SELINA Solution Class 9 Expansions Chapter 4 Exercise 4D

Question 1

If x + 2y + 3z = 0 and x³ + 4y³ + 9z³ = 18xyz ; evaluate : 
${\displaystyle \frac{{(x+2y)}^2}{xy}+\frac{{(2y+3z)}^2}{yz}+\frac{{(3z+x)}^2}{zx}}$

Sol:

Given that x³ + 4y³ + 9z³ = 18xyz and x + 2y + 3z = 0
x + 2y = - 3z, 2y + 3z = -x and 3z + x = -2y
Now
${\displaystyle \frac{{(x+2y)}^2}{xy}+\frac{{(2y+3z)}^2}{yz}+\frac{{(3x+x)}^2}{zx}}$

= ${\displaystyle \frac{{(-3z)}^2}{xy}+\frac{{(-x)}^2}{yz}+\frac{{(-2y)}^2}{zx}}$

= ${\displaystyle \frac{9z^2}{xy}+\frac{x^2}{yz}+\frac{4y^2}{zx}}$

= ${\displaystyle \frac{x^3+4y^3+9z^3}{xyz}}$

Given that x³ + 4y³ + 9z³ = 18xyz

∴ ${\displaystyle \frac{{(x+2y)}^2}{xy}+\frac{{(2y+3z)}^2}{yz}+\frac{{(3z+x)}^2}{zx}=\frac{18xyz}{xyz}=18}$

Question 2.1

If a + ${\displaystyle \frac{1}{a}}$ = m and a ≠ 0 ; find in terms of 'm'; the value of :
${\displaystyle a-\frac{1}{a}}$

Sol:

Given that a + ${\displaystyle \frac{1}{a}}$ = m
Now consider the expansion of ${\displaystyle {(a+\frac{1}{a})}^2}$ :
${\displaystyle {(a+\frac{1}{a})}^2=a^2+\frac{1}{a^2}+2}$

⇒ m² = a² + ${\displaystyle \frac{1}{a^2}}$ + 2
⇒ a² + ${\displaystyle \frac{1}{a^2}}$ = m² - 2
Now consider the expansion of ${\displaystyle {(a-\frac{1}{a})}^2}$ :
${\displaystyle {(a-\frac{1}{a})}^2=a^2+\frac{1}{a^2}-2}$

⇒ ${\displaystyle {(a-\frac{1}{a})}^2=m^2-2-2}$

⇒ ${\displaystyle {(a-\frac{1}{a})}^2=m^2-4}$

⇒ ${\displaystyle {(a-\frac{1}{a})}^2=\pm \sqrt{m^2-4}}$

Question 2.2

If a + ${\displaystyle \frac{1}{a}}$ = m and a ≠ 0 ; find in terms of 'm'; the value of : 
${\displaystyle a^2-\frac{1}{a^2}}$

Sol:

${\displaystyle a^2-\frac{1}{a^2}=(a+\frac{1}{a})(a-\frac{1}{a})}$         
                                               [Since a² - b² = ( a + b)( a - b )]
                      = ${\displaystyle m(\pm \sqrt{m^2-4})}$
                      = ${\displaystyle \pm m\sqrt{m^2-4}}$

Question 3.1

In the expansion of (2x² - 8) (x - 4)²; find the value of coefficient of x³.

Sol;

( 2x² - 8 )( x - 4 )²
= ( 2x² - 8 )( x² - 8x + 16 )

= 2x²( x² - 8x + 16 ) - 8( x² - 8x + 16 )
= 4x⁴ - 16x³ + 32x² - 8x² + 64x -128
= 4x⁴ - 16x³ + 24x² + 64x - 128
Hence,
Coefficient of x³ = - 16

Question 3.2

In the expansion of (2x² - 8) (x - 4)²; find the value of coefficient of x²

Sol:

( 2x² - 8 )( x - 4 )²
= ( 2x² - 8 )( x² - 8x + 16 )
= 4x⁴ - 16x³ + 32x² - 8x² + 64x -128
= 4x⁴ - 16x³ + 24x² + 64x - 128
Hence,
Coefficient of x² = 24

Question 3.3

In the expansion of (2x² - 8) (x - 4)²; find the value of constant term.

