Exploring Algebraic Identities – NCERT Solutions Class 9 Maths Ganita Manjari includes all the questions with solutions given in NCERT Class 9 Ganita Manjari textbook.
NCERT Solutions Class 9 Exploring Algebraic Identities – NCERT Solutions Q.1:
What can you say about a and b if (a + b)2 < a2 + b2 ?
Solution:
Given:
( a + b ) 2 < a 2 + b 2 (a+b)^2<a^2+b^2 ( a + b ) 2 < a 2 + b 2
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
a 2 + 2 a b + b 2 < a 2 + b 2 a^2+2 a b+b^2<a^2+b^2 a 2 + 2 ab + b 2 < a 2 + b 2 a 2 + b 2 a^2+b^2 a 2 + b 2
2 a b < 0 2 a b <0 2 ab < 0 a b < 0 a b <0 ab < 0 a a a b b b
Q.2:
What can you say about a and b if (a + b)2 > a2 + b2 ?
Solution:
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
a = 8( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 12 2 = 64 + 32 + 32 + 16 = 144 12^2=64+32+32+16=144 1 2 2 = 64 + 32 + 32 + 16 = 144 a 2 + b 2 a^2+b^2 a 2 + b 2
( a + b ) 2 = a 2 + b 2 + 2 a b (a+b)^2=a^2+b^2+2 a b ( a + b ) 2 = a 2 + b 2 + 2 ab
( a + b ) 2 > a 2 + b 2 ⇒ 2 a b > 0 ⇒ a b > 0 (a+b)^2>a^2+b^2 \Rightarrow 2 a b>0 \Rightarrow a b>0 ( a + b ) 2 > a 2 + b 2 ⇒ 2 ab > 0 ⇒ ab > 0 a a a b b b
Therefore, when a a a b b b ( a + b ) 2 > a 2 + b 2 (a+b)^2>a^2+b^2 ( a + b ) 2 > a 2 + b 2
Q.3:
When will (a + b)2 be equal to a2 + b2 ?
Solution:
For ( a + b ) 2 (a+b)^2 ( a + b ) 2 a 2 + b 2 a^2+b^2 a 2 + b 2
a 2 + 2 a b + b 2 = a 2 + b 2 a^2+2 a b+b^2=a^2+b^2 a 2 + 2 ab + b 2 = a 2 + b 2 ⇒ 2 a b = 0 ⇒ a b = 0 \Rightarrow 2 a b=0 \Rightarrow a b=0 ⇒ 2 ab = 0 ⇒ ab = 0 a = 0 a=0 a = 0 b = 0 b=0 b = 0
Therefore, ( a + b ) 2 = a 2 + b 2 (a+b)^2=a^2+b^2 ( a + b ) 2 = a 2 + b 2
Q.4:
Did you observe that ( a + b ) 2 (a+b)^2 ( a + b ) 2 a 2 + b 2 a^2+b^2 a 2 + b 2 ( a + b ) 2 (a+b)^2 ( a + b ) 2
Solution:
Both ( a + b ) 2 (a+b)^2 ( a + b ) 2 a 2 + b 2 a^2+b^2 a 2 + b 2
( a + b ) 2 = a 2 + b 2 + 2 a b (a+b)^2=a^2+b^2+2 a b ( a + b ) 2 = a 2 + b 2 + 2 ab 2 a b 2 a b 2 ab
If 2 a b > 0 2 a b>0 2 ab > 0 ( a + b ) 2 > a 2 + b 2 (a+b)^2>a^2+b^2 ( a + b ) 2 > a 2 + b 2 If 2 a b < 0 2 a b<0 2 ab < 0 ( a + b ) 2 < a 2 + b 2 (a+b)^2<a^2+b^2 ( a + b ) 2 < a 2 + b 2 If 2 a b = 0 2 a b=0 2 ab = 0 ( a + b ) 2 = a 2 + b 2 (a+b)^2=a^2+b^2 ( a + b ) 2 = a 2 + b 2 Therefore, the deciding term is 2 a b 2 a b 2 ab
Q.5:
What if we replace b b b − b -b − b ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
Solution:
Replace b by −b in the identity:
So,
( a + ( − b ) ) 2 = a 2 + 2 a ( − b ) + ( − b ) 2 (a+(-b))^2=a^2+2 a(-b)+(-b)^2 ( a + ( − b ) ) 2 = a 2 + 2 a ( − b ) + ( − b ) 2 = a 2 − 2 a b + b 2 =a^2-2 a b+b^2 = a 2 − 2 ab + b 2 b b b − b -b − b
( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2=a^2-2 a b+b^2 ( a − b ) 2 = a 2 − 2 ab + b 2
Q.6:
Label the squares and rectangles in Fig. 4.4 so that it represents the identity ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a (a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ab + 2 b c + 2 c a
Solution:
Q.7:
Try to evaluate the following using a suitable identity:
352 652 852 1052 Do you observe any interesting pattern?
Solution:
Use the identity:
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
35 2 = ( 30 + 5 ) 2 = 30 2 + 2 ⋅ 30 ⋅ 5 + 5 2 35^2=(30+5)^2=30^2+2 \cdot 30 \cdot 5+5^2 3 5 2 = ( 30 + 5 ) 2 = 3 0 2 + 2 ⋅ 30 ⋅ 5 + 5 2 = 900 + 300 + 25 = =900+300+25= = 900 + 300 + 25 = 65 2 = ( 60 + 5 ) 2 = 60 2 + 2 ⋅ 60 ⋅ 5 + 5 2 65^2=(60+5)^2=60^2+2 \cdot 60 \cdot 5+5^2 6 5 2 = ( 60 + 5 ) 2 = 6 0 2 + 2 ⋅ 60 ⋅ 5 + 5 2 = 3600 + 600 + 25 = =3600+600+25= = 3600 + 600 + 25 = 85 2 = ( 80 + 5 ) 2 = 80 2 + 2 ⋅ 80 ⋅ 5 + 5 2 85^2=(80+5)^2=80^2+2 \cdot 80 \cdot 5+5^2 8 5 2 = ( 80 + 5 ) 2 = 8 0 2 + 2 ⋅ 80 ⋅ 5 + 5 2 = 6400 + 800 + 25 = =6400+800+25= = 6400 + 800 + 25 = 105 2 = ( 100 + 5 ) 2 = 100 2 + 2 ⋅ 100 ⋅ 5 + 5 2 105^2=(100+5)^2=100^2+2 \cdot 100 \cdot 5+5^2 10 5 2 = ( 100 + 5 ) 2 = 10 0 2 + 2 ⋅ 100 ⋅ 5 + 5 2 = 10000 + 1000 + 25 = 11025 =10000+ 1000+25=11025 = 10000 + 1000 + 25 = 11025 Interesting pattern:
All numbers end with 5 Their squares always end with 25 The remaining digits come from multiplying the number before 5 with its next number (e.g., 3 × 4 = 12 , 6 × 7 = 42 , etc.) \text { (e.g., } 3 \times 4=12,6 \times 7=42 \text {, etc.) } (e.g., 3 × 4 = 12 , 6 × 7 = 42 , etc.) So, numbers ending in 5 follow a special squaring pattern.
Q.8:
Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.
Solution:
From the figure, the same area is represented in two different ways.
a + b + c a+b+c a + b + c a + b − c a+b-c a + b − c a − b + c a-b+c a − b + c a − b − c a-b-c a − b − c So, total area:
( a + b + c ) ( a + b − c ) ( a − b + c ) ( a − b − c ) (a+b+c)(a+b-c)(a-b+c)(a-b-c) ( a + b + c ) ( a + b − c ) ( a − b + c ) ( a − b − c )
a square of side 2 a → 2 a \rightarrow 2 a → 4 a 2 4 a^2 4 a 2 a square of side 2 b → 2 b \rightarrow 2 b → 4 b 2 4 b^2 4 b 2 a square of side 2 c → 2 c \rightarrow 2 c → 4 c 2 4 c^2 4 c 2 So, total area:
4 a 2 − 4 b 2 − 4 c 2 + 4 b c (after simplification) 4 a^2-4 b^2-4 c^2+4 b c \text { (after simplification) } 4 a 2 − 4 b 2 − 4 c 2 + 4 b c (after simplification)
( a + b + c ) ( a + b − c ) ( a − b + c ) ( a − b − c ) (a+b+c)(a+b-c)(a-b+c)(a-b-c) ( a + b + c ) ( a + b − c ) ( a − b + c ) ( a − b − c ) = ( a 2 − b 2 − c 2 ) 2 − ( 2 b c ) 2 =\left(a^2-b^2-c^2\right)^2-(2 b c)^2 = ( a 2 − b 2 − c 2 ) 2 − ( 2 b c ) 2 = a 4 + b 4 + c 4 − 2 a 2 b 2 − 2 b 2 c 2 − 2 c 2 a 2 =a^4+b^4+c^4-2 a^2 b^2-2 b^2 c^2-2 c^2 a^2 = a 4 + b 4 + c 4 − 2 a 2 b 2 − 2 b 2 c 2 − 2 c 2 a 2
( a + b + c ) ( a + b − c ) ( a − b + c ) ( a − b − c ) (a+b+c)(a+b-c)(a-b+c)(a-b-c) ( a + b + c ) ( a + b − c ) ( a − b + c ) ( a − b − c ) = a 4 + b 4 + c 4 − 2 a 2 b 2 − 2 b 2 c 2 − 2 c 2 a 2 =a^4+b^4+c^4-2 a^2 b^2-2 b^2 c^2-2 c^2 a^2 = a 4 + b 4 + c 4 − 2 a 2 b 2 − 2 b 2 c 2 − 2 c 2 a 2
Q.9:
Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.
