Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4A Solutions
EXERCISE – 4A
(1) Find the square of
(i) 2a + b
= Solution:
We have to square of (2a + b)
Now, we have to use formula
(a + b)2 = a2 + 2ab + b2
a = 2a and b = b
(2a + b)² = (2a) + 2 × 2a × b + b2
(2a + b)2 = 4a2 + 4ab + b2
(ii) 3a + 7b
= Solution :
We have to square of (3a + 7b)
Now, we have to use formula,
(a + b)2 = a2 + 2ab + b2
a = 3a and b = 7b
(3a + 7b) = (3a)2 + 2 × (3a) × (7b) + (7b)2
(3a + 7b)2 = 9a2 + 42ab + 49b2
(iii) 3a – 4b
= Solution:-
We have to square of (3a – 4b)
Now, we have to use formula
(a – b)2 = a2 – 2ab + b2
a = 3a and b = 4b
(3a – 4b)2 = (3a)2 – 2 × (3a) × (4b) + (4b)2
(3a – 4b)2 = 9a2 – 24ab + 16b2
(iv) 3a/2a – 2b/3a
= Solution:
we have to square of (3a/2b – 2b/3a)
Now, we have to use formula
(a – b)2 = a2 – 2ab + b2
a = 3a/2b and b = 2b/3a
(3a/2b – 2b/3a)2
= (3a/2b)2 – 2 × (3a/2b) x (2b/3a) + (2b/3a)
= 9a2/4b2 – 12ab/6ab + 4b2/9a2
= 9a2/4b2 – 2 + 4b2/9a2
(2) Use identities to evaluate
(i) (101)2
= Solution:
We have to evaluate (101)2 first we have to convert in to (a + b)2 so, we have write (101) into
(101)2 = (100 +)2
a = 100 and b = 1
Now, we have to use identity
(a + b)2 = a2 + 2ab + b2
(100 + 1)2 = (100)2 + 2 × (100) × (1)2
= 10,000 + 200 + 1
∴ (100 + 1)2 = 10,201
∴ (101)2 = 10,201
(ii) (502)2
= Solution:
We have to evaluate (502)2 first we have to convert in to (a + b)2 so we have write (502) in to
(502)2 = (500 + 2)2
a = 500 and b = 2
Now, we have to use identity
(a + b)2 = a2 + 2ab + b2
(502)2 = (500 + 2)2
= (500)2 + 2 × (500) × (2) + (2)2
= 25,0000 + 20,00 + 4
(502)2 = 25,2004
(iii) (97)2
= Solution
We have to evaluate (97)2
First we have to convert into (a + b)2
So, We have write (97) into
(97)2 = (90 + 7)2
a = 90 and b = 7
Now we have to use identity
(a + b)2 = a2 + 2ab + b2
(97)2 = (90 + 7)2
= (90)2 + 2 × 90 × 7 + (7)2
= 81,00 + 1260 + 49
(97)2 = 9409
(iv) (998)2
= solution:
we have to evaluate 998)2 first we have to convert into (a + b)2
So, we have to write (998) into,
(998)2 = (1000 – 2)2
a = 1000 and b = 2
Now, we have to use identity
(a – b)2 = a2 – 2ab + b2