Sol:

( 2x² - 8 )( x - 4 )²
= ( 2x² - 8 )( x² - 8x + 16 )
= 4x⁴ - 16x³ + 32x² - 8x² + 64x -128
= 4x⁴ - 16x³ + 24x² + 64x - 128
Hence,
Constant term = -128

Question 4

If x > 0 and ${\displaystyle x^2+\frac{1}{9x^2}=\frac{25}{36},\text{Find} x^3+\frac{1}{27x^3}}$ 

Sol:

Given that
${\displaystyle x^2+\frac{1}{9x^2}=\frac{25}{36}}$
⇒ ${\displaystyle x^2+\frac{1}{{\left(3x\right)}^2}=\frac{25}{36}}$             ...(1)
Now consider the expansion of ${\displaystyle {(x+\frac{1}{3x})}^2}$ :
${\displaystyle {(x+\frac{1}{3x})}^2=x^2+\frac{1}{{\left(3x\right)}^2}+2\times x\times \frac{1}{3x}}$

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

⇒                       = ${\displaystyle \frac{25}{36}+\frac{2}{3}}$          [From(1)]

⇒                       = ${\displaystyle \frac{25+24}{36}}$

⇒                       = ${\displaystyle \frac{49}{36}}$

⇒ ${\displaystyle x+\frac{1}{3x}=\pm \sqrt{\frac{49}{36}}}$

⇒ ${\displaystyle x+\frac{1}{3x}=\pm \frac{7}{6}}$
 
We need to find ${\displaystyle x^3+\frac{1}{27x^3}}$ :
Let us consider the expansion of ${\displaystyle {(x+\frac{1}{3x})}^3}$ :
${\displaystyle {(x+\frac{1}{3x})}^3=x^3+\frac{1}{27x^3}+3\times x\times \frac{1}{3x}\times (x+\frac{1}{3x})}$

⇒ ${\displaystyle {\left(\frac{7}{6}\right)}^3=x^3+\frac{1}{27x^3}+3\times x\times \frac{1}{3x}\times (x+\frac{1}{3x})}$

⇒ ${\displaystyle \frac{343}{216}=x^3+\frac{1}{27x^3}+x+\frac{1}{3x}}$

⇒ ${\displaystyle \frac{343}{216}=x^3+\frac{1}{27x^3}+\frac{7}{6}}$       [∵ ${\displaystyle x+\frac{1}{3}x=\frac{7}{6}}$ ]

⇒ ${\displaystyle (\frac{343}{216}-\frac{7}{6})=x^3+\frac{1}{27x^3}}$

⇒ ${\displaystyle x^3+\frac{1}{27x^3}=\left[\frac{343-252}{216}\right]}$

⇒ ${\displaystyle x^3+\frac{1}{27x^3}=\left(\frac{91}{216}\right)}$

Question 5

If 2( x² + 1 ) = 5x, find :
(i) ${\displaystyle x-\frac{1}{x}}$

(ii) ${\displaystyle x^3-\frac{1}{x^3}}$

Sol:

(i) 2( x² + 1 ) = 5x
( x² + 1 ) = ${\displaystyle \frac{5}{2}}$x

Dividing by x, we have
${\displaystyle \frac{x^2+1}{x}=\frac{5}{2}}$

⇒ ${\displaystyle (x+\frac{1}{x})=\frac{5}{2}}$                           .....(1)
Now consider the expansion of  ${\displaystyle {(x+\frac{1}{x})}^2}$ :
${\displaystyle {(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2}$

⇒ ${\displaystyle {\left(\frac{5}{2}\right)}^2=x^2+\frac{1}{x^2}+2}$                 [From(1)]

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

⇒ ${\displaystyle \frac{25}{4}-2=x^2+\frac{1}{x^2}}$

⇒ ${\displaystyle x^2+\frac{1}{x^2}=\frac{25-8}{4}}$

⇒ ${\displaystyle x^2+\frac{1}{x^2}=\frac{17}{4}}$                     ....(2)
Now consider the expansion of ${\displaystyle {(x-\frac{1}{x})}^2}$ :
${\displaystyle {(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2}$

⇒ ${\displaystyle {(x-\frac{1}{x})}^2=\frac{17}{4}-2}$                 [from(2)]

⇒ ${\displaystyle {(x-\frac{1}{x})}^2=\frac{17-8}{4}}$

⇒ ${\displaystyle {(x-\frac{1}{x})}^2=\frac{9}{4}}$

⇒ ${\displaystyle {(x-\frac{1}{x})}^2=\pm \frac{3}{2}}$                  ....(3)

