x = 1/(x-5) and x ≠ 5; find x2 – 1/x2
Solution:
Given that,
x = 1/x-5 and x ≠ 5
∴ x (x – 5) = 1
x2 – 5x = 1
Dividing by x2 on both sides,
x2/x – 5x/x = 1/x
x – 5 = 1/x
(x – 1/x) = 5
Squaring on both sides,
(x – 1/x)2 = (5)2
x2 – 2 × x × 1/x + (1/x2) = 25
x2 – 2 + 1/x2 = 25
x2 + 1/x2 = 25 + 2
x2 + 1/x2 = 27
We know that,
(a2 – b2) = (a – b) (a + b)
∴ (x2 – 1/x2) = (x – 1/x) (x + 1/x) —- (A)
Now, we have to use,
(a + b)2 = a2 + 2ab + b2
∴ (x + 1/x)2 = x2 + 2×x×1/x + 1/x2
= x2 + 2 + 1/x2
= (x2 + 1/x2) + 2
= 27 + 2
(x + 1/x)2 = 29
Taking square root on both sides,
√(x + 1/x)2 = √29
x + 1/x = ± √29
Now, equation (A) => x2 – 1/x2 = (x – 1/x) (x + 1/x)
(x2 – 1/x2) = (5) (±√29)
= ± 5√29
(Q10) If x = 1/(5-x) and x ≠ 5 I find x3 + 1/x3
Solution:
Given that,
x = 1/(5-x) and x ≠ 5
= (5-x) = 1
= 5x – x2 = 1
Dividing x on both sides,
5x/x – x2/x = 1/x
5 – x = 1/x
1/x + x = 5 —- (1)
Cubing on both sides,
(x + 1/x)3 = (5)3
Now, we have to use,
(a + b)3 = a3 + 3ab (a + b) + b3
∴ (x + 1/x)3 = x3 + 3x × 1/x (x+ 1/x)+ 1/x3
(x + 1/x)3 = x3 + 1/x3 + 3 (x + 1/x)
x3 + 1/x3 + 3 (x + 1/x)= 125
x3 + 1/x3 + 3 (5) = 125 (∵ from (1))
x3 + 1/x3 + 15 = 125
x3 + 1/x3 = 125 – 15
x3 + 1/x3 = 110
(Q11) If 3a + 5b + 4c = 10, show that:
27a3 + 125 b3 + 64 c3 = 180 abc,
Solution:
Given that,
3a + 5b + 4c = 0
3a + 5b = -4c
Cubing on both sides,
(3a + 5b)3 = (-4c)3
We have to use,
(a + b)3 = a3 + 3ab (a + b) + b3
(3a)3 + 3 × (3a) (5b) (3a + 5b) + (5b)3 = -64c3
27a + 45ab (3a + 5b) + 125b3 = -64c3
27a3 + 45ab (-4c) + 125b3 = -64c3
27a3 – 180abc + 125b3 = -64c3
27a3 + 125b3 + 64c3 = 180 abc (Hence Proved)
(Q12) The sum of two numbers is 7 and the sum of their cubes is 133, find the sum of their cubes.
Solution:
Let, the two numbers are x and y.
Given that, the sum of two numbers is 7
∴ x + y = 7 —- (1)
and also the sum of their cubes is 133
x3 + y3 = 133 —– (2)
To find, the sum of their squares.
x2 + y2 =?
Squaring on both sides of equation (1) we get,
(x + y)2 = (7)2
x2 + 2xy + y2 = 49
x2 + y2 + 2xy = 49 —— (A)
Now, cubing on both sides of equation (1)
(x + y)3 = (7)3
x3 + 3xy (x + y) + y3 = 343
x3 + y3 + 3xy (x + y) = 343
(133) + 3xy (7) = 343
21xy = 343- 133 (∵ from (1) and (2))
21xy = 210
xy = 210/21
xy = 10 —- (B)
Equation (A) =>
x2 + y2 + 2xy = 49
x2 + y2 + 2×10 = 49 (∵ from B)
x2 + y2 + 20 = 49
x2 + y2 = 49 – 20
x2 + y2 = 29
∴ The sum of their squares are 29.