Solution:
No, a similar rectangular arrangement cannot be formed if 7 x 7 x 7 x 2 x + 5 x 2 x+5 x 2 x + 5 x
The split must allow tiles to arrange into equal rows and columns The numbers should correspond to factors of the constant term For x 2 + 7 x + 12 x^2+7 x+12 x 2 + 7 x + 12 7 x = 3 x + 4 x 7 x=3 x+4 x 7 x = 3 x + 4 x 3 × 4 = 12 3 \times 4=12 3 × 4 = 12
→ forms rectangle ( x + 3 ) ( x + 4 ) (x+3)(x+4) ( x + 3 ) ( x + 4 ) 7 x = 2 x + 5 x 7 x=2 x+5 x 7 x = 2 x + 5 x
→ 2 × 5 = 10 ≠ 12 \rightarrow 2 \times 5=10 \neq 12 → 2 × 5 = 10 = 12 3 x + 4 x 3 x+4 x 3 x + 4 x 2 x + 5 x 2 x+5 x 2 x + 5 x
Q.10:
Algebra tiles can be used to represent products and find factors. Figure out the product of x + 2 and x + 3 using algebra tiles.
Solution:
Product of x + 2 x+2 x + 2 x + 3 x+3 x + 3
Using algebra tiles:
x × x = x 2 x \times x=x^2 x × x = x 2 x × 3 = 3 x x \times 3=3 x x × 3 = 3 x 2 × x = 2 x 2 \times x=2 x 2 × x = 2 x 2 × 3 = 6 2 \times 3=6 2 × 3 = 6 Adding:
x 2 + 3 x + 2 x + 6 = x 2 + 5 x + 6 x^2+3 x+2 x+6=x^2+5 x+6 x 2 + 3 x + 2 x + 6 = x 2 + 5 x + 6
( x + 2 ) ( x + 3 ) = x 2 + 5 x + 6 (x+2)(x+3)=x^2+5 x+6 ( x + 2 ) ( x + 3 ) = x 2 + 5 x + 6
Q.11:
Algebra tiles can be used to represent products and find factors. Lay out algebra tiles for x2 + 11x + 30 in such a way that you will see its factors.
Solution:
Factorisation of x 2 + 11 x + 30 x^2+11 x+30 x 2 + 11 x + 30
To arrange algebra tiles into a rectangle:11 x 11 x 11 x
11 x = 5 x + 6 x and 5 × 6 = 30 11 x=5 x+6 x \text { and } 5 \times 6=30 11 x = 5 x + 6 x and 5 × 6 = 30
x 2 x^2 x 2 5 x 5 x 5 x 6 x 6 x 6 x 30 unit tiles forming a rectangle This forms a rectangle with sides:
( x + 5 ) and ( x + 6 ) (x+5) \text { and }(x+6) ( x + 5 ) and ( x + 6 )
Q.12:
James and Reshma were talking about algebraic identities they learnt in school.
James: ( a − b ) 2 ( a + b ) = ( a 2 − 2 a b + b 2 ) ( a + b ) (a-b)^2(a+b)=\left(a^2-2 a b+b^2\right)(a+b) ( a − b ) 2 ( a + b ) = ( a 2 − 2 ab + b 2 ) ( a + b ) ( a − b ) 2 ( a + b ) = ( a − b ) [ ( a − b ) ( a + b ) ] (a-b)^2(a+b)=(a-b)[(a-b)(a+b)] ( a − b ) 2 ( a + b ) = ( a − b ) [( a − b ) ( a + b )] = ( a − b ) ( a 2 − b 2 ) =(a-b)\left(a^2-b^2\right) = ( a − b ) ( a 2 − b 2 )
Solution:
James’s method:
( a − b ) 2 ( a + b ) = ( a 2 − 2 a b + b 2 ) ( a + b ) (a-b)^2(a+b)=\left(a^2-2 a b+b^2\right)(a+b) ( a − b ) 2 ( a + b ) = ( a 2 − 2 ab + b 2 ) ( a + b )
Reshma’s method:
( a − b ) 2 ( a + b ) = ( a − b ) [ ( a − b ) ( a + b ) ] (a-b)^2(a+b)=(a-b)[(a-b)(a+b)] ( a − b ) 2 ( a + b ) = ( a − b ) [( a − b ) ( a + b )]
( a − b ) ( a + b ) = a 2 − b 2 (a-b)(a+b)=a^2-b^2 ( a − b ) ( a + b ) = a 2 − b 2
= ( a − b ) ( a 2 − b 2 ) =(a-b)\left(a^2-b^2\right) = ( a − b ) ( a 2 − b 2 )
Conclusion:
Both methods give the same final result Reshma’s method is better because it uses identities cleverly and reduces steps New identity formed:
( a − b ) 2 ( a + b ) = ( a − b ) ( a 2 − b 2 ) (a-b)^2(a+b)=(a-b)\left(a^2-b^2\right) ( a − b ) 2 ( a + b ) = ( a − b ) ( a 2 − b 2 )
= a 3 − a 2 b − a b 2 + b 3 =a^3-a^2 b-a b^2+b^3 = a 3 − a 2 b − a b 2 + b 3
( a − b ) 2 (a-b)^2 ( a − b ) 2 ( a − b ) ( a + b ) (a-b)(a+b) ( a − b ) ( a + b ) helps create new identities and faster methods for solving problems.
Q.13:
Try to simplify the following rational expression:
36 s 2 − 12 s t + t 2 t 2 + 2 t s − 48 s 2 = ( 6 s − t ) 2 ( _ _ _ _ _ _ _ _ + _ _ _ _ _ _ _ _ ) ( _ _ _ _ _ _ _ _ + _ _ _ _ _ _ _ _ ) \frac{36 s^2-12 s t+t^2}{t^2+2 t s-48 s^2}=\frac{(6 s-t)^2}{(\_\_\_\_\_\_\_\_+\_\_\_\_\_\_\_\_)(\_\_\_\_\_\_\_\_+\_\_\_\_\_\_\_\_)} t 2 + 2 t s − 48 s 2 36 s 2 − 12 s t + t 2 = ( ________ + ________ ) ( ________ + ________ ) ( 6 s − t ) 2
Solution:
First factor both numerator and denominator.
36 s 2 − 12 s t + t 2 = ( 6 s − t ) 2 36 s^2-12 s t+t^2=(6 s-t)^2 36 s 2 − 12 s t + t 2 = ( 6 s − t ) 2 t 2 + 2 t s − 48 s 2 t^2+2 t s-48 s^2 t 2 + 2 t s − 48 s 2
sum = 2 =2 = 2 product = − 48 =-48 = − 48 These are 8 and -6 .
t 2 + 2 t s − 48 s 2 = ( t + 8 s ) ( t − 6 s ) t^2+2 t s-48 s^2=(t+8 s)(t-6 s) t 2 + 2 t s − 48 s 2 = ( t + 8 s ) ( t − 6 s )
36 s 2 − 12 s t + t 2 t 2 + 2 t s − 48 s 2 = ( 6 s − t ) 2 ( t + 8 s ) ( t − 6 s ) \frac{36 s^2-12 s t+t^2}{t^2+2 t s-48 s^2}=\frac{(6 s-t)^2}{(t+8 s)(t-6 s)} t 2 + 2 t s − 48 s 2 36 s 2 − 12 s t + t 2 = ( t + 8 s ) ( t − 6 s ) ( 6 s − t ) 2
( 6 s − t ) = − ( t − 6 s ) (6 s-t)=-(t-6 s) ( 6 s − t ) = − ( t − 6 s )
( 6 s − t ) 2 = ( t − 6 s ) 2 (6 s-t)^2=(t-6 s)^2 ( 6 s − t ) 2 = ( t − 6 s ) 2
= t − 6 s t + 8 s =\frac{t-6 s}{t+8 s} = t + 8 s t − 6 s
Q.14:
Using the identity ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 ( 7 x + 4 y ) 2 (7 x+4 y)^2 ( 7 x + 4 y ) 2
Solution:
Using the identity
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
we expand ( 7 x + 4 y ) 2 (7 x+4 y)^2 ( 7 x + 4 y ) 2
Here a = 7 x , b = 4 y a=7 x, b=4 y a = 7 x , b = 4 y
= ( 7 x ) 2 + 2 ( 7 x ) ( 4 y ) + ( 4 y ) 2 = 49 x 2 + 56 x y + 16 y 2 =(7 x)^2+2(7 x)(4 y)+(4 y)^2=49 x^2+56 x y+16 y^2 = ( 7 x ) 2 + 2 ( 7 x ) ( 4 y ) + ( 4 y ) 2 = 49 x 2 + 56 x y + 16 y 2
Q.15:
Using the identity ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 ( 7 5 x + 3 2 y ) 2 \left(\frac{7}{5} x+\frac{3}{2} y\right)^2 ( 5 7 x + 2 3 y ) 2
Solution:
Using the identity
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
we expand ( 7 5 x + 3 2 y ) 2 \left(\frac{7}{5} x+\frac{3}{2} y\right)^2 ( 5 7 x + 2 3 y ) 2
= ( 7 5 x ) 2 + 2 ( 7 5 x ) ( 3 2 y ) + ( 3 2 y ) 2 =\left(\frac{7}{5} x\right)^2+2\left(\frac{7}{5} x\right)\left(\frac{3}{2} y\right)+\left(\frac{3}{2} y\right)^2 = ( 5 7 x ) 2 + 2 ( 5 7 x ) ( 2 3 y ) + ( 2 3 y ) 2 = 49 25 x 2 + 21 5 x y + 9 4 y 2 =\frac{49}{25} x^2+\frac{21}{5} x y+\frac{9}{4} y^2 = 25 49 x 2 + 5 21 x y + 4 9 y 2
Q.16:
Using the identity ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 ( 2.5 p + 1.5 q ) 2 (2.5 p+1.5 q)^2 ( 2.5 p + 1.5 q ) 2
Solution:
Using the identity
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