(ii) We know that,
${\displaystyle (x^3-\frac{1}{x^3})={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})}$

∴ ${\displaystyle (x^3-\frac{1}{x^3})={(\pm \frac{3}{2})}^3+3(\pm \frac{3}{2})}$         [from(3)]

                             = ${\displaystyle \pm \frac{27}{8}+\frac{9}{2}}$

⇒ ${\displaystyle (x^3-\frac{1}{x^3})=\pm \frac{27+36}{8}}$

⇒ ${\displaystyle (x^3-\frac{1}{x^3})=\pm \frac{63}{8}}$

Question 6.1

If a² + b² = 34 and ab = 12; find : 3(a + b)² + 5(a - b)²

Sol:

a² + b² = 34, ab= 12

(a + b)² = a² + b² + 2ab
             = 34 + 2 x 12 = 34 + 24 = 58  

(a - b)² = a² + b² - 2ab
            = 34 - 2 x 12 = 34- 24 = 10

3(a + b)² + 5(a - b)² = 3 x 58 + 5 x 10 = 174 + 50 = 224

Question 6.2

If a² + b² = 34 and ab = 12; find : 7(a - b)² - 2(a + b)²

Sol:

a² + b² = 34, ab= 12

(a + b)² = a² + b² + 2ab
             = 34 + 2 x 12 = 34 + 24 = 58  

(a - b)² = a² + b² - 2ab
            = 34 - 2 x 12 = 34- 24 = 10\

7(a - b)² - 2(a + b)²  = 7 x 10 - 2 x 58 = 70 - 116 = - 46

Question 7

If 3x - ${\displaystyle \frac{4}{x}}$ = 4; and x ≠ 0 find : 27x³ - ${\displaystyle \frac{64}{x^3}}$

Sol:

3x - ${\displaystyle \frac{4}{x}}$ = 4;

We need to find 27x³ - ${\displaystyle \frac{64}{x^3}}$

Let us now consider the expansion of ${\displaystyle {(3x-\frac{4}{x})}^3}$ :
${\displaystyle {(3x-\frac{4}{x})}^3=27x^3-\frac{64}{x^3}-3\times 3x\times \frac{4}{x}(3x-\frac{4}{x})}$

⇒ ${\displaystyle {\left(4\right)}^3=27x^3-\frac{64}{x^3}-144 [\text{Given :}3x-\frac{4}{x}=4]}$

⇒ 64 + 144 = 27x³ - ${\displaystyle \frac{64}{x^3}}$

⇒ 27x3 - ${\displaystyle \frac{64}{x^3}}$ = 208

Question 8

If x² + ${\displaystyle x^{\frac{1}{2}}}$= 7 and  x ≠ 0; find the value of : 
7x³ + 8x - ${\displaystyle \frac{7}{x^3}-\frac{8}{x}}$

Sol:

Given that ${\displaystyle x^2+\frac{1}{x^2}=7}$

We need to find the value of  7x³ + 8x - ${\displaystyle \frac{7}{x^3}-\frac{8}{x}}$

Consider the given equation :
x2 + ${\displaystyle \frac{1}{x^2}}$ - 2 = 7 - 2  [ subtract 2 from both the sides ]

⇒ ${\displaystyle {(x-\frac{1}{x})}^2=5}$

⇒ ${\displaystyle (x-\frac{1}{x})=\pm \sqrt{5}}$                               ....(1)
∴ ${\displaystyle 7x^3+8x-\frac{7}{x^3}-\frac{8}{x}=7x^3-\frac{7}{x^3}+8x-\frac{8}{x}}$

= ${\displaystyle 7(x^3-\frac{1}{x^3})+8(x-\frac{1}{x})}$                 .....(2)
Now consider the expansion of ${\displaystyle {(x-\frac{1}{x})}^3}$ :

${\displaystyle {(x-\frac{1}{x})}^3=x^3-\frac{1}{x^3}-3(x-\frac{1}{x})}$

⇒ ${\displaystyle x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})}$

⇒ ${\displaystyle x^3-\frac{1}{x^3}={\left(\sqrt{5}\right)}^3+3\left(\sqrt{5}\right)}$         ....(3)
Now substitute the value of  ${\displaystyle x^3- \frac{1}{x^3}}$ in equation (2), we have
${\displaystyle 7x^3+8x-\frac{7}{x^3}-\frac{8}{x}=7(x^3-\frac{1}{x^3})+8(x-\frac{1}{x})}$

