SELINA Solution Class 9 Expansions Chapter 4 Exercise 4C

Question 1.1

Expand : ( x + 8 ) ( x + 10 )

Sol:

( x + 8 )( x + 10 ) = x² + ( 8 + 10 )x + 8 x 10
                            = x² + 18x + 80

Question 1.2

Expand : ( x + 8 )( x - 10 )

Sol:

( x + 8 )( x - 10 ) = x² + ( 8 - 10 )x + 8 x (-10)
                           = x² - 2x - 80

Question 1.3

Expand : ( X - 8 ) ( X + 10 )

Sol:

( X - 8 ) ( X + 10 ) = X² - ( 8 - 10 )X - 8 x 10
                            = X² + 2X - 80

Question 1.4

Expand : ( x - 8 )( x - 10 )

Sol:

( x - 8 )( x - 10 )

= x(x - 10) - 8(x - 10)

= x² - 10x - 8x + 80

= x² - 18x + 80

Question 2.1

Expand : ${\displaystyle (2x-\frac{1}{x})(3x+\frac{2}{x})}$

Sol:

${\displaystyle (2x-\frac{1}{x})(3x+\frac{2}{x})=\left(2x\right)\left(3x\right)-\left(\frac{1}{x}\right)\left(3x\right)+\left(\frac{2}{x}\right)\left(2x\right)-\left(\frac{1}{x}\right)\left(\frac{2}{x}\right)}$

= ${\displaystyle 6x^2-(3-2)-\frac{2}{x^2}}$

= ${\displaystyle 6x^2-(-1)-\frac{2}{x^2}}$

= ${\displaystyle 6x^2+1-\frac{2}{x^2}}$

Question 2.2

Expand : ${\displaystyle (3a+\frac{2}{b})(2a-\frac{3}{b})}$

Sol:

${\displaystyle (3a+\frac{2}{b})(2a-\frac{3}{b})=\left(3a\right)\left(2a\right)+\left(\frac{2}{b}\right)\left(2a\right)-\left(\frac{3}{b}\right)\left(3a\right)-\left(\frac{2}{b}\right)\left(\frac{3}{b}\right)}$

= ${\displaystyle 6a^2+(\frac{4}{b}-\frac{9}{b})a-\frac{6}{b^2}}$

= ${\displaystyle 6a^2+(-\frac{5}{b})a-\frac{6}{b^2}}$

= ${\displaystyle 6a^2-\frac{5a}{b}-\frac{6}{b^2}}$

Question 3.1

Expand : ( x + y - z )²

Sol:

( x + y - z )² = x² + y² + z² + 2(x)(y) - 2(y)(z) - 2(z)(x)
= x² + y² + z² + 2xy - 2yz - 2zx

Question 3.2

Expand : ( x - 2y + 2 )² 

Sol:

( x - 2y + 2 )²  = x² + (2y)² + (2)² - 2(x)(2y) - 2(2y)(2) + 2(2)(x)
= x² + 4y² + 4 - 4xy - 8y + 4x

Question 3.3

Expand : ( 5a - 3b + c )²

Sol:

( 5a - 3b + c)² = (5a)² + (3b)² + (c)² - 2(5a)(3b) - 2(3b)(c) + 2(c)(5a)

= 25a² + 9b² + c² - 30ab - 6bc + 10ca

Question 3.4

Expand : ( 5x - 3y - 2 )²

Sol:

( 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

Question 3.5

Expand : ${\displaystyle {(x-\frac{1}{x}+5)}^2}$

Sol:

${\displaystyle {(x-\frac{1}{x}+5)}^2={\left(x\right)}^2+{\left(\frac{1}{x}\right)}^2+{\left(5\right)}^2-2\left(x\right)\left(\frac{1}{x}\right)-2\left(\frac{1}{x}\right)\left(5\right)+2\left(5\right)\left(x\right)}$ 

=${\displaystyle x^2+\frac{1}{x^2}+25-2-\frac{10}{x}+10x}$ 

=${\displaystyle x^2+\frac{1}{x^2}+23-\frac{10}{x}+10x}$

Question 4

If a + b + c = 12 and a² + b² + c² = 50; find ab + bc + ca.

Sol:

We know that
( a + b + c )² = a² + b² + c² + 2( ab + bc + ca )       .......(1)
Given that, a² + b² + c² = 50 and a + b + c = 12.
We need to find ab + bc + ca :
Substitute the values of  (a² + b² + c² ) and ( a + b + c )
in the identity (1), we have
(12)² = 50 + 2( ab + bc + ca )

⇒ 144 = 50 + 2( ab + bc + ca )
⇒ 94 = 2( ab + bc + ca)
⇒ ab + bc + ca = ${\displaystyle \frac{94}{2}}$
⇒ ab + bc + ca = 47

Question 5

If a² + b² + c² = 35 and ab + bc + ca = 23; find a + b + c.

Sol:

We know that
( a + b + c )² = a² + b² + c² + 2( ab + bc + ca )     ....(1)
Given that, a² + b² + c² = 35 and ab + bc + ca = 23
We need to find a + b + c :
Substitute the values of ( a² + b² + c² ) and ( ab + bc + ca )
in the identity (1), we have
( a + b + c )² = 35 + 2(23)
⇒ ( a + b + c )² = 81
⇒ a + b + c = ${\displaystyle \pm \sqrt{81}}$ 
⇒ a + b + c = ${\displaystyle \pm 9}$

Question 6

If a + b + c = p and ab + bc + ca = q ; find a² + b² + c².

Sol:

We know that
( a + b + c )² = a² + b² + c² + 2( ab + bc + ca )    .....(1)
Given that, a + b + c = p and ab + bc + ca = q
We need to find a² + b² + c² :
Substitute the values of ( ab + bc + ca ) and ( a + b + c )
in the identity (1), we have
(p)² = a² + b² + c² + 2q
⇒ p² = a² + b² + c² + 2q
⇒ a² + b² + c² = p² - 2q

Question 7

If a² + b² + c² = 50 and ab + bc + ca = 47, find a + b + c.

Sol:

a² + b² + c² = 50 and ab + bc + ca = 47
Since ( a + b + c )² = a² + b² + c² + 2( ab + bc + ca )
∴ ( a + b + c )² = 50 + 2(47)
⇒ ( a + b + c )² = 50 + 94 = 144
⇒ a + b +c = ${\displaystyle \sqrt{144}=\pm 12}$
∴ a + b + c = ${\displaystyle \pm 12}$

Question 8

If x+ y - z = 4 and x² + y² + z² = 30, then find the value of xy - yz - zx.

Sol:

x + y - z = 4 and x² + y² + z² = 30
Since ( x + y - z)2 = x2 + y2 + z2 + 2( xy - yz - zx ), we have
(4)2 = 30 + 2( xy - yz - zx )
⇒ 16 = 30 + 2( xy - yz - zx )
⇒ 2( xy - yz - zx ) = -14
⇒ xy - yz - zx = ${\displaystyle -\frac{14}{2}}$ = -7
∴ xy - yz - zx = -7