SELINA Solution Class 9 Expansions Chapter 4 Exercise 4a

Question 1.1

Find the square of : 2a + b

Sol:

We Know that
( a + b )² = a² + b² + 2ab
(2a + b)²  = 4a² + b² + 2 x 2a x b
                 = 4a² + b² + 4ab

Question 1.2

Find the square of : 3a + 7b

Sol:

We know that
( a + b )² = a² + b² + 2ab
( 3a + 7b )² = 9a² + 49b² + 2 x 3a x 7b
                   = 9a² + 49b² + 42ab

Question 1.3

Find the square of : 3a - 4b

Sol:

We know that
( a - b )² = a² + b² - 2ab
( 3a - 4b )² = 9a² + 16b² - 2 x 3a x 4b
                  = 9a² + 16b² - 24ab

Question 1.4

Find the square of : ${\displaystyle \frac{3a}{2b}-\frac{2b}{3a}}$

Sol:

We know that,
( a - b )2 = a2 + b2 - 2ab
${\displaystyle \frac{3a}{2b}-\frac{2b}{3a}={\left[\frac{3a}{2b}\right]}^2+{\left[\frac{2b}{3a}\right]}^2-2\times \frac{3a}{2b}\times \frac{2b}{3a}}$

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

Question 2.1

Use identities to evaluate : (101)²

Sol:

(101)² 
(101)² = (100 + 1)²
We know that,
(a + b)² = a² + b² + 2ab
∴ (100 + 1)² = 100² + 1² + 2 x 100 x 1
                     = 10,000 + 1 + 200
                     = 10,201

Question 2.2

Use identities to evaluate : (502)²

Sol:

(502)²
(502)² = (500 + 2)²
We know that,
( a + b )² = a² + b² + 2ab
∴ ( 500 + 2 )² = 500² + 2² + 2 x 500 x 2
                        = 250000 + 4 + 2000
                        = 2,52,004

Question 2.3

Use identities to evaluate : (97)²

Sol:

(97)²
(97)² = (100 - 3)²
We know that,
( a - b )² = a² + b² - 2ab
∴ (100 - 3)² = 100² + 3² - 2 x 100 x 3
                    = 10000 + 9 - 600
                    = 9,409

Quesiton 2.4

Use identities to evaluate : (998)²

Sol:

(998)²
(998)² = (1000 - 2)²
We know that
( a - b )² = a² + b² - 2ab
∴ (1000 - 2)² = 1000² + 2² - 2 x 1000 x 2
                      = 1000000 + 4 - 4000
                      = 9,96,004

Quesiton 3.1

Evalute : ${\displaystyle {(\frac{7}{8}x+\frac{4}{5}y)}^2}$

Sol:

${\displaystyle {(\frac{7}{8}x+\frac{4}{5}y)}^2}$

We know that
( a + b )² = a² + 2ab + b² 

∴ ${\displaystyle {(\frac{7}{8}x+\frac{4}{5}y)}^2={\left(\frac{7}{8}x\right)}^2+{\left(\frac{4}{5}y\right)}^2+2\times \frac{7}{8}x\times \frac{4}{5}y}$

                                = ${\displaystyle \frac{49x^2}{64}+\frac{16y^2}{25}+\frac{7xy}{5}}$

Question 3.2

Evalute : ${\displaystyle {(\frac{2x}{7}-\frac{7y}{4})}^2}$

Sol:

${\displaystyle {(\frac{2x}{7}-\frac{7y}{4})}^2}$

We know that
( a - b )² = a² - 2ab + b² 

∴ ${\displaystyle {(\frac{2x}{7}-\frac{7y}{4})}^2={\left(\frac{2x}{7}\right)}^2+{\left(\frac{7y}{4}\right)}^2-2\times \frac{2x}{7}\times \frac{7y}{4}}$

                                = ${\displaystyle \frac{4x^2}{49}+\frac{49y^2}{16}-xy}$

Question 4.1

Evaluate : ${\displaystyle {(\frac{a}{2b}+\frac{2b}{a})}^2-{(\frac{a}{2b}-\frac{2b}{a})}^2-4}$

Sol:

Consider the given expression :
Let us expand the first term : ${\displaystyle {[\frac{a}{2b}+\frac{2b}{a}]}^2}$
We know that,
( a + b )² = a2 + b² + 2ab