we expand ( 2.5 p + 1.5 q ) 2 (2.5 p+1.5 q)^2 ( 2.5 p + 1.5 q ) 2 = ( 2.5 p ) 2 + 2 ( 2.5 p ) ( 1.5 q ) + ( 1.5 q ) 2 = (2.5 p)^2+2(2.5 p)(1.5 q)+(1.5 q)^2 = ( 2.5 p ) 2 + 2 ( 2.5 p ) ( 1.5 q ) + ( 1.5 q ) 2 = 6.25 p 2 + 7.5 p q + 2.25 q 2 =6.25 p^2+7.5 p q+2.25 q^2 = 6.25 p 2 + 7.5 pq + 2.25 q 2
Q.17:
Using the identity ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 ( 3 4 s + 8 t ) 2 \left(\frac{3}{4} s+8 t\right)^2 ( 4 3 s + 8 t ) 2
Solution:
Using the identity
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
we expand ( 3 4 s + 8 t ) 2 \left(\frac{3}{4} s+8 t\right)^2 ( 4 3 s + 8 t ) 2 = ( 3 4 s ) 2 + 2 ( 3 4 s ) ( 8 t ) + ( 8 t ) 2 =\left(\frac{3}{4} s\right)^2+2\left(\frac{3}{4} s\right)(8 t)+(8 t)^2 = ( 4 3 s ) 2 + 2 ( 4 3 s ) ( 8 t ) + ( 8 t ) 2 = 9 16 s 2 + 12 s t + 64 t 2 =\frac{9}{16} s^2+12 s t+64 t^2 = 16 9 s 2 + 12 s t + 64 t 2
Q.18:
Using the identity ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 ( x + 1 2 y ) 2 \left(x+\frac{1}{2 y}\right)^2 ( x + 2 y 1 ) 2
Solution:
Using the identity
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
we expand ( x + 1 2 y ) 2 = x 2 + 2 ( x ⋅ 1 2 y ) + ( 1 2 y ) 2 \left(x+\frac{1}{2 y}\right)^2 =x^2+2\left(x \cdot \frac{1}{2 y}\right)+\left(\frac{1}{2 y}\right)^2 ( x + 2 y 1 ) 2 = x 2 + 2 ( x ⋅ 2 y 1 ) + ( 2 y 1 ) 2 = x 2 + x y + 1 4 y 2 =x^2+\frac{x}{y}+\frac{1}{4 y^2} = x 2 + y x + 4 y 2 1
Q.19:
Using the identity ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 ( 1 x + 1 y ) 2 \left(\frac{1}{x}+\frac{1}{y}\right)^2 ( x 1 + y 1 ) 2
Solution:
Using the identity
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
we expand ( 1 x + 1 y ) 2 \left(\frac{1}{x}+\frac{1}{y}\right)^2 ( x 1 + y 1 ) 2 = ( 1 x ) 2 + 2 ( 1 x ⋅ 1 y ) + ( 1 y ) 2 =\left(\frac{1}{x}\right)^2+2\left(\frac{1}{x} \cdot \frac{1}{y}\right)+\left(\frac{1}{y}\right)^2 = ( x 1 ) 2 + 2 ( x 1 ⋅ y 1 ) + ( y 1 ) 2 = 1 x 2 + 2 x y + 1 y 2 =\frac{1}{x^2}+\frac{2}{x y}+\frac{1}{y^2} = x 2 1 + x y 2 + y 2 1
Q.20:
Using the identity ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 ( 64 ) 2 (64)^2 ( 64 ) 2
Solution:
Using the identity
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
Write 64 = 60 + 4 64=60+4 64 = 60 + 4
( 60 + 4 ) 2 = 60 2 + 2 ( 60 ) ( 4 ) + 4 2 (60+4)^2=60^2+2(60)(4)+4^2 ( 60 + 4 ) 2 = 6 0 2 + 2 ( 60 ) ( 4 ) + 4 2 = 3600 + 480 + 16 = 4096 =3600+480+16=4096 = 3600 + 480 + 16 = 4096
Q.21:
Using the identity ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 ( 105 ) 2 (105)^2 ( 105 ) 2
Solution:
Using the identity
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
Write 105 = 100 + 5 105=100+5 105 = 100 + 5
( 100 + 5 ) 2 = 100 2 + 2 ( 100 ) ( 5 ) + 5 2 (100+5)^2=100^2+2(100)(5)+5^2 ( 100 + 5 ) 2 = 10 0 2 + 2 ( 100 ) ( 5 ) + 5 2
Q.22:
Using the identity ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 ( 205 ) 2 (205)^2 ( 205 ) 2
Solution:
Using the identity
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
Write 205 = 200 + 5 205=200+5 205 = 200 + 5
( 200 + 5 ) 2 = 200 2 + 2 ( 200 ) ( 5 ) + 5 2 (200+5)^2=200^2+2(200)(5)+5^2 ( 200 + 5 ) 2 = 20 0 2 + 2 ( 200 ) ( 5 ) + 5 2 = 40000 + 2000 + 25 = 42025 =40000+2000+25=42025 = 40000 + 2000 + 25 = 42025
Q.23:
Factor 9 x 2 + 24 x y + 16 y 2 9 x^2+24 x y+16 y^2 9 x 2 + 24 x y + 16 y 2
Solution:
We use the identity
a 2 + 2 a b + b 2 = ( a + b ) 2 a^2+2 a b+b^2=(a+b)^2 a 2 + 2 ab + b 2 = ( a + b ) 2
9 x 2 + 24 x y + 16 y 2 9 x^2+24 x y+16 y^2 9 x 2 + 24 x y + 16 y 2 = ( 3 x ) 2 + 2 ( 3 x ) ( 4 y ) + ( 4 y ) 2 = ( 3 x + 4 y ) 2 =(3 x)^2+2(3 x)(4 y)+(4 y)^2=(3 x+4 y)^2 = ( 3 x ) 2 + 2 ( 3 x ) ( 4 y ) + ( 4 y ) 2 = ( 3 x + 4 y ) 2
Q.24:
Factor 4 s 2 + 20 s t + 25 t 2 4 s^2+20 s t+25 t^2 4 s 2 + 20 s t + 25 t 2
Solution:
We use the identity
a 2 + 2 a b + b 2 = ( a + b ) 2 a^2+2 a b+b^2=(a+b)^2 a 2 + 2 ab + b 2 = ( a + b ) 2
4 s 2 + 20 s t + 25 t 2 4 s^2+20 s t+25 t^2 4 s 2 + 20 s t + 25 t 2 = ( 2 s ) 2 + 2 ( 2 s ) ( 5 t ) + ( 5 t ) 2 = ( 2 s + 5 t ) 2 =(2 s)^2+2(2 s)(5 t)+(5 t)^2=(2 s+5 t)^2 = ( 2 s ) 2 + 2 ( 2 s ) ( 5 t ) + ( 5 t ) 2 = ( 2 s + 5 t ) 2
Q.25:
Factor 49 x 2 + 28 x y + 4 y 2 49 x^2+28 x y+4 y^2 49 x 2 + 28 x y + 4 y 2
Solution:
We use the identity
a 2 + 2 a b + b 2 = ( a + b ) 2 a^2+2 a b+b^2=(a+b)^2 a 2 + 2 ab + b 2 = ( a + b ) 2
49 x 2 + 28 x y + 4 y 2 49 x^2+28 x y+ 4 y^2 49 x 2 + 28 x y + 4 y 2 = ( 7 x ) 2 + 2 ( 7 x ) ( 2 y ) + ( 2 y ) 2 = ( 7 x + 2 y ) 2 =(7 x)^2+2(7 x)(2 y)+(2 y)^2=(7 x+2 y)^2 = ( 7 x ) 2 + 2 ( 7 x ) ( 2 y ) + ( 2 y ) 2 = ( 7 x + 2 y ) 2
Q.26:
Factor 64 p 2 + 32 3 p q + 4 9 q 2 64 p^2+\frac{32}{3} p q+\frac{4}{9} q^2 64 p 2 + 3 32 pq + 9 4 q 2
Solution:
We use the identity
a 2 + 2 a b + b 2 = ( a + b ) 2 a^2+2 a b+b^2=(a+b)^2 a 2 + 2 ab + b 2 = ( a + b ) 2
64 p 2 + 32 3 p q + 4 9 q 2 64 p^2+\frac{32}{3} p q +\frac{4}{9} q^2 64 p 2 + 3 32 pq + 9 4 q 2 = ( 8 p ) 2 + 2 ( 8 p ) ( 2 3 q ) ↓ + ( 2 3 q ) 2 = ( 8 p + 2 3 q ) 2 = (8 p)^2+2(8 p)\left(\frac{2}{3} q\right)^{\downarrow}+\left(\frac{2}{3} q\right)^2=\left(8 p+\frac{2}{3} q\right)^2 = ( 8 p ) 2 + 2 ( 8 p ) ( 3 2 q ) ↓ + ( 3 2 q ) 2 = ( 8 p + 3 2 q ) 2
Q.27:
Factor 3 a 2 + 4 a b + 4 3 b 2 3 a^2+4 a b+\frac{4}{3} b^2 3 a 2 + 4 ab + 3 4 b 2
Solution:
3 a 2 + 4 a b + 4 3 b 2 3 a^2+4 a b+\frac{4}{3} b^2 3 a 2 + 4 ab + 3 4 b 2
Take 3 common:
= 3 ( a 2 + 4 3 a b + 4 9 b 2 ) =3\left(a^2+\frac{4}{3} a b+\frac{4}{9} b^2\right) = 3 ( a 2 + 3 4 ab + 9 4 b 2 ) = 3 ( a + 2 3 b ) 2 =3\left(a+\frac{2}{3} b\right)^2 = 3 ( a + 3 2 b ) 2
Q.28:
Factor 9 5 s 2 + 6 s v + 5 v 2 \frac{9}{5} s^2+6 s v+5 v^2 5 9 s 2 + 6 s v + 5 v 2
Solution:
9 5 s 2 + 6 s v + 5 v 2 \frac{9}{5} s^2+6 s v+5 v^2 5 9 s 2 + 6 s v + 5 v 2
Take 1 5 \frac{1}{5} 5 1
= 1 5 ( 9 s 2 + 30 s v + 25 v 2 ) =\frac{1}{5}\left(9 s^2+30 s v+25 v^2\right) = 5 1 ( 9 s 2 + 30 s v + 25 v 2 ) = 1 5 ( 3 s + 5 v ) 2 =\frac{1}{5}(3 s+5 v)^2 = 5 1 ( 3 s + 5 v ) 2
Q.29:
Find the value of (79)2 using the identity (a – b)2 = a2 – 2ab + b2 .