⇒ ${\displaystyle 7x^3+8x-\frac{7}{x^3}-\frac{8}{x}=7[{\left(\sqrt{5}\right)}^3+3\left(\sqrt{5}\right)]+8\left[\sqrt{5}\right]}$ [From(3)]

⇒ ${\displaystyle 7x^3+8x-\frac{7}{x^3}-\frac{8}{x}=7[5\left(\sqrt{5}\right)+3\left(\sqrt{5}\right)]+8\left[\sqrt{5}\right]]}$

⇒ ${\displaystyle 7x^3+8x-\frac{7}{x^3}-\frac{8}{x}=64\left[\sqrt{5}\right]}$

Question 9

If x = ${\displaystyle \frac{1}{x-5}\text{and x ≠ 5. Find}:x^2-\frac{1}{x^2}}$

Sol:

Given x = ${\displaystyle \frac{1}{x-5}}$;
By cross multiplication,
⇒ x (x - 5) = 1
⇒ x² - 5x = 1
⇒ x² - 1 = 5x                                     ....(1)

Dividing both sides by x,
${\displaystyle {(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2}$
⇒ ${\displaystyle {\left(5\right)}^2=x^2+\frac{1}{x^2}-2}$
⇒ ${\displaystyle x^2+\frac{1}{x^2}=25+2=27}$           ....(2)

Let us consider the expansion of ${\displaystyle {(x+\frac{1}{x})}^2}$ :
${\displaystyle {(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2}$

⇒ ${\displaystyle {(x+\frac{1}{x})}^2=27+2}$                   [from(2)]

⇒ ${\displaystyle {(x+\frac{1}{x})}^2=29}$

⇒ ${\displaystyle (x+\frac{1}{x})=\pm \sqrt{29}}$               ....(3)
We know that,

${\displaystyle x^2-\frac{1}{x^2}=(x+\frac{1}{x})(x-\frac{1}{x})}$= ${\displaystyle (\pm \sqrt{29})\left(5\right)}$  [From equation (1) and (3)] 

${\displaystyle x^2-\frac{1}{x^2}=\pm 5\sqrt{29}}$

Question 10

If x = ${\displaystyle \frac{1}{5-x}\text{and x ≠ 5 find}{x}^3+\frac{1}{x^3}}$

Sol:

Given x = ${\displaystyle \frac{1}{5-x}}$;
By cross multiplication
⇒ x( 5 - x ) = 1
⇒ x² - 5x = -1
⇒ x² + 1 = 5x

${\displaystyle \frac{x^2+1}{x}=5}$

⇒ ${\displaystyle [x+\frac{1}{x}]=5}$          ....(1)
We know that

${\displaystyle (x^3+\frac{1}{x^3})={(x+\frac{1}{x})}^3-3(x+\frac{1}{x})}$

= ${\displaystyle {\left(5\right)}^3-3\left(5\right)}$                   [ from equation (1) ]
⇒ ` x^3 + 1/x^3 = 125 - 15 = 110                    

Question 11

If 3a + 5b + 4c = 0, show that : 27a³ + 125b³ + 64c³ = 180 abc

Sol:

Given that 3a + 5b + 4c = 0
3a + 5b = - 4c
Cubing both sides,
(3a + 5b)³ = (-4c)³
⇒ (3a)³ + (5b)³ + 3 x 3a x 5b (3a + 5b) = -64c³
⇒ 27a³ + 125b³ + 45ab x (-4c) = -64c³
⇒ 27a³ + 125b³ - 180abc = -64c³
⇒ 27a³ + 125b³ + 64c³ = 180abc             
Hence proved.

Question 12

The sum of two numbers is 7 and the sum of their cubes is 133, find the sum of their square.