∴ ${\displaystyle {[\frac{a}{2b}+\frac{2b}{a}]}^2={\left(\frac{a}{2b}\right)}^2+{\left(\frac{2b}{a}\right)}^2+2\times \frac{a}{2b}\times \frac{2b}{a}}$

                                      = ${\displaystyle \frac{a^2}{{\left(4b\right)}^2}+\frac{{\left(4b\right)}^2}{a^2}+2}$       ...(1)

Let us expand the second term : ${\displaystyle {[\frac{a}{2b}-\frac{2b}{a}]}^2}$
We know that,
( a - b )² = a² + b² - 2ab

∴ ${\displaystyle {[\frac{a}{2b}-\frac{2b}{a}]}^2={\left(\frac{a}{2b}\right)}^2+{\left(\frac{2b}{a}\right)}^2-2\times \frac{a}{2b}\times \frac{2b}{a}}$

                                    = ${\displaystyle \frac{a^2}{{\left(4b\right)}^2}+\frac{{\left(4b\right)}^2}{a^2}-2}$       ...(2)
Thus from (1) and (2), the given expression is

${\displaystyle {[\frac{a}{2b}+\frac{2b}{a}]}^2-{[\frac{a}{2b}-\frac{2b}{a}]}^2-4=\frac{a^2}{{\left(4b\right)}^2}+\frac{{\left(4b\right)}^2}{a^2}+2-\frac{a^2}{{\left(4b\right)}^2}-\frac{{\left(4b\right)}^2}{a^2}+2-4}$
                                                                         = 0.

Question 4.2

Evaluate : (4a +3b)² - (4a - 3b)² + 48ab.

Sol:

(4a +3b)² - (4a - 3b)² + 48ab.
Consider the given expression:
Let us expand the first term : (4a +3b)²
We know that,
( a + b )² = a2 + b² + 2ab
∴ (4a +3b)² = (4a)2 + (3b)2 + 2 x 4a x 3b
                    = 16a2 + 9b2 + 24ab                     ....(1) 

Let us expand the second term : (4a - 3b)²
We know that,
( a + b )² = a² + b² + 2ab
∴ (4a - 3b)² = (4a)² + (3b)² - 2 x 4a x 3b
                    = 16a² + 9b² - 24ab                         ...(2)

Thus from (1) and (2), the given expression is
(4a +3b)² - (4a - 3b)² + 48ab 
= 16a2 + 9b2 + 24ab - 16a² - 9b² + 24ab + 48ab
= 96ab

Question 5

If a + b = 7 and ab = 10; find a - b.

Sol:

We know that,
( a + b )² = a² + 2ab + b²
and
( a - b )² = a² - 2ab + b²
Rewrite the above equation, we have
( a - b )² = a²  + b² - 2ab + 4ab
               = ( a + b )² - 4ab              ...(1)
Given that a + b = 7; ab = 10
Substitute the values of ( a + b ) and (ab)
in equation (1), we have
( a - b )² = (7)² - 4(10)
              = 49 - 40 = 9
⇒ a - b = ${\displaystyle \pm \sqrt{9}}$
⇒ a - b = ${\displaystyle \pm 3}$

Question 6

If a - b = 7 and ab = 18; find a + b.

Sol:

We know that,
( a - b )² = a² - 2ab + b²
and
( a + b )² = a² + 2ab + b²
Rewrite the above equation, we have
( a + b )² = a²  + b² - 2ab + 4ab
               = ( a + b )² + 4ab              ...(1)
Given that a - b = 7; ab = 18
Substitute the values of ( a - b ) and (ab)
in equation (1), we have
( a + b )² = (7)² + 4(18)
              = 49 + 72 = 121
⇒ a + b = ${\displaystyle \pm \sqrt{121}}$
⇒ a + b = ${\displaystyle \pm 11}$

Question 7

If x + y = ${\displaystyle \frac{7}{2} \text{and xy}=\frac{5}{2}}$  ; find :  x - y  and x² - y²

Sol:

We know that,
( x + y )² = x² + 2xy + y²
and
( x - y )² = x² - 2xy + y²
Rewrite the above equation, we have
( x - y )² = x²  + y² + 2xy - 4xy 
               = ( x + y )² - 4xy              ...(1)