(998)2 = (1000 – 2)2
= (1000)2 – 2× (1000) × (2) + (2)2
= 10,0000 – 4000 + 4
(998)2 = 99,6004
(3) Evaluate
(i) (7x/8 + 4y/5)2
= Solution:
we have to evaluate (7x/8 + 4y/5)2
we have to use identity
(a + b)2 = a2 + 2ab + b2
∴ a = 7x/8 and b = 4y/5
∴ (7x/8 + 4y/5)2
= (7x/8)2 + 2 × (7x/8) × (4y/5) + (4y/5)2
(7x/8 + 4y/5)2 = 49x2/64 + 56xy/40 + 16y2/25
(7x/8 + 4y/5)2 = 49x2/64 + 7xy/5 + 16xy2/25
(ii) (2x/7 – 7y/4)2
= Solution:
We have to evaluate (2x/7 – 7y/4)
we have to use identity
(a – b)2 = a2 – 2ab + b2
a = 2x/7 and b = 7y/4
(2x/7 – 7y/4)2
= (2x/7)2 – 2× (2x/7) (7y/4) + (7y/4)2
= 4x2/49 – 28x/28 + 49y2/16
(2x/7 – 7y/4)2 = 4x2/49 – xy + 49y2/16
(4) Evaluate
(i) (a/2b + 2b/a)2 – (a/2b – 2b/a)2 = 4
= Solution:
We have to Evaluate
(a/2b + 2b/a)2 – (a/2b – 2b/a)2 – 4
Firstly we have to expand the first term of given expansion
So we have to use identity
(a + b)2 = a2 + 2ab + b2
(a/2b + 2b/a)2 = (a/2b)2 + 2 × (a/2b) × (2b/a) + (2b/a)2
= a2/4b2 + 4ab/2ab + 4b2/a2
(a/2b + 2b/a)2 = a2/4b2 + 2 + 4b2/a2
Now, we have to expand second term
(a/2b – 2b/a)2
So, we have to use identity
(a – b)2 = a2 – 2ab + b2
a = a/2b and b = 2b/a
(a/2b – 2b/a)2 = (a/2b)2 – 2 × (a/2b) × (2b/a) + (2b/a)2
= a2/4b2 – 4ab/2ab + 4b2/a2
(a/2b – 2b/a)2 = a2/4b2 – 2 + 4b2/a2
put equation (A) and (B) in given expansion
(a2/4b2 + 2 + 4b2/a2) – (a2/4b2 – 2 + 4b2/a2) – 4
= a2/4b2 + 2 + 4b2/a2 – a2/4b2 + 2 – 4b2/a2 – 4
= 0
(ii) (4a + 3b)2 – (4a – 3b)2 + 48ab
= Solution:
We have to Evaluate
(4a + 3b)2 – (4a – 3b)2 + 48ab
Firstly we have to expand the first term of given expansion
So, we have to use identity
(a + b)2 = a2 + 2ab + b2
(4a + 3b)2 = (4a)2 + 2 × (4a) x (3b) + (3b)2
(4a + 3b)2 = 16a2 – 24ab + 9b2
Now, we have to expand second term
(4a – 3b)2 = (4a)2 – 2 × (4a) × (3b) + (3b)2
(4a – 3b)2 = 16a2 – 24ab + 9b2
(a – b)2 = a2 – 2ab + b2
Put equation (A) and (B) in given expansion
(16a2 + 24ab + 9b2) – (16a2 – 24ab + 9b2) + 48ab
= 16a2 + 24ab + 9b2 – 16a2 + 24ab – 9b2 + 48ab
= 48ab + 48ab
= 96ab
(5) If a + b = 7 and 10, find a – b
= Solution:
given that, a + b = 7 and ab = 7
To find = (a – b) = ?