Sol:

Let a, b be the two numbers.
.'. a + b = 7 and a³ + b³ = 133
(a + b)³ = a³ + b³ + 3ab (a + b)

⇒ (7)³ = 133 + 3ab (7)
⇒ 343 = 133 + 21ab
⇒  21ab = 343 - 133 = 210
⇒ 21ab = 210
⇒ ab= 10

Now a² + b² = (a + b)² - 2ab
                     = 7² - 2 x 10 = 49 - 20 = 29

Question 13.1

Find the value of 'a':  4x² + ax + 9 = (2x + 3)²

Sol:

4x² + ax + 9 = (2x + 3)²
Comparing coefficients of x terms, we get
ax = 12x
so, a = 12

Question 13.2

Find the value of 'a': 4x² + ax + 9 = (2x - 3)²

Sol:

4x² + ax + 9 = (2x - 3)²
Comparing coefficients of x terms, we get
ax = -12x
so, a = -12

Question 13.3

Find the value of 'a': 9x² + (7a - 5)x + 25 = (3x + 5)²

Sol:

9x² + (7a - 5)x + 25 = (3x + 5)²
Comparing coefficients of x terms, we get
(7a - 5)x = 30x
7a - 5 = 30
7a = 35
a = 5

9x² + (7a - 5)x + 25 = (3x + 5)²
Comparing coefficients of x terms, we get
(7a - 5)x = 30x
7a - 5 = 30
7a = 35
a = 5

Question 14.1

If ${\displaystyle \frac{x^2+1}{x}=3\frac{1}{3} \text{and x > 1; Find If}x-\frac{1}{x}}$

Sol:

Given ${\displaystyle \frac{x^2+1}{x}=3\frac{1}{3}}$

${\displaystyle \frac{x^2+1}{x}=\frac{10}{3}}$

${\displaystyle [x+\frac{1}{x}]=\frac{10}{3}}$

Squaring on both sides, we get

${\displaystyle x^2+\frac{1}{x^2}+2=\frac{100}{9}}$

${\displaystyle x^2+\frac{1}{x^2}=\frac{100-18}{9}=\frac{82}{9}}$

${\displaystyle x-\frac{1}{x}=\sqrt{{(x+\frac{1}{x})}^2-4}=\sqrt{\frac{100}{9}-4}=\sqrt{\frac{64}{9}}=\frac{8}{3}}$

∴ ${\displaystyle x-\frac{1}{x}=\frac{8}{3}}$

Question 14.2

If ${\displaystyle \frac{x^2+1}{x}=3\frac{1}{3} \text{and x > 1; Find If}{x}^3-\frac{1}{x^3}}$

Sol:

Given ${\displaystyle \frac{x^2+1}{x}=3\frac{1}{3}}$

${\displaystyle \frac{x^2+1}{x}=\frac{10}{3}}$

${\displaystyle [x+\frac{1}{x}]=\frac{10}{3}}$

Squaring on both sides, we get

${\displaystyle x^2+\frac{1}{x^2}+2=\frac{100}{9}}$

${\displaystyle x^2+\frac{1}{x^2}=\frac{100-18}{9}=\frac{82}{9}}$

${\displaystyle x-\frac{1}{x}=\sqrt{{(x+\frac{1}{x})}^2-4}=\sqrt{\frac{100}{9}-4}=\sqrt{\frac{64}{9}}=\frac{8}{3}}$

∴ ${\displaystyle x-\frac{1}{x}=\frac{8}{3}}$

Cubing both sides, we get

${\displaystyle {(x-\frac{1}{x})}^3=\frac{512}{27}}$

${\displaystyle x^3-\frac{1}{x^3}-3(x-\frac{1}{x})=\frac{512}{27}}$

${\displaystyle x^3-\frac{1}{x^3}=\frac{512}{27}+8=\frac{512+216}{27}=\frac{728}{27}}$

Question 15.1

The difference between two positive numbers is 4 and the difference between their cubes is 316.
Find : Their product

Sol:

Given difference between two positive numbers is 4 and difference between their cubes is 316.
Let the positive numbers be a and b
a - b = 4
a³ - b³ = 316
Cubing both sides,
(a - b)³ = 64
a³ - b³ - 3ab(a - b) = 64

Given a³ - b³ = 316
So 316 - 64 = 3ab(4)
252 = 12ab
So ab = 21; product of numbers is 21

Question 15.2

The difference between two positive numbers is 4 and the difference between their cubes is 316.
Find : The sum of their squares

Sol:

Given difference between two positive numbers is 4 and difference between their cubes is 316.
Let the positive numbers be a and b
a - b = 4                         .....(1)
a³ - b³ = 316                  .....(2)

Squaring(eq 1) both sides, we get
(a - b)² = 16
a² + b² - 2ab = 16
a² + b² = 16 + 42 = 58
Sum of their squares is 58.