(Q13) In each of the following, find the value of ‘a’:
(i) 4x2 + ax + 9 = (2x + 3)2
(ii) 4x2 + ax + 9 = (2x – 3)2
(iii) 9x2 + (7a – 5) x + 25 = (3x + 5)2
Solution:
(i) 4x2 + ax + 9 = (2x + 3)2
4x2 + ax + 9 = (2x)2 + 2 × (2x) (3) + 32
(∵ (a + b)2 = a2 + 2ab + b2)
4x2 + ax + 9 = 4x2 + 12x + 9
Comparing on both sides,
ax = 12x
∴ a = 12
(ii) 4x2 + ax + 9 = (2x – 3)2
4x2 + ax + 9 = (2x)2 – 2×(2x) (3) + (-3)2
4x2 + ax + 9 = 4x2 – 12x+ 9
Comparing on both sides,
ax = -12x
∴ a = -12
(iii) 9x2 + (7a – 5) x + 25 = (3x + 5)2
9x2 + (7a – 5) x + 25 = (3x)2 + 2 ×(3x) × (5) + 52
9x2 + 7ax – 5x + 25 = 9x2 + 30x + 25
9x2 + 7ax + 25 = 9x2 + 5x + 30x + 25
9x2 + 7ax + 25 = 9x2 + 35x + 25
Comparing on both sides,
7ax = 35x
7a = 35
a = 35/7
a = 5
(Q14) If (x2 + 1)/x = 31/3 and x > 1; find
(1) x – 1/x
(ii) x3 – 1/x3
Solution:
Given that,
x2+1/x = 3 1/3
= 9+1/3
x2+1/x = 10/3
3 (x2 + 1) = 10x
3x2 + 3 = 10x
Dividing by x on both sides,
3x2/x + 3/x = 10x/x
3x + 3/x = 10
3 (x + 1/x) = 10
x + 1/x = 10/3
Squaring on both sides,
(x + 1/x)2 = (10/3)2
x2 + 2 × x × 1/x + 1/x2 = 100/9 {∵ (a + b)2 = a2 + 2ab + b2}
x2 + 2 + 1/x2 = 100/9
x2 + 1/x2 = 100/9 – 2
x2 + 1/x2 = 100-18/9
x2 + 1/x2 = 82/9
(i) Now, we have to find,
(x – 1/x)
We have to use,
(a – b)2 = a2 – 2ab + b2
∴ (x – 1/x)2 = x2 – 2 × x × 1/x + (-1/x)2
(x – 1/x)2 = x2 – 2 + 1/x2
= (x2 + 1/x2) – 2
= 82 – 2
= 82-18/9
(x – 1/x)2 = 64/9
Taking square root on both sides,
√(x – 1/x)2 = √64/9
x – 1/x = ± 8/3
(ii) Now, we have to find –
(x3 – 1/x3) –
We have to find –
(a – b)3 = a3 – 3ab (a – b) + b3
∴ (x – 1/x)3 = x3 – 3×x×1/x (x – 1/x)- 1/x3
(x – 1/x)3 = x3 – 3 (x – 1/x) – 1/x3
= (x3 – 1/x3) – 3 (x – 1/x)
(8/3)3 = (x3 – 1/x3) – 3 (8/3)
(8/3)3 = (x3 – 1/x3) – 3 (8/3)
(8/3)3 = (x3 – 1/x3) – 24/3
512/27 + 24/3 = x3 – 1/x3
216+512/27 = x3 – 1/x3
x3– 1/x3 = 728/27
x3 – 1/x3 = 728/27
(Q15) The difference between two positive numbers is 4 and the difference between their cubes is 316 Find
(i) Their product
(ii) The sum of their squares
Solution:
Let, the two positive numbers are x and y.
Given that, The difference between two positive numbers is 4.
∴ x – y = 4 —- (1)
And also given that, the difference between their cubes is 316.
x3 – y3 = 316 —– (2)
Squaring on both sides of equation (1),
(x – y)2 = (4)2
x2 – 2xy + y2 = 16 {∵ (a – b)2 = a2 – 2ab + b2)
x2 + y2 – 2xy = 16 —- (A)
Cubing on both sides of equation (1),
(x – y)3 = (4)3
x3 – 3xy (x – y) + y3 = 64 {∵ (a – b)3 = a3 – 3ab (a – b) + b2
x3 – 3xy (4) – y3 = 64
(x3 – y3) – 12xy = 64
316 – 12xy = 64
-12xy = 64 – 316
-12xy = -252
12xy = 252
xy = 252/12
xy = 21 — (B)
Equation (A) =>
x2 + y2 – 2 × (21) = 16 (∵ from B)
x2 + y2 – 42 = 16
x2 + y2 = 16 + 42
x2 + y2 = 58
Hence, The product of two positive numbers is 21 and their sum is 58.
Here is your solution of Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4D
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