Now, we know that,
(a- b)2 = a2 – 2ab + b2
= a2 + b2 – 2ab
= a2 + b2 + 2ab – 4ab {- 2ab = 2ab – 4ab}
(a – b)2 = (a + b)2 – 4ab
Substitute a = b = 7 and ab = 10 in (A)
(a – b)2 = (7)2 – 4 × 10)
= 49 – 40
(a –b)2 = 9
Taking square root on both sides,
a – b = ± 3
(6) If a – b = 7 and ab = 18, find a + b
= Solution:
Given that, a – b = 7 and ab = 18
To find = (a + b) = 2
Now, we know that,
(a + b)2 = a2 + 2ab + b2
= a2 + b2 – 2ab + 4ab
(a +b)2 = (a – b)2 + 4ab
Substitute a – b = 7 and ab = 18 in (A)
(a + b)2 = (7)2 + 4 x 18
= 49 + 72
(a + b)2 = 121
Taking square root on both sides
√(a + b)2 = √121
a + b = ± 11
(7) If x + y = 7/2 and xy = 5/2 find
(i) x – y
(ii) x2 – y2
= Solution:
given that, x + y = 7/2 and xy = 5/2
(i) To find = x – y
we have to use identity
(x – y)2 = x2 – 2xy + y2
= x2 + y2 + 2xy – 4xy
(x – y)2 = (x + y)2 – 4xy
Now, we have to substitute
x + y = 7/2 and xy = 5/2
(x – y)2 = (7/2)2 – 4 × 5/2
= 49/4 – 10
= 49 – 40/4
(x – y)2 = 9/4
Taking square root on both side
√(x – y)2 = √9/4
[(x – y) = ± 3/2
(ii) To find = x2 – y2
We know that,
a2 – b2 = (a + b) (a – b)
x2 – y2 = (x + y) (x – y)
= (7/2) (+3/2) from A and from given
x2 – y2 = ±21/4
(8) If a – b = 0.9 and ab = 0.36; find
(i) a + b
(ii) a2 – b2
= Solution:
given that, a – b = 0.9 and ab = 0.36
(i) To find = (i) a + b
We know that,
(a + b)2 = a2 + 2ab + b2
= a2 + b2 – 2ab + 4ab
(a + b)2 = (a – b)2 + 4ab
Substitute a – b = 0.9 and ab = 0.36
(a + b)2 = (0.9)2 + 4(0.36)
= 0.81 + 1.44
(a + b)2 = 2.25
Taking square root on both sides
√(a + b)2 = √2.25
(a + b) = ± 1.5
(ii) To find = a2 – b2
We know that,
a2 – b2 = (a + b) (a – b)
= (± 1.5) (.09) { from (A) and from given}
a2 – b2 = ± 1.35
(9) If a – b = 4 and a + b = 6, find
(i) a2 + b2
(ii) ab
= Solution:
given that, a – b = 4 and a + b = 6
To find = (ii) ab
Now, we know that,
(a + b)2 = a2 + b2 + 2ab
= a2 + b2 – 2ab + 4ab
(a + b)2 = (a – b)2 + 4ab
Substitute, a – b = 4 and (a + b) = 6
(6)2 = (4)2 + 4ab
36 = 16 + 4ab
36 – 16 = 4ab
20 = 4ab
20/4 = ab
5 = ab
∴ ab = 5
(ii) To find = a2 + b2
Now, we know that
(a + b)2 = a2 + 2ab + b2
(a + b)2 = a2 + b2 + 2ab
(a + b)2 = (a2 + b2) + 2ab
(6)2 = (a2 + b2) + 2 × 5
36 = (a2 + b2) + 10 {from given and from (A)}
36 – 10 = a2 + b2
26 = a2 + b2
a2 + b2 = 26
(10) If a + 1/a = 6 and a ≠ o find:
(i) a – 1/a
(ii) a2 – 1/a2
= solution:
given that, a + 1/a = 6 and a ≠ o
(i) To find = a – 1/a
We have to use identity
(a – b)2 = a2 – 2ab + b2
Now,
(a – 1/a)2 = a2 – 2 × a × (1/a) + (1/a)2
= a2 – 2 + 1/a2