Given that ${\displaystyle \text{x + y}=\frac{7}{2} \text{and xy}=\frac{5}{2}}$ 
Substitute the values of ( x + y ) and (xy)
in equation (1), we have
( x - y )² =${\displaystyle {\left(\frac{7}{2}\right)}^2-4\left(\frac{5}{2}\right)}$

= ${\displaystyle \frac{49}{4}-10=\frac{9}{4}}$

⇒ x - y = ${\displaystyle \pm \sqrt{\frac{9}{4}}}$

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

We know that,
x² - y² = ( x + y )( x - y )             ...(3)
From equation (2) we have,

x - y = ${\displaystyle \pm \frac{3}{2}}$

Thus equation (3) becomes,

x² - y² = ${\displaystyle \left(\frac{7}{2}\right)(\pm \frac{3}{2})}$            [ given x + y = ${\displaystyle \frac{7}{2}}$ ]

⇒ x² - y² = ${\displaystyle \pm \frac{21}{4}}$

Question 8

If a - b = 0.9 and ab = 0.36; find:
(i) a + b
(ii) a² - b².

Sol:

(i) We know that,
( a - b )² = a² - 2ab + b²
and
( a + b )² = a² + 2ab + b²
Rewrite the above equation, we have
( a + b )² = a²  + b² - 2ab + 4ab
               = ( a - b )² + 4ab              ...(1)
Given that a - b = 0.9 ; ab = 0.36
Substitute the values of ( a - b ) and (ab)
in equation (1), we have
( a + b )² = ( 0.9 )² + 4( 0.36 )
              = 0.81 + 1.44 = 2.25
⇒ a + b = ${\displaystyle \pm \sqrt{2.25}}$
⇒ a + b = ${\displaystyle \pm 1.5}$                        ..(2)

(ii) We know that,
a² - b² = ( a + b )( a - b )             ....(3)
From equation (2) we have,
a + b = ${\displaystyle \pm }$1.5
Thus equation (3) becomes,
a² - b² = ${\displaystyle (\pm 1.5)\left(0.9\right)}$                [ given a - b = 0.9 ]
⇒ a² - b² = ${\displaystyle \pm }$1.35

Question 9

If a - b = 4 and a + b = 6; find
(i) a² + b²
(ii) ab

Sol:

(i) We know that,
( a - b )² = a² - 2ab + b² 
Rewrite the above identity as,
a²  + b² = ( a - b ) + 2ab           ....(1)
Similarly, we know that,
( a + b )² = a² + 2ab + b²
Rewrite the above identity as,
 a²  + b² = ( a + b )² - 2ab                                     .....(2)
Adding the equations (1) and (2), we have
2( a² + b² ) = ( a - b )² + 2ab + ( a + b )² - 2ab
⇒ 2( a² + b² ) = ( a - b )²  + ( a + b )²
⇒ ( a² + b² ) = ${\displaystyle \frac{1}{2}[{(a-b)}^2 +{(a+b)}^2]}$          ....(3)

Given that a + b = 6 ; a - b = 4
Substitute the values of ( a + b ) and (a - b)
in equation (3), we have
a² + b² = ${\displaystyle \frac{1}{2}[{\left(4\right)}^2+{\left(6\right)}^2]}$

= ${\displaystyle \frac{1}{2}[16+36]}$

= ${\displaystyle \frac{52}{2}}$
⇒ ( a² + b² ) = 26                                            .....(4)

From equation (4), we have
a2 + b2 = 26
Consider the identity,
( a - b )2 = a2 + b2 - 2ab                                ....(5)
Substitute the value a - b = 4 and a2 + b2 = 26
in the above equation, we have
(4)2 = 26 - 2ab
⇒ 2ab = 26 - 16
⇒  2ab = 10
⇒  ab = 5

Question 10

If a + ${\displaystyle \frac{1}{a}}$= 6 and  a ≠ 0 find :
(i) ${\displaystyle a-\frac{1}{a} \left(ii\right) a^2-\frac{1}{a^2}}$

Sol:

We know that,
( a + b )² = a² + 2ab + b²
and
( a - b )² = a² - 2ab + b² 
Thus,
${\displaystyle {(a+\frac{1}{a})}^2=a^2+\frac{1}{a^2}+2\times a\times \frac{1}{a}}$

                       = ${\displaystyle a^2+\frac{1}{a^2}+2}$           .....(1)
Given that ${\displaystyle a+\frac{1}{a}}$ = 6; Substitute in equation (1), we have
${\displaystyle {\left(6\right)}^2=a^2+\frac{1}{a^2}+2}$

