Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4C Solutions
Exercise – 4C
(1) Expand
(i) (x + 8) (x + 10)
=> Solution:-
given that,
(x + 8) (x + 10) = x × (x + 10) + 8 × (x + 10)
= x2 + 10x + 8x + 80
= x2 + 18x + 80
(ii) ( x + 8) (x – 10)
=> solution:-
given that,
(x + 8) (x – 10) = x × (x -10) + 8 × (x – 10)
= x2 – 10x + 8x – 80
= x2 – 2x – 80
(iii) (x -8) (x + 10)
=> solution:-
given that,
(x – 8) (x + 10) = x × (x + 10) – 8 × (x + 10)
= x2 + 10x – 8x – 80
= x2 + 2x – 80
(iv) (x – 8) (x – 10)
=> solution:-
given that,
(x – 8) (x -10) = x × (x -10) + 8 × (x – 10)
= x2 – 10x – 8x + 80
= x2 – 18x + 80
(2) Expand:
(i) (2x – 1/x) (3x + 2/x)
=> solution:-
we have to expand,
(2x – 1/x) (3x + 2/x) = 2x × (3x + 2/x) – 1/x (3x + 2/x)
= 6x2 + 2x × 2/x – 1/x × 3x – 1/x × 2/x
= 6x2 + 4 – 3 – 2/x2
= 6x2 – 2/x2 + 1
(ii) (3a + 2/b) (2a – 3/b)
=> Solution:-
we have to expand,
(3a + 2/b ) (2a – 3/b) = 3a × (2a – 3/b) + 2/b × (2a – 3/b)
= 3a × 2a – 3a × 3/b + 2/b × 2a – 2/b × 3/b
= 6a2 – 9a/b + 4a/b – 6/b2
= 6a2 – 5a/b – 6/b2
(3) Expand:
(i) (x + y – z)2
=> Solution:-
We have to use formula,
(a + b + c)2 = a2 + b2 + c2 + 2 (a) (b) + 2bc + 2ca
∴ (x + y – z)2 = x2 + y2 + z2 + 2xy + 2yz + 2 (-z) (x)
= x2 + y2 + z2 + 2xy + 2yz – 2zx
(ii) (x – 2y + 2)2
=> solution:-
we have to use formula,
(a + b + c)2= a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (x – 2y + 2)2 = x2 + (- 2y)2 + (2)2 + 2 (x) (-2y) + 2 ( -2y) (2) + 2 (2) (x)
= x2 + 4y2 + 4 + ( -4xy) – 8y + 4x
= x2 + 4y2 + 4 – 4xy – 8y + 4x
(iii) (5a – 3b + c)2
=> solution:-
we have to use formula,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (5a – 3b + c)2 = (5a)2 + (-3b)2 + (c)2 + 2(5a) (-3b) + 2 (-3b)c + 2 c(5a)
(5a – 3b + c)2 = 25a2 + 9b2 + c2 – 30ba – 6bc + 10ca
(iv) (5x – 3y – 2)2
=> solution:-
we have to use formula,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (5x – 3y – 2)2 = (5x)2 + (-3y)2 + (-2)2 + 2(5x) (-3y) + 2(-3y)(-2) + 2(-2) (5x)
= 25x2 + 9y2 + 4 + (-30xy) + 12y – 20x
= 25x2 + 9y2 + 4 – 30xy + 12y – 20x
(v) (x -1/x + 5)2
=> solution:-
we have to use formula,
(a +b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ ( x -1/x + 5)2 = x2 + (-1/x)2 + (5)2 + 2 × x × (-1/x) + 2 × (-1/x) × 5 + 2 (5) (x)
(x -1/x + 5)2 = x2 + 1/x2 + 25 – 2 – 10/x + 10x
= x2 + 1/x2 + 23 – 10/x + 10x
(4) If a + b + c = 12 and a2 + b2 + c2 = 50, find ab + bc + ca
=> solution:-
given that, a + b + c = 12 and
a2 + b2 + c2 = 50
Now, we have to use formula,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(12)2 = 50 + 2 (ab + bc + ca)
144 = 50 + 2 (ab + bc + ca)
144 – 50 = 2 (ab +bc + ca)
94 = 2 (ab + bc + ca)
94/2 = ab + bc + ca
47 = ab + bc + ca
∴ ab + bc + ca = 47
(5) If a2 + b2 + c2 = 35 and ab + bc + ca = 23, find a + b + c
=> solution:-
given that, a2 + b2 + c2 = 35 and
ab + bc + ca = 23
Now, we have to use formula,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
= 35 + 2 (23)
= 35 + 46
(a + b + c)2 = 81
Taking square root on both sides,
√ (a + b + c)2 = √81
a + b + c = ±9
(6) If a + b + c = p and ab + bc + ca = q, find a2 + b2 + c2
=> solution:-
given that, a + b + c = p and
ab + bc + ca = q
Now, we have to use formula,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
p2 = a2 + b2+ c2 + 2q
p2 – 2q = a2 + b2 + c2
∴ a2 + b2 + c2 = p2 – 2q
(7) If a2 + b2 + c2 = 50 and ab + bc + ca = 47, find a + b + c
=> solution:- given that,
a2 + b2 + c2 = 50 and ab + bc + ca = 47
Now, we have to use formula,
(a + b + c )2 = a2 + b2 + c2 + 2 (ab + bc + ca)
(a + b + c)2 = 50 + 2(47)
= 50 + 94
(a + b + c)2 = 144
Taking square root on both side,
√(a + b + c)2 = √144
a + b + c = ± 12
(8) If x + y – z = 4 and x2 + y2 + z2 = 30, then find the value of xy – yz – zx
=> solution:- given that,
x + y – z = 4 and x2 + y2 + z2 = 30
Now, we have to use formula,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
∴ (x + y – z)2 = x2 + y2 + (-z)2 + 2(xy + y(-z) + (-z)x)
= x2 + y2 + z2 + 2 (xy – yz – zx)
(4)2 = 30 + 2 (xy – yz – zx)
16 = 30 + 2 (xy – yz – zx)
16 – 30 = 2(xy – yz – zx)
– 14 = 2 (xy – yz – zx)
-14/2 = xy – yz – zx
– 7 = xy – yz – zx
∴ xy – yz – zx = -7
Here is your solution of Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4C
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