(a – 1/a)2 = (a2 + 1/a2) – 2
Now, we use another identity,
(a + 1/a)2 = a2 + 2a × 1/a + 1/a2
(a + 1/a)2 = (a2 + 1/a2) + 2
(6)2 = (a2 + 1/a2) + 2 {from given}
36 = (a2 + 1/a2) + 2
36 – 2 = a2 + 1/a2
34 = a2 + 1/a2
a2 + 1/a2 = 34 put in equation (A)
(a – 1/a)2 = (34) – 2
(a – 1/a)2 = 32
Taking square root on both side,
√(a – 1/a)2 = √32
(a – 1/a) = √8 × 4
= 2√8
= 2√4 × 2
(a – 1/a) = 4√2 (B)
(ii) To find = a2 – 1/a2
We know that,
(a2 – 1/a2) = (a + 1/a) (a – 1/a)
{ ∴ (a2 – b2) (a – b)}
a2 – 1/a2 = (6) (4√2) {∴ from given and equation (B)
a2 – 1/a2 = 24√2
(11) If a – 1/a = 8 and a ≠ 0 and
(i) a + 1/a
(ii) a2 – 1/a2
= Solution :
given that, a – 1/a = 8 and a ≠ 0
(i) To find = a + 1/a
We have to use identity,
(a + b)2 = a2 + 2ab + b2
∴ (a + 1/a)2 = a2 + 2 × a × 1/a + (1/a)2
= a2 + 2 + 1/a2
Now, we have to use another identity,
(a – b)2 = a2 – 2ab + b2
∴ (a – 1/a)2 = a2 – 2 × a × 1/a + (1/a)2
(a – 1/a)2 = (a2 + 1/a2) – 2
(8)2 = (a2 + 1/a2) – 2 (from given)
64 = (a2 + 1/a2) – 2
64 + 2 = (a2 + 1/a2)
66 = a2 = 1/a2
a2 + 1/a2 = 66 put in equation (A)
(a + 1/a)2 = 66 + 2
(a + 1/a)2 = 68
Taking square root on both sides,
√(a + 1/a)2 = √68
a + 1/a = √4 × 17
(a + 1/a) = ± 2√17 (B)
(ii) To find = a2 – 1/a
We know that,
(a2 – b2) = (a + b) (a – b)
∴ (a2 – 1/a2) = (a + 1/a) (a – 1/a)
= (±2√17) (8) {∴ from given and from equation}
(a2 – 1/a2) = ± 16√17
(12) If a² – 3a + 1 = o and a ≠ o; find:
(i) a + 1/a
(ii) a2 = 1/a2
= Solution:
given that, a2 – 3a + 1 = 0 and a ≠ o
(i) To find = a + 1/a
given equation:
a2 – 3a + 1 = 0
divide ‘a’ on both sides
a2/a – 3a/a + 1/a = 0/a
a – 3 + 1/a = 0
(a + 1/a) – 3 = 0
a + 1/a = 3 (A)
(ii) To find = a2 + 1/a2
We know that,
(a + b)2 = a2 + 2ab + b2
(a + 1/a)2 = a2 + 2 × a × 1/a + 1/a2
(a + 1/a)2 = (a2 + 1/a2) + 2
(3)2 = (a2 + 1/a2) + 2 {from A}
9 = (a2 + 1/a2) + 2
9 – 2 = a2 + 1/a2
7 = a2 + 1/a2
a2 + 1/a2 = 7
(13) If a2 – 5a – 1 = 0 and a ≠ 0; find
(i) a – 1/a
(ii) a + 1/a
(iii) a2 – 1/a2
= Solution:
given that, a2 – 5a – 1 = 0 and a ≠ 0
(i) To find = a – 1/a
given equation:
a2 – 5a – 1 = 0
divide ‘a’ on both sides;
a2/a – 5a/a – 1/a = 0
a – 5 – 1/a = 0
(a – 1/a) – 5 = 0
(a – 1/a) = 5
(ii) To find = a + 1/a
We know that,
(a – b)2 = a2 – 2ab + b2
∴ (a – 1/a)2 = a2 – 2 × a × 1/a + (1/a)2
= a2 – 2 + 1/a2
(a – 1/a)2 = (a2 + 1/a2) – 2
(5)2 = (a2 + 1/a2) – 2
25 = (a2 + 1/a2) – 2
25 + 2 = (a2 + 1/a2)
27 = a2 + 1/a2
∴ a2 + 1/a2 = 27 (A)
Now, we have to use another identity,
(a + b)2 = a2 + 2ab + b2
(a + 1/a)2 = a2 + × a × 1/a + 1/a2
= a2 + 2 + 1/a2