Solution:
Using the identity
( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2=a^2-2 a b+b^2 ( a − b ) 2 = a 2 − 2 ab + b 2
Write 79 = 80 − 1 79=80-1 79 = 80 − 1
( 80 − 1 ) 2 = 80 2 − 2 ( 80 ) ( 1 ) + 1 2 (80-1)^2=80^2-2(80)(1)+1^2 ( 80 − 1 ) 2 = 8 0 2 − 2 ( 80 ) ( 1 ) + 1 2 = 6400 − 160 + 1 = 6241 =6400-160+1=6241 = 6400 − 160 + 1 = 6241
Q.30:
Find the value of ( 193 ) 2 (193)^2 ( 193 ) 2 2 = a2 – 2ab + b2 .
Solution:
Using the identity
( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2=a^2-2 a b+b^2 ( a − b ) 2 = a 2 − 2 ab + b 2
Write 193 = 200 − 7 193=200-7 193 = 200 − 7
( 200 − 7 ) 2 = 200 2 − 2 ( 200 ) ( 7 ) + 7 2 (200-7)^2=200^2-2(200)(7)+7^2 ( 200 − 7 ) 2 = 20 0 2 − 2 ( 200 ) ( 7 ) + 7 2 = 40000 − 2800 + 49 = 37249 =40000-2800+49=37249 = 40000 − 2800 + 49 = 37249
Q.31:
Find the value of ( 299 ) 2 (299)^2 ( 299 ) 2 2 = a2 – 2ab + b2 .
Solution:
Using the identity
( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2=a^2-2 a b+b^2 ( a − b ) 2 = a 2 − 2 ab + b 2
Write 299 = 300 − 1 299=300-1 299 = 300 − 1
( 300 − 1 ) 2 = 300 2 − 2 ( 300 ) ( 1 ) + 1 2 (300-1)^2=300^2-2(300)(1)+1^2 ( 300 − 1 ) 2 = 30 0 2 − 2 ( 300 ) ( 1 ) + 1 2 = 90000 − 600 + 1 = 89401 =90000-600+1=89401 = 90000 − 600 + 1 = 89401
Q.32:
Find 117 2 117^2 11 7 2
Solution:
We use identity:
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
117 2 = ( 100 + 17 ) 2 117^2= (100+17)^2 11 7 2 = ( 100 + 17 ) 2 = 100 2 + 2 ( 100 ) ( 17 ) + 17 2 =100^2+2(100)(17)+17^2 = 10 0 2 + 2 ( 100 ) ( 17 ) + 1 7 2 = 10000 + 3400 + 289 = 13689 =10000+3400+289=13689 = 10000 + 3400 + 289 = 13689
Q.33:
Find 78 2 78^2 7 8 2
Solution:
We use identity:
( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2=a^2-2 a b+b^2 ( a − b ) 2 = a 2 − 2 ab + b 2
78 2 = ( 80 − 2 ) 2 78^2=(80-2)^2 7 8 2 = ( 80 − 2 ) 2 = 80 2 − 2 ( 80 ) ( 2 ) + 2 2 = 6400 − 320 + 4 = 6084 =80^2-2(80)(2)+2^2=6400-320+4=6084 = 8 0 2 − 2 ( 80 ) ( 2 ) + 2 2 = 6400 − 320 + 4 = 6084
Q.34:
Find 198 2 198^2 19 8 2
Solution:
We use identity:
( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2=a^2-2 a b+b^2 ( a − b ) 2 = a 2 − 2 ab + b 2
198 2 = ( 200 − 2 ) 2 198^2= (200-2)^2 19 8 2 = ( 200 − 2 ) 2 = 200 2 − 2 ( 200 ) ( 2 ) + 2 2 =200^2-2(200)(2)+2^2 = 20 0 2 − 2 ( 200 ) ( 2 ) + 2 2 = 40000 − 800 + 4 = 39204 =40000-800+4=39204 = 40000 − 800 + 4 = 39204
Q.35:
Find 214 2 214^2 21 4 2
Solution:
We use identity:
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
214 2 = ( 200 + 14 ) 2 214^2= (200+14)^2 21 4 2 = ( 200 + 14 ) 2 = 200 2 + 2 ( 200 ) ( 14 ) + 14 2 =200^2+2(200)(14)+14^2 = 20 0 2 + 2 ( 200 ) ( 14 ) + 1 4 2 = 40000 + 5600 + 196 = 45796 =40000+5600+196=45796 = 40000 + 5600 + 196 = 45796
Q.36:
Find 1104 2 1104^2 110 4 2
Solution:
We use identity:
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
1104 2 = ( 1100 + 4 ) 2 1104^2=(1100+4)^2 110 4 2 = ( 1100 + 4 ) 2 = 1100 2 + 2 ( 1100 ) ( 4 ) + 4 2 = =1100^2+2(1100)(4)+4^2= = 110 0 2 + 2 ( 1100 ) ( 4 ) + 4 2 = 1210000 + 8800 + 16 = 1218816 1210000+8800+16=1218816 1210000 + 8800 + 16 = 1218816
Q.37:
Find 1120 2 1120^2 112 0 2
Solution:
We use identity:
( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
1120 2 = ( 1100 + 20 ) 2 1120^2=(1100+20)^2 112 0 2 = ( 1100 + 20 ) 2 = 1100 2 + 2 ( 1100 ) ( 20 ) + 20 2 = 1210000 + 44000 + 400 = 1254400 =1100^2+2(1100)(20)+20^2=1210000+44000+400=1254400 = 110 0 2 + 2 ( 1100 ) ( 20 ) + 2 0 2 = 1210000 + 44000 + 400 = 1254400
Q.38:
Factor 16 y 2 − 24 y + 9 16 y^2-24 y+9 16 y 2 − 24 y + 9
Solution:
We use identity:
( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2=a^2-2 a b+b^2 ( a − b ) 2 = a 2 − 2 ab + b 2
16 y 2 − 24 y + 9 16 y^2-24 y+9 16 y 2 − 24 y + 9
= ( 4 y ) 2 − 2 ( 4 y ) ( 3 ) + 3 2 = ( 4 y − 3 ) 2 =(4 y)^2-2(4 y)(3)+3^2=(4 y-3)^2 = ( 4 y ) 2 − 2 ( 4 y ) ( 3 ) + 3 2 = ( 4 y − 3 ) 2
Q.39:
Factor 9 4 s 2 + 6 s t + 4 t 2 \frac{9}{4} s^2+6 s t+4 t^2 4 9 s 2 + 6 s t + 4 t 2
Solution:
We use identity:( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2 a b+b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
9 4 s 2 + 6 s t + 4 t 2 \frac{9}{4} s^2+6 s t+ 4 t^2 4 9 s 2 + 6 s t + 4 t 2 = ( 3 2 s ) 2 + 2 ( 3 2 s ) ( 2 t ) + ( 2 t ) 2 = ( 3 2 s + 2 t ) 2 =\left(\frac{3}{2} s\right)^2+2\left(\frac{3}{2} s\right)(2 t)+(2 t)^2=\left(\frac{3}{2} s+2 t\right)^2 = ( 2 3 s ) 2 + 2 ( 2 3 s ) ( 2 t ) + ( 2 t ) 2 = ( 2 3 s + 2 t ) 2
Q.40:
Factor m 2 9 + m k 3 + k 2 4 + 3 n k + 2 m n + 9 n 2 \frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}+3 n k+2 m n+9 n^2 9 m 2 + 3 mk + 4 k 2 + 3 nk + 2 mn + 9 n 2
Solution:
We use identity: ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a \text { }(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ab + 2 b c + 2 c a
m 2 9 + m k 3 + k 2 4 + 3 n k + 2 m n + 9 n 2 \frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}+3 n k+2 m n+9 n^2 9 m 2 + 3 mk + 4 k 2 + 3 nk + 2 mn + 9 n 2
Group terms:
= ( m 3 ) 2 + ( k 2 ) 2 + ( 3 n ) 2 + 2 ( m 3 ⋅ k 2 ) =\left(\frac{m}{3}\right)^2+\left(\frac{k}{2}\right)^2+(3 n)^2+2\left(\frac{m}{3} \cdot \frac{k}{2}\right) = ( 3 m ) 2 + ( 2 k ) 2 + ( 3 n ) 2 + 2 ( 3 m ⋅ 2 k ) + 2 ( k 2 ⋅ 3 n ) + 2 ( m 3 ⋅ 3 n ) +2\left(\frac{k}{2} \cdot 3 n\right)+2\left(\frac{m}{3} \cdot 3 n\right) + 2 ( 2 k ⋅ 3 n ) + 2 ( 3 m ⋅ 3 n ) = ( m 3 + k 2 + 3 n ) 2 =\left(\frac{m}{3}+\frac{k}{2}+3 n\right)^2 = ( 3 m + 2 k + 3 n ) 2
Q.41:
Factor p 2 16 − 2 + 16 p 2 \frac{p^2}{16}-2+\frac{16}{p^2} 16 p 2 − 2 + p 2 16
Solution:
We use identity:
( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2=a^2-2 a b+b^2 ( a − b ) 2 = a 2 − 2 ab + b 2
p 2 16 − 2 + 16 p 2 \frac{p^2}{16}-2+\frac{16}{p^2} 16 p 2 − 2 + p 2 16
= ( p 4 ) 2 − 2 ( p 4 ⋅ 4 p ) + ( 4 p ) 2 = ( p 4 − 4 p ) 2 =\left(\frac{p}{4}\right)^2-2\left(\frac{p}{4} \cdot \frac{4}{p}\right)+\left(\frac{4}{p}\right)^2=\left(\frac{p}{4}-\frac{4}{p}\right)^2 = ( 4 p ) 2 − 2 ( 4 p ⋅ p 4 ) + ( p 4 ) 2 = ( 4 p − p 4 ) 2
Q.42:
Factor 9 a 2 + 4 b 2 + c 2 − 12 a b + 6 a c − 4 b c 9 a^2+4 b^2+c^2-12 a b+6 a c-4 b c 9 a 2 + 4 b 2 + c 2 − 12 ab + 6 a c − 4 b c
Solution:
We use identity:
( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a (a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ab + 2 b c + 2 c a
9 a 2 + 4 b 2 + c 2 − 12 a b + 6 a c − 4 b c 9 a^2+4 b^2+c^2-12 a b+6 a c -4 b c 9 a 2 + 4 b 2 + c 2 − 12 ab + 6 a c − 4 b c = ( 3 a ) 2 + ( − 2 b ) 2 + c 2 =(3 a)^2+(-2 b)^2+c^2 = ( 3 a ) 2 + ( − 2 b ) 2 + c 2 + 2 ( 3 a ) ( − 2 b ) + 2 ( 3 a ) ( c ) + 2 ( − 2 b ) ( c ) +2(3 a)(-2 b)+2(3 a)(c)+2(-2 b)(c) + 2 ( 3 a ) ( − 2 b ) + 2 ( 3 a ) ( c ) + 2 ( − 2 b ) ( c ) = ( 3 a − 2 b + c ) 2 = (3 a-2 b+c)^2 = ( 3 a − 2 b + c ) 2
Q.43:
Expand ( p + 3 q + 7 r ) 2 (p+3 q+7 r)^2 ( p + 3 q + 7 r ) 2 ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a (a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ab + 2 b c + 2 c a
Solution:
Using the identity
( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a (a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ab + 2 b c + 2 c a
( p + 3 q + 7 r ) 2 (p+3 q+7 r)^2 ( p + 3 q + 7 r ) 2
Here a = p , b = 3 q , c = 7 r a=p, b=3 q, c=7 r a = p , b = 3 q , c = 7 r
= p 2 + ( 3 q ) 2 + ( 7 r ) 2 + 2 ( p ) ( 3 q ) =p^2+(3 q)^2+(7 r)^2+2(p)(3 q) = p 2 + ( 3 q ) 2 + ( 7 r ) 2 + 2 ( p ) ( 3 q ) + 2 ( 3 q ) ( 7 r ) + 2 ( 7 r ) ( p ) +2(3 q)(7 r)+2(7 r)(p) + 2 ( 3 q ) ( 7 r ) + 2 ( 7 r ) ( p ) = p 2 + 9 q 2 + 49 r 2 + 6 p q + 42 q r + 14 p r =p^2+9 q^2+49 r^2+6 p q+42 q r+14 p r = p 2 + 9 q 2 + 49 r 2 + 6 pq + 42 q r + 14 p r
Q.44:
Expand ( 3 x − 2 y + 4 z ) 2 (3 x-2 y+4 z)^2 ( 3 x − 2 y + 4 z ) 2 ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a (a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ab + 2 b c + 2 c a
Solution:
Using the identity
( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a (a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ab + 2 b c + 2 c a
( 3 x − 2 y + 4 z ) 2 (3 x-2 y+4 z)^2 ( 3 x − 2 y + 4 z ) 2
Here a = 3 x , b = − 2 y , c = 4 z a=3 x, b=-2 y, c=4 z a = 3 x , b = − 2 y , c = 4 z
= ( 3 x ) 2 + ( − 2 y ) 2 + ( 4 z ) 2 + 2 ( 3 x ) ( − 2 y ) =(3 x)^2+(-2 y)^2+(4 z)^2+2(3 x)(-2 y) = ( 3 x ) 2 + ( − 2 y ) 2 + ( 4 z ) 2 + 2 ( 3 x ) ( − 2 y ) + 2 ( − 2 y ) ( 4 z ) + 2 ( 4 z ) ( 3 x ) +2(-2 y)(4 z)+2(4 z)(3 x) + 2 ( − 2 y ) ( 4 z ) + 2 ( 4 z ) ( 3 x ) = 9 x 2 + 4 y 2 + 16 z 2 − 12 x y − 16 y z + 24 x z =9 x^2+4 y^2+16 z^2-12 x y-16 y z+24 x z = 9 x 2 + 4 y 2 + 16 z 2 − 12 x y − 16 y z + 24 x z
Q.45:
Is this an identity?
( a + b − c ) 2 + ( a − b + c ) 2 + ( a − b − c ) 2 (a+b-c)^2+(a-b+c)^2+(a-b-c)^2 ( a + b − c ) 2 + ( a − b + c ) 2 + ( a − b − c ) 2 = 2 a 2 + 2 b 2 + 2 c 2 =2 a^2+2 b^2+2 c^2 = 2 a 2 + 2 b 2 + 2 c 2
Solution:
Let us verify by expanding the LHS.
( a + b − c ) 2 = a 2 + b 2 + c 2 + 2 a b − 2 a c − 2 b c (a+b-c)^2=a^2+b^2+c^2+2 a b-2 a c-2 b c ( a + b − c ) 2 = a 2 + b 2 + c 2 + 2 ab − 2 a c − 2 b c ( a − b + c ) 2 = a 2 + b 2 + c 2 − 2 a b + 2 a c − 2 b c (a-b+c)^2=a^2+b^2+c^2-2 a b+2 a c-2 b c ( a − b + c ) 2 = a 2 + b 2 + c 2 − 2 ab + 2 a c − 2 b c ( a − b − c ) 2 = a 2 + b 2 + c 2 − 2 a b − 2 a c + 2 b c (a-b-c)^2=a^2+b^2+c^2-2 a b-2 a c+2 b c ( a − b − c ) 2 = a 2 + b 2 + c 2 − 2 ab − 2 a c + 2 b c
L H S = 3 a 2 + 3 b 2 + 3 c 2 + ( 2 a b − 2 a b − 2 a b ) {LHS}=3 a^2+3 b^2+3 c^2+(2 a b-2 a b-2 a b) L H S = 3 a 2 + 3 b 2 + 3 c 2 + ( 2 ab − 2 ab − 2 ab ) + ( − 2 a c + 2 a c − 2 a c ) + ( − 2 b c − 2 b c + 2 b c ) +(-2 a c+2 a c-2 a c)+(-2 b c-2 b c+2 b c) + ( − 2 a c + 2 a c − 2 a c ) + ( − 2 b c − 2 b c + 2 b c ) = 3 a 2 + 3 b 2 + 3 c 2 − 2 a b − 2 a c − 2 b c =3 a^2+3 b^2+3 c^2-2 a b-2 a c-2 b c = 3 a 2 + 3 b 2 + 3 c 2 − 2 ab − 2 a c − 2 b c
2 a 2 + 2 b 2 + 2 c 2 2 a^2+2 b^2+2 c^2 2 a 2 + 2 b 2 + 2 c 2 ≠ \neq =
Q.46:
s 2 − 11 s + 24 s^2-11 s+24 s 2 − 11 s + 24
Solution:
We factor expression using the method x 2 + ( a + b ) x + a b = ( x + a ) ( x + b ) x^2+(a+b) x+a b=(x+a)(x+b) x 2 + ( a + b ) x + ab = ( x + a ) ( x + b ) s 2 − 11 s + 24 s^2-11 s+24 s 2 − 11 s + 24
We need two numbers whose sum = − 11 =-11 = − 11 = 24 : − 3 , − 8 =24:-3,-8 = 24 : − 3 , − 8
= ( s − 3 ) ( s − 8 ) =(s-3)(s-8) = ( s − 3 ) ( s − 8 )
Q.47:
(________) ( x + 1 ) = ( 3 x 2 − 4 x − 7 ) (x+1)=\left(3 x^2-4 x-7\right) ( x + 1 ) = ( 3 x 2 − 4 x − 7 )
Solution:
We factor expression using the method x 2 + ( a + b ) x + a b = ( x + a ) ( x + b ) x^2+(a+b) x+a b=(x+a)(x+b) x 2 + ( a + b ) x + ab = ( x + a ) ( x + b ) ( x + 1 ) = ( 3 x 2 − 4 x − 7 ) (x+1)=\left(3 x^2-4 x-7\right) ( x + 1 ) = ( 3 x 2 − 4 x − 7 )
Factor RHS:
3 x 2 − 4 x − 7 = 3 x 2 − 7 x + 3 x − 7 = ( 3 x − 7 ) ( x + 1 ) 3 x^2-4 x-7=3 x^2-7 x+3 x-7=(3 x-7)(x+1) 3 x 2 − 4 x − 7 = 3 x 2 − 7 x + 3 x − 7 = ( 3 x − 7 ) ( x + 1 ) = 3 x − 7 =3 x-7 = 3 x − 7
Q.48:
10x2 – 11x – 6 = (2x – ________) (________ + 2)
Solution:
We factor expression using the method x 2 + ( a + b ) x + a b = ( x + a ) ( x + b ) x^2+(a+b) x+a b=(x+a)(x+b) x 2 + ( a + b ) x + ab = ( x + a ) ( x + b ) 2 – 11x – 6 = (2x – ________) (________ + 2)
Try factors of 10 x 2 10 x^2 10 x 2
= ( 2 x − 3 ) ( 5 x + 2 ) =(2 x-3)(5 x+2) = ( 2 x − 3 ) ( 5 x + 2 ) = 3 =3 = 3 5 x 5 x 5 x
Q.49:
6 x 2 + 7 x + 2 = 6 x^2+7 x+2= 6 x 2 + 7 x + 2 =
Solution:
We factor expression using the method x 2 + ( a + b ) x + a b = ( x + a ) ( x + b ) x^2+(a+b) x+a b=(x+a)(x+b) x 2 + ( a + b ) x + ab = ( x + a ) ( x + b ) 6 x 2 + 7 x + 2 6 x^2+7 x+2 6 x 2 + 7 x + 2
Find numbers with sum 7, product 12: 3,4
= 6 x 2 + 3 x + 4 x + 2 = 3 x ( 2 x + 1 ) =6 x^2+3 x+4 x+2=3 x(2 x+1) = 6 x 2 + 3 x + 4 x + 2 = 3 x ( 2 x + 1 ) + 2 ( 2 x + 1 ) = ( 3 x + 2 ) ( 2 x + 1 ) +2(2 x+1)=(3 x+2)(2 x+1) + 2 ( 2 x + 1 ) = ( 3 x + 2 ) ( 2 x + 1 )
Q.50:
Select and use the identity that will help you to find ( 41 ) 2 (41)^2 ( 41 ) 2
Solution:
We use identity (a + b)2
41 2 = ( 40 + 1 ) 2 41^2=(40+1)^2 4 1 2 = ( 40 + 1 ) 2 = 1600 + 80 + 1 = 1681 =1600+80+1=1681 = 1600 + 80 + 1 = 1681
Q.51:
Select and use the identity that will help you to find ( 27 ) 2 (27)^2 ( 27 ) 2
Solution:
We use identity (a – b)2
27 2 = ( 30 − 3 ) 2 27^2=(30-3)^2 2 7 2 = ( 30 − 3 ) 2 = 900 − 180 + 9 = 729 =900-180+9=729 = 900 − 180 + 9 = 729
Q.52:
Select and use the identity that will help you to find 23 × 17 23 \times 17 23 × 17
Solution:
We use identity (a + b)(a – b)
23 × 17 = ( 20 + 3 ) ( 20 − 3 ) 23 \times 17=(20+3)(20 -3) 23 × 17 = ( 20 + 3 ) ( 20 − 3 ) = 20 2 − 3 2 = 400 − 9 = 391 =20^2-3^2=400-9=391 = 2 0 2 − 3 2 = 400 − 9 = 391
Q.53:
Select and use the identity that will help you to find ( 135 ) 2 (135)^2 ( 135 ) 2
Solution:
We use identity (a + b)2
135 2 = ( 100 + 35 ) 2 135^2=(100+35)^2 13 5 2 = ( 100 + 35 ) 2 = 10000 + 7000 + 1225 = 18225 =10000+7000+1225=18225 = 10000 + 7000 + 1225 = 18225
Q.54:
Select and use the identity that will help you to find ( 97 ) 2 (97)^2 ( 97 ) 2
Solution:
We use identity (a – b)2
97 2 = ( 100 − 3 ) 2 97^2=(100-3)^2 9 7 2 = ( 100 − 3 ) 2 = 10000 − 600 + 9 = 9409 =10000-600+9=9409 = 10000 − 600 + 9 = 9409
Q.55:
Select and use the identity that will help you to find 18 × 29 18 \times 29 18 × 29
Solution:
18 × 29 = ( 23 − 5 ) ( 23 + 6 ) 18 \times 29=(23-5)(23+6) 18 × 29 = ( 23 − 5 ) ( 23 + 6 ) 2
18 × 29 = ( ( 18 + 29 ) / 2 ) 2 − ( ( 29 − 18 ) / 2 ) 2 18 \times 29=((18+29) / 2)^2-((29-18) / 2)^2 18 × 29 = (( 18 + 29 ) /2 ) 2 − (( 29 − 18 ) /2 ) 2
= ( 20 − 2 ) ( 20 + 9 ) =(20-2)(20+9) = ( 20 − 2 ) ( 20 + 9 )
= 522 (direct) =522 \text { (direct) } = 522 (direct)
Q.56:
Select and use the identity that will help you to find ( 34 × 43 ) (34 \times 43) ( 34 × 43 )
Solution:
We use identity (a – b)2 34 × 43 = ( ( 34 + 43 ) / 2 ) 2 − ( ( 43 − 34 ) / 2 ) 2 34 \times 43=((34+43) / 2)^2-((43-34) / 2)^2 34 × 43 = (( 34 + 43 ) /2 ) 2 − (( 43 − 34 ) /2 ) 2 = 77 / 2 , 9 / 2 ⇒ = ( 77 2 ) 2 − ( 9 2 ) 2 =77 / 2,9 / 2 \Rightarrow=\left(\frac{77}{2}\right)^2-\left(\frac{9}{2}\right)^2 = 77/2 , 9/2 ⇒= ( 2 77 ) 2 − ( 2 9 ) 2 = 5929 − 81 4 = 5848 4 = 1462 =\frac{5929-81}{4}=\frac{5848}{4}=1462 = 4 5929 − 81 = 4 5848 = 1462
Q.57:
Select and use the identity that will help you to find ( 205 ) 2 (205)^2 ( 205 ) 2
Solution:
We use identity (a + b)2
205 2 = ( 200 + 5 ) 2 205^2=(200+5)^2 20 5 2 = ( 200 + 5 ) 2 = 40000 + 2000 + 25 = 42025 =40000+2000+25=42025 = 40000 + 2000 + 25 = 42025
Q.58:
Factor: 9 a 2 + b 2 + 4 c 2 − 6 a b + 12 a c − 4 b c 9 a^2+b^2+4 c^2-6 a b+12 a c-4 b c 9 a 2 + b 2 + 4 c 2 − 6 ab + 12 a c − 4 b c
Solution:
9 a 2 + b 2 + 4 c 2 − 6 a b + 12 a c − 4 b c 9 a^2+b^2+4 c^2-6 a b+12 a c -4 b c 9 a 2 + b 2 + 4 c 2 − 6 ab + 12 a c − 4 b c = ( 3 a ) 2 + ( − b ) 2 + ( 2 c ) 2 + 2 ( 3 a ) ( − b ) =(3 a)^2+(-b)^2+(2 c)^2 +2(3 a)(-b) = ( 3 a ) 2 + ( − b ) 2 + ( 2 c ) 2 + 2 ( 3 a ) ( − b ) + 2 ( 3 a ) ( 2 c ) + 2 ( − b ) ( 2 c ) +2(3 a)(2 c)+2(-b)(2 c) + 2 ( 3 a ) ( 2 c ) + 2 ( − b ) ( 2 c ) = ( 3 a − b + 2 c ) 2 = (3 a-b+2 c)^2 = ( 3 a − b + 2 c ) 2
Q.59:
Factor: 16 s 2 + 25 t 2 − 40 s t 16 s^2+25 t^2-40 s t 16 s 2 + 25 t 2 − 40 s t
Solution:
16 s 2 + 25 t 2 − 40 s t 16 s^2+25 t^2-40 s t 16 s 2 + 25 t 2 − 40 s t = ( 4 s ) 2 + ( 5 t ) 2 − 2 ( 4 s ) ( 5 t ) = ( 4 s − 5 t ) 2 =(4 s)^2+(5 t)^2-2(4 s)(5 t)=(4 s-5 t)^2 = ( 4 s ) 2 + ( 5 t ) 2 − 2 ( 4 s ) ( 5 t ) = ( 4 s − 5 t ) 2
Q.60:
Factor: r 2 − r − 42 r^2-r-42 r 2 − r − 42
Solution:
r 2 − r − 42 r^2-r-42 r 2 − r − 42
Find numbers: product = − 42 =-42 = − 42 = − 1 → − 7 , 6 =-1 \rightarrow-7,6 = − 1 → − 7 , 6
= ( r − 7 ) ( r + 6 ) =(r-7)(r+6) = ( r − 7 ) ( r + 6 )
Q.61:
Factor: 49 g 2 + 14 g h + h 2 49 g^2+14 g h+h^2 49 g 2 + 14 g h + h 2
Solution:
49 g 2 + 14 g h + h 2 49 g^2+14 g h+h^2 49 g 2 + 14 g h + h 2 = ( 7 g ) 2 + 2 ( 7 g ) ( h ) + h 2 = ( 7 g + h ) 2 =(7 g)^2+2(7 g)(h)+h^2=(7 g+h)^2 = ( 7 g ) 2 + 2 ( 7 g ) ( h ) + h 2 = ( 7 g + h ) 2
Q.62:
Factor: 64 u 2 + 121 v 2 + 4 w 2 − 176 u v − 32 u w + 44 v w \text {} 64 u^2+121 v^2+4 w^2-176 u v-32 u w+44 v w 64 u 2 + 121 v 2 + 4 w 2 − 176 uv − 32 u w + 44 v w
Solution:
64 u 2 + 121 v 2 + 4 w 2 − 176 u v − 32 u w + 44 v w \text {} 64 u^2+121 v^2+4 w^2-176 u v-32 u w+44 v w 64 u 2 + 121 v 2 + 4 w 2 − 176 uv − 32 u w + 44 v w = ( 8 u ) 2 + ( − 11 v ) 2 + ( − 2 w ) 2 + 2 ( 8 u ) ( − 11 v ) =(8 u)^2+(-11 v)^2+(-2 w)^2 +2(8 u)(-11 v) = ( 8 u ) 2 + ( − 11 v ) 2 + ( − 2 w ) 2 + 2 ( 8 u ) ( − 11 v ) + 2 ( 8 u ) ( − 2 w ) + 2 ( − 11 v ) ( − 2 w ) +2(8 u)(-2 w)+2(-11 v)(-2 w) + 2 ( 8 u ) ( − 2 w ) + 2 ( − 11 v ) ( − 2 w ) = ( 8 u − 11 v − 2 w ) 2 =(8 u-11 v-2 w)^2 = ( 8 u − 11 v − 2 w ) 2