⇒ ${\displaystyle a^2+\frac{1}{a^2}=36-2}$

⇒ ${\displaystyle a^2+\frac{1}{a^2}=34}$                                ....(2)

Similarly, consider
${\displaystyle {(a-\frac{1}{a})}^2=a^2+\frac{1}{a^2}-2\times a\times \frac{1}{a}}$

= ${\displaystyle a^2+\frac{1}{a^2}-2}$   
= 34 - 2                                [ from (2) ]

⇒ ${\displaystyle {(a-\frac{1}{a})}^2}$ = 32

⇒ ${\displaystyle (a-\frac{1}{a})=\pm \sqrt{32}}$

⇒ ${\displaystyle (a-\frac{1}{a})=\pm 4\sqrt{2}}$                       ....(3)

(ii) We need to find  ${\displaystyle a^2-\frac{1}{a^2}}$

We know that, ${\displaystyle a^2-\frac{1}{a^2}=(a-\frac{1}{a})(a+\frac{1}{a})}$ 

${\displaystyle a-\frac{1}{a}=\pm 4\sqrt{2};a+\frac{1}{a}=6}$

Thus,
${\displaystyle a^2-\frac{1}{a^2}=(\pm 4\sqrt{2})\left(6\right)}$

⇒ ${\displaystyle a^2-\frac{1}{a^2}=(\pm 24\sqrt{2})}$

Question 11

If a - ${\displaystyle \frac{1}{a}}$= 8 and  a ≠ 0 find :
(i) ${\displaystyle a+\frac{1}{a} \left(ii\right) a^2-\frac{1}{a^2}}$

Sol:

We know that,
( a + b )² = a² + 2ab + b²
and
( a - b )² = a² - 2ab + b² 
Thus,
${\displaystyle {(a-\frac{1}{a})}^2=a^2+\frac{1}{a^2}-2\times a\times \frac{1}{a}}$

                       = ${\displaystyle a^2+\frac{1}{a^2}-2}$           .....(1)
Given that ${\displaystyle a-\frac{1}{a}}$ = 8 ; Substitute in equation (1), we have
${\displaystyle {\left(8\right)}^2=a^2+\frac{1}{a^2}-2}$

⇒ ${\displaystyle a^2+\frac{1}{a^2}=64+2}$

⇒ ${\displaystyle a^2+\frac{1}{a^2}=66}$                                ....(2)

Similarly, consider
${\displaystyle {(a-\frac{1}{a})}^2=a^2+\frac{1}{a^2}+2\times a\times \frac{1}{a}}$

= ${\displaystyle a^2+\frac{1}{a^2}+2}$   
= 66 + 2                                [ from (2) ]

⇒ ${\displaystyle {(a-\frac{1}{a})}^2}$ = 68

⇒ ${\displaystyle (a-\frac{1}{a})=\pm \sqrt{68}}$

⇒ ${\displaystyle (a-\frac{1}{a})=\pm 2\sqrt{17}}$                       ....(3)

(ii) We need to find  ${\displaystyle a^2-\frac{1}{a^2}}$

We know that, ${\displaystyle a^2-\frac{1}{a^2}=(a-\frac{1}{a})(a+\frac{1}{a})}$ 

${\displaystyle a-\frac{1}{a}=8 ;a+\frac{1}{a}=\pm 2\sqrt{17}}$

Thus,
${\displaystyle a^2-\frac{1}{a^2}=(\pm 2\sqrt{17})\left(8\right)}$

⇒ ${\displaystyle a^2-\frac{1}{a^2}=(\pm 16\sqrt{17})}$

Question 12

If a² - 3a + 1 = 0, and a ≠ 0; find : 
(i) ${\displaystyle a+\frac{1}{a} \left(ii\right)a^2+\frac{1}{a^2}}$

Sol:

(i) Consider the given equation
a² - 3a + 1 = 0
Rewrite the given equation, we have
a² + 1 = 3a
⇒ ${\displaystyle \frac{a^2+1}{a}=3}$

⇒ ${\displaystyle [\frac{a^2}{a}+\frac{1}{a}]=3}$

⇒ ${\displaystyle [a+\frac{1}{a}]=3}$             ...(1)

(ii) We need to find ${\displaystyle a^2+\frac{1}{a^2}}$ :