(a + 1/a)2 = (a2 + 1/a2) + 2
= 27 + 2 { from equation (A)}
(a + 1/a)2 = 29
Taking square root on both side,
√ (a + 1/a) = √29
a + 1/a = ± √29 (B)
(iii) To find = a2 – 1/a2
We know that,
(a2 – b2) = (a + b) (a – b)
∴ (a2 – 1/a2) = (a + 1/a) (a – 1/a)
(a2 – 1/a2) = (± √29) (5) {∴ from (B) and from given}
(a2 = 1/a2) = ± 5√29
(14) If 3a + 4b = 16 and ab = 4; and the value of 9a2 + 16b2
= Solution:
given that, 3a + 4b = 16 and ab = 4
To find = 9a2 + 16b2
∴ We have to write expansion of (3a + 4b)
(3a + 4b)2 = (3a)2 + 2 × (3a) × (4b) + (4b)2
= 9a2 + 24ab + 16b2
(3a + 4b)2 = (9a2 + 16b2) + 24ab
(16)2 = (9a2 + 16b2) + 24 x (4)
{∴ from given}
256 = (9a2 + 16b2) + 96
256 – 96 = 9a2 + 16b2
160 = 9a2 + 16b2
9a2 + 16b2 = 16
(15) The number a is 2 more than the number b. If the sum of the squares of and b is 34, then find the product of a and b.
= Solution:
given that, The number a is 2 more that the number b
a = b + 2 (A)
and also given that, the sum of the squares of a and b is 34
a2 + b2 = 34
(b + 2)2 = b2 = 34 (from (A))
b2 + 2 × b × 2 + (2)2 + b2 = 34
b2 + 4b + 4 + b2 = 34 {∴ (a + b)2 = a2 + b2 + 2ab}
2b2 + 4b + 4 = 34
2b2 + 4b = 34 – 4
2b2 + 4b = 30
2b2 + 4b – 30 = 0
2b2 + 4b – 30 = 0
divide ‘2’ on both sides,
2b2/2 + 4b/2 – 30/2 = 0
b2 + 2b – 15 = 0
b2 + 5b – 3b – 15 = 0
b (b + 5) – 3 (b + 5) = 0
(b + 5) (b – 3) = 0
b + 5 = 0 or b – 3 = 0
b = – 5 or b = 3
∴ We have to find the product of a and b
If a = -5 then a = – 5 + 2 = -3
If b = 3 then a = 3 + 2 = 5
The product is
ab = (- 3) × (- 5) = ± 15 and
ab = 5 × 3 = 15
(16) The difference between two positive numbers is 5 and the sum of their squares is 73 find the product of these numbers.
= Solution:
Let the two positive numbers are x and y
given that, the difference between two positive numbers is 5
x – y = 5 (i)
and also given that the sum of their squares is 73.
x2 + y2 = 73 (2)
equation (i) = x – y = 5
x = 5 + y put in equation (2)
(5 + y)2 + y2 = 73
(5)2 + 2 × 5 × y + y2 + y2 = 73 {∴ (a + b)2 = a2 + b2 + 2ab)
25 + 10y + 2y2 = 73
2y2 + 10y = 73 – 25
2y2 + 10y = 48
2y2 + 10y – 48 = 0
divide ‘2’ on both sides,
2y2/2 + 10y/2 – 48/2 = 0
y2 + 5y – 24 = 0
y2 + 8y – 3y – 24 = 0
y (y + 8) – 3 (y + 8) = 0
(y + 8) (y – 3) = 0
y + 8 = 0 or y – 3 = 0
y = 8 or y = 3
If y = – 8 then
If y = 3 then
x = 5 – 8 = -3
x = 5 + 3 = 8
Now, we have to find product of these two numbers:
xy = 8 × 3 = 24
∴ xy = 24
Here is your solution of Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4A
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