Q.63:
Simplify the rational expression 3 p 2 − 3 p q − 18 q 2 p 2 + 3 p q − 10 q 2 \frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2} p 2 + 3 pq − 10 q 2 3 p 2 − 3 pq − 18 q 2
Solution:
3 p 2 − 3 p q − 18 q 2 p 2 + 3 p q − 10 q 2 \frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2} p 2 + 3 pq − 10 q 2 3 p 2 − 3 pq − 18 q 2
Factor:
= 3 ( p 2 − p q − 6 q 2 ) ( p 2 + 3 p q − 10 q 2 ) = 3 ( p − 3 q ) ( p + 2 q ) ( p + 5 q ) ( p − 2 q ) =\frac{3\left(p^2-p q-6 q^2\right)}{\left(p^2+3 p q-10 q^2\right)}=\frac{3(p-3 q)(p+2 q)}{(p+5 q)(p-2 q)} = ( p 2 + 3 pq − 10 q 2 ) 3 ( p 2 − pq − 6 q 2 ) = ( p + 5 q ) ( p − 2 q ) 3 ( p − 3 q ) ( p + 2 q )
Q.64:
Simplify the rational expression n 3 − 3 n 2 m + 3 n m 2 − m 3 5 m 2 − 10 m n + 5 n 2 \frac{n^3-3 n^2 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2} 5 m 2 − 10 mn + 5 n 2 n 3 − 3 n 2 m + 3 n m 2 − m 3
Solution:
n 3 − 3 n 2 m + 3 n m 2 − m 3 5 m 2 − 10 m n + 5 n 2 \frac{n^3-3 n^2 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2} 5 m 2 − 10 mn + 5 n 2 n 3 − 3 n 2 m + 3 n m 2 − m 3 = ( n − m ) 3 5 ( m − n ) 2 = ( n − m ) 3 5 ( n − m ) 2 = n − m 5 =\frac{(n-m)^3}{5(m-n)^2}=\frac{(n-m)^3}{5(n-m)^2}=\frac{n-m}{5} = 5 ( m − n ) 2 ( n − m ) 3 = 5 ( n − m ) 2 ( n − m ) 3 = 5 n − m
Q.65:
Simplify the rational expression w 3 − v 3 + x 3 + 3 w v x w 2 + v 2 + x 2 − 2 w v − 2 v x + 2 w x \frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x} w 2 + v 2 + x 2 − 2 w v − 2 v x + 2 w x w 3 − v 3 + x 3 + 3 w v x
Solution:
w 3 − v 3 + x 3 + 3 w v x w 2 + v 2 + x 2 − 2 w v − 2 v x + 2 w x \frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x} w 2 + v 2 + x 2 − 2 w v − 2 v x + 2 w x w 3 − v 3 + x 3 + 3 w v x
Use identity:
w 3 + x 3 − v 3 + 3 w v x = w^3+x^3-v^3+3 w v x= w 3 + x 3 − v 3 + 3 w v x = ( w + x − v ) ( w 2 + v 2 + x 2 − 2 w v − 2 v x + 2 w x ) (w+x-v)\left(w^2+v^2+x^2-2 w v-2 v x+2 w x\right) ( w + x − v ) ( w 2 + v 2 + x 2 − 2 w v − 2 v x + 2 w x )
= w + x − v =w+x-v = w + x − v
Q.66:
Simplify the rational expression 4 y 2 − 20 y z + 25 z 2 25 z 2 − 4 y 2 \frac{4 y^2-20 y z+25 z^2}{25 z^2-4 y^2} 25 z 2 − 4 y 2 4 y 2 − 20 y z + 25 z 2
Solution:
4 y 2 − 20 y z + 25 z 2 25 z 2 − 4 y 2 \frac{4 y^2-20 y z+25 z^2}{25 z^2-4 y^2} 25 z 2 − 4 y 2 4 y 2 − 20 y z + 25 z 2 = ( 2 y − 5 z ) 2 ( 5 z − 2 y ) ( 5 z + 2 y ) =\frac{(2 y-5 z)^2}{(5 z-2 y)(5 z+2 y)} = ( 5 z − 2 y ) ( 5 z + 2 y ) ( 2 y − 5 z ) 2 = ( 2 y − 5 z ) 2 − ( 2 y − 5 z ) ( 5 z + 2 y ) = 5 z − 2 y 5 z + 2 y =\frac{(2 y-5 z)^2}{-(2 y-5 z)(5 z+2 y)}=\frac{5 z-2 y}{5 z+2 y} = − ( 2 y − 5 z ) ( 5 z + 2 y ) ( 2 y − 5 z ) 2 = 5 z + 2 y 5 z − 2 y
Q.67:
Simplify the rational expression ( x 2 + x − 6 ) ( x 2 − 7 x + 12 ) ( x 2 − 6 x + 8 ) ( x 2 − 9 ) \frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)} ( x 2 − 6 x + 8 ) ( x 2 − 9 ) ( x 2 + x − 6 ) ( x 2 − 7 x + 12 )
Solution:
( x 2 + x − 6 ) ( x 2 − 7 x + 12 ) ( x 2 − 6 x + 8 ) ( x 2 − 9 ) \frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)} ( x 2 − 6 x + 8 ) ( x 2 − 9 ) ( x 2 + x − 6 ) ( x 2 − 7 x + 12 )
Factor:
= ( x + 3 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 2 ) ( x − 4 ) ( x − 3 ) ( x + 3 ) =\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)} = ( x − 2 ) ( x − 4 ) ( x − 3 ) ( x + 3 ) ( x + 3 ) ( x − 2 ) ( x − 3 ) ( x − 4 )
= 1 =1 = 1
Q.68:
Simplify the rational expression p 4 − 16 p 2 − 4 p + 4 \frac{p^4-16}{p^2-4 p+4} p 2 − 4 p + 4 p 4 − 16
Solution:
p 4 − 16 p 2 − 4 p + 4 \frac{p^4-16}{p^2-4 p+4} p 2 − 4 p + 4 p 4 − 16 = ( p 2 − 4 ) ( p 2 + 4 ) ( p − 2 ) 2 = ( p − 2 ) ( p + 2 ) ( p 2 + 4 ) ( p − 2 ) 2 =\frac{\left(p^2-4\right)\left(p^2+4\right)}{(p-2)^2}=\frac{(p-2)(p+2)\left(p^2+4\right)}{(p-2)^2} = ( p − 2 ) 2 ( p 2 − 4 ) ( p 2 + 4 ) = ( p − 2 ) 2 ( p − 2 ) ( p + 2 ) ( p 2 + 4 ) p − 2 p-2 p − 2
= ( p + 2 ) ( p 2 + 4 ) p − 2 =\frac{(p+2)\left(p^2+4\right)}{p-2} = p − 2 ( p + 2 ) ( p 2 + 4 )
Q.69:
Use suitable identity to find the product ( − 3 x + 4 ) 2 (-3 x+4)^2 ( − 3 x + 4 ) 2
Solution:
( − 3 x + 4 ) 2 (-3 x+4)^2 ( − 3 x + 4 ) 2
Using ( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2=a^2-2 a b+b^2 ( a − b ) 2 = a 2 − 2 ab + b 2
= ( − 3 x ) 2 + 2 ( − 3 x ) ( 4 ) + 4 2 = 9 x 2 − 24 x + 16 =(-3 x)^2+2(-3 x)(4)+4^2=9 x^2-24 x+16 = ( − 3 x ) 2 + 2 ( − 3 x ) ( 4 ) + 4 2 = 9 x 2 − 24 x + 16
Q.70:
Use suitable identity to find the product ( 2 s + 7 ) ( 2 s − 7 ) (2 s+7)(2 s-7) ( 2 s + 7 ) ( 2 s − 7 )
Solution:
( 2 s + 7 ) ( 2 s − 7 ) (2 s+7)(2 s-7) ( 2 s + 7 ) ( 2 s − 7 )
Using a 2 − b 2 = ( a + b ) ( a − b ) a^2-b^2=(a+b)(a-b) a 2 − b 2 = ( a + b ) ( a − b )
= ( 2 s ) 2 − 7 2 = 4 s 2 − 49 =(2 s)^2-7^2=4 s^2-49 = ( 2 s ) 2 − 7 2 = 4 s 2 − 49
Q.71:
Use suitable identity to find the product ( p 2 + 1 2 ) ( p 2 − 1 2 ) \left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right) ( p 2 + 2 1 ) ( p 2 − 2 1 )
Solution:
( p 2 + 1 2 ) ( p 2 − 1 2 ) \left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right) ( p 2 + 2 1 ) ( p 2 − 2 1 ) = ( p 2 ) 2 − ( 1 2 ) 2 = p 4 − 1 4 =\left(p^2\right)^2-\left(\frac{1}{2}\right)^2=p^4-\frac{1}{4} = ( p 2 ) 2 − ( 2 1 ) 2 = p 4 − 4 1
Q.72:
Use suitable identity to find the product ( 2 n + 7 ) ( 2 n − 7 ) (2 n+7)(2 n-7) ( 2 n + 7 ) ( 2 n − 7 )
