Selina Concise Class 9 Maths Chapter 4 Expansions 4C Solutions

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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