We know the identity, ( a + b )2 = a2 + b2 + 2ab

∴ ${\displaystyle {(a+\frac{1}{a})}^2=a^2+\frac{1}{a^2}+2}$           ....(2)

From equation (1), we have,
${\displaystyle a+\frac{1}{a}}$ = 3

Thus equation (2), becomes,

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

⇒ 9 = ${\displaystyle a^2+\frac{1}{a^2}+2}$

⇒ ${\displaystyle a^2+\frac{1}{a^2}=7}$

Question 13

If a² - 5a - 1 = 0 and a ≠ 0 ; find :
(i) ${\displaystyle a-\frac{1}{a}}$  
(ii) ${\displaystyle a+\frac{1}{a}}$
(iii) ${\displaystyle a^2-\frac{1}{a^2}}$

Sol:

(i) Consider the given equation
   a2 - 5a - 1 = 0
Rewrite the given equation, we have
a2 - 1 = 5a
⇒ ${\displaystyle \frac{a^2-1}{a}=5}$

⇒ ${\displaystyle [\frac{a^2}{a}-\frac{1}{a}]=5}$

⇒ ${\displaystyle a-\frac{1}{a}=5}$                       .....(1)

(ii) We need to find ${\displaystyle a+\frac{1}{a}}$ :
We know the identity, ( a - b )² = a² + b² - 2ab
∴ ${\displaystyle {(a-\frac{1}{a})}^2=a^2+\frac{1}{a^2}-2}$

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

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

⇒ ${\displaystyle a^2+\frac{1}{a^2}=27}$                ....(2)

Now consider the identity ( a + b )² = a² + b² + 2ab
∴ ${\displaystyle {(a+\frac{1}{a})}^2=a^2+\frac{1}{a^2}+2}$

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

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

⇒ ${\displaystyle a+\frac{1}{a}=\pm \sqrt{29}}$              .....(3)

(iii) We need to find ${\displaystyle a^2-\frac{1}{a^2}}$

We know the identity, a² - b² = ( a + b )( a - b )
∴ ${\displaystyle a^2-\frac{1}{a^2}=(a+\frac{1}{a})(a-\frac{1}{a})}$              ....(4)

From equation (3), we have,

${\displaystyle a+\frac{1}{a}=\pm \sqrt{29}}$

From equation (1), we have,
${\displaystyle a-\frac{1}{a}=5}$;

Thus identity (4), becomes,
`a^2 - 1/a^2 = (+- sqrt29)(5)

⇒ ${\displaystyle a^2-\frac{1}{a^2}=5(\pm \sqrt{29})}$

Question 14

If 3x + 4y = 16 and xy = 4; find the value of 9x² + 16y².

Sol:

Given that ( 3x + 4y ) = 16 and xy = 4
We need to find 9x² + 16y².
We know that
( a + b )² = a² + b² + 2ab
Consider the square of 3x + 4y :
∴ ( 3x + 4y )² = (3x)² + (4y)² + 2 x 3x x 4y
⇒ ( 3x + 4y )² = 9x² + 16y² + 24xy         .....(1)

Substitute the values of ( 3x + 4y ) and xy
in the above equation (1), we have
( 3x + 4y )² = 9x² + 16y² + 24xy

⇒ (16)² = 9x² + 16y² + 24(4)

⇒ 256 = 9x² + 16y² + 96

⇒ 9x² + 16y² = 160

Question 15

The number x is 2 more than the number y. If the sum of the squares of x and y is 34, then find the product of x and y.

Sol:

Given x is 2 more than y, so x = y + 2
Sum of squares of x and y is 34, so x² + y² = 34.
Replace x = y + 2 in the above equation and solve for y.
We get (y + 2)² + y² = 34

2y² + 4y - 30 = 0
y² + 2y - 15 = 0
(y + 5)(y - 3) = 0
So y = -5 or 3
For y = -5, x =-3
For y = 3, x = 5
Product of x and y is 15 in both cases.

Question 16

The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.

Sol:

Let the two positive numbers be a and b.
Given difference between them is 5 and sum of squares is 73.

So a - b = 5, a² + b² = 73
Squaring on both sides gives
(a - b)² = 5²
a² + b² - 2ab = 25
but a² + b² = 73
so 2ab = 73 - 25 = 48
ab = 24
So, the product of numbers is 24.