Solution:
Here, a = 2 n a=2 n a = 2 n b = 7 b=7 b = 7
( 2 n + 7 ) ( 2 n − 7 ) = ( 2 n ) 2 − 7 2 (2 n+7)(2 n-7)=(2 n)^2-7^2 ( 2 n + 7 ) ( 2 n − 7 ) = ( 2 n ) 2 − 7 2 = 4 n 2 − 49 =4 n^2-49 = 4 n 2 − 49
Q.73:
Use suitable identity to find the product ( s − 2 t ) ( s 2 + 2 s t + 4 t 2 ) (s-2 t)\left(s^2+2 s t+4 t^2\right) ( s − 2 t ) ( s 2 + 2 s t + 4 t 2 )
Solution:
( s − 2 t ) ( s 2 + 2 s t + 4 t 2 ) (s-2 t)\left(s^2+2 s t+4 t^2\right) ( s − 2 t ) ( s 2 + 2 s t + 4 t 2 ) Using a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) \text { Using } a^3-b^3=(a-b)\left(a^2+a b+b^2\right) Using a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 ) = s 3 − ( 2 t ) 3 = s 3 − 8 t 3 =s^3-(2 t)^3=s^3-8 t^3 = s 3 − ( 2 t ) 3 = s 3 − 8 t 3
Q.74:
Use suitable identity to find the product ( 1 2 r − 4 r ) 2 \left(\frac{1}{2 r}-4 r\right)^2 ( 2 r 1 − 4 r ) 2
Solution:
( 1 2 r − 4 r ) 2 \left(\frac{1}{2 r}-4 r\right)^2 ( 2 r 1 − 4 r ) 2 = ( 1 2 r ) 2 − 2 ( 1 2 r ⋅ 4 r ) + ( 4 r ) 2 =\left(\frac{1}{2 r}\right)^2-2\left(\frac{1}{2 r} \cdot 4 r\right)+(4 r)^2 = ( 2 r 1 ) 2 − 2 ( 2 r 1 ⋅ 4 r ) + ( 4 r ) 2 = 1 4 r 2 − 4 + 16 r 2 =\frac{1}{4 r^2}-4+16 r^2 = 4 r 2 1 − 4 + 16 r 2
Q.75:
Use suitable identity to find the product ( − 3 m + 4 k − l ) 2 (-3 m+4 k-l)^2 ( − 3 m + 4 k − l ) 2
Solution:
( − 3 m + 4 k − l ) 2 (-3 m+4 k-l)^2 ( − 3 m + 4 k − l ) 2
Using ( a + b + c ) 2 (a+b+c)^2 ( a + b + c ) 2
= 9 m 2 + 16 k 2 + l 2 − 24 m k + 6 m l − 8 k l =9 m^2+16 k^2+l^2-24 m k+6 m l-8 k l = 9 m 2 + 16 k 2 + l 2 − 24 mk + 6 m l − 8 k l
Q.76:
Use suitable identity to find the product ( x − 1 3 y ) 3 \left(x-\frac{1}{3} y\right)^3 ( x − 3 1 y ) 3
Solution:
( x − 1 3 y ) 3 \left(x-\frac{1}{3} y\right)^3 ( x − 3 1 y ) 3 Using ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 \text { Using }(a-b)^3=a^3-3 a^2 b +3 a b^2-b^3 Using ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 = x 3 − x 2 y + 1 3 x y 2 − 1 27 y 3 =x^3-x^2 y+\frac{1}{3} x y^2-\frac{1}{27} y^3 = x 3 − x 2 y + 3 1 x y 2 − 27 1 y 3
Q.77:
Use suitable identity to find the product ( 7 2 k − 2 3 m ) 3 \left(\frac{7}{2} k-\frac{2}{3} m\right)^3 ( 2 7 k − 3 2 m ) 3
Solution:
( 7 2 k − 2 3 m ) 3 \left(\frac{7}{2} k-\frac{2}{3} m\right)^3 ( 2 7 k − 3 2 m ) 3 = ( 7 2 k ) 3 − 3 ( 7 2 k ) 2 ( 2 3 m ) + 3 ( 7 2 k ) ( 2 3 m ) 2 − ( 2 3 m ) 3 =\left(\frac{7}{2} k\right)^3-3\left(\frac{7}{2} k\right)^2\left(\frac{2}{3} m\right)+3\left(\frac{7}{2} k\right)\left(\frac{2}{3} m\right)^2-\left(\frac{2}{3} m\right)^3 = ( 2 7 k ) 3 − 3 ( 2 7 k ) 2 ( 3 2 m ) + 3 ( 2 7 k ) ( 3 2 m ) 2 − ( 3 2 m ) 3 = 343 8 k 3 − 49 2 k 2 m + 14 3 k m 2 − 8 27 m 3 =\frac{343}{8} k^3-\frac{49}{2} k^2 m+\frac{14}{3} k m^2-\frac{8}{27} m^3 = 8 343 k 3 − 2 49 k 2 m + 3 14 k m 2 − 27 8 m 3
Q.78:
Find the value of 17 × 21 17 \times 21 17 × 21
Solution:
17 × 21 17 \times 21 17 × 21
Use ( a − b ) ( a + b ) = a 2 − b 2 (a-b)(a+b)=a^2-b^2 ( a − b ) ( a + b ) = a 2 − b 2
= ( 19 − 2 ) ( 19 + 2 ) = 19 2 − 2 2 = 361 − 4 = 357 =(19-2)(19+2)=19^2-2^2=361-4=357 = ( 19 − 2 ) ( 19 + 2 ) = 1 9 2 − 2 2 = 361 − 4 = 357
Q.79:
Find the value of 104 × 96 104 \times 96 104 × 96
Solution:
104 × 96 104 \times 96 104 × 96 = ( 100 + 4 ) ( 100 − 4 ) = 100 2 − 4 2 = 10000 − 16 = 9984 =(100+4)(100-4)=100^2-4^2=10000-16=9984 = ( 100 + 4 ) ( 100 − 4 ) = 10 0 2 − 4 2 = 10000 − 16 = 9984
Q.80:
Find the value of 24 × 16 24 \times 16 24 × 16
Solution:
24 × 16 24 \times 16 24 × 16 = ( 20 + 4 ) ( 20 − 4 ) = 20 2 − 4 2 = 400 − 16 = 384 =(20+4)(20-4)=20^2-4^2=400-16=384 = ( 20 + 4 ) ( 20 − 4 ) = 2 0 2 − 4 2 = 400 − 16 = 384
Q.81:
Find the value of 147 3 147^3 14 7 3
Solution:
Use( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 (a-b)^3= a^3-3 a^2 b+3 a b^2-b^3 ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 = ( 150 − 3 ) 3 = = (150-3)^3= = ( 150 − 3 ) 3 = 150 3 − 3 ( 150 2 ) ( 3 ) + 3 ( 150 ) ( 9 ) − 27 150^3-3\left(150^2\right)(3)+3(150)(9)-27 15 0 3 − 3 ( 15 0 2 ) ( 3 ) + 3 ( 150 ) ( 9 ) − 27 = 3375000 − 202500 + 4050 − 27 = 3176523 =3375000-202500+4050-27=3176523 = 3375000 − 202500 + 4050 − 27 = 3176523
Q.82:
Find the value of 199 3 199^3 19 9 3
Solution:
199 3 199^3 19 9 3 = ( 200 − 1 ) 3 = 200 3 − 3 ( 200 2 ) ( 1 ) + 3 ( 200 ) ( 1 ) − 1 =(200-1)^3=200^3-3\left(200^2\right)(1)+3(200)(1)-1 = ( 200 − 1 ) 3 = 20 0 3 − 3 ( 20 0 2 ) ( 1 ) + 3 ( 200 ) ( 1 ) − 1 = 8000000 − 120000 + 600 − 1 = 7880599 =8000000-120000+600-1=7880599 = 8000000 − 120000 + 600 − 1 = 7880599
Q.83:
Find the value of 127 3 127^3 12 7 3
Solution:
127 3 127^3 12 7 3 = ( 100 + 27 ) 3 = 100 3 + 3 ( 100 2 ) ( 27 ) + 3 ( 100 ) ( 27 2 ) + 27 3 =(100+27)^3=100^3+3\left(100^2\right)(27)+3(100)\left(27^2\right)+27^3 = ( 100 + 27 ) 3 = 10 0 3 + 3 ( 10 0 2 ) ( 27 ) + 3 ( 100 ) ( 2 7 2 ) + 2 7 3 = 1000000 + 810000 + 218700 + 19683 = 2048383 =1000000+810000+218700+19683=2048383 = 1000000 + 810000 + 218700 + 19683 = 2048383
Q.84:
Find the value of ( − 107 ) 3 (-107)^3 ( − 107 ) 3
Solution:
( − 107 ) 3 (-107)^3 ( − 107 ) 3 = − ( 107 3 ) =-\left(107^3\right) = − ( 10 7 3 ) 107 3 = ( 100 + 7 ) 3 = 1000000 + 210000 + 14700 + 343 = 1225043 107^3=(100+7)^3=1000000+210000+14700+343=1225043 10 7 3 = ( 100 + 7 ) 3 = 1000000 + 210000 + 14700 + 343 = 1225043 ⇒ ( − 107 ) 3 = − 1225043 \Rightarrow(-107)^3=-1225043 ⇒ ( − 107 ) 3 = − 1225043
Q.85:
Find the value of ( − 299 ) 3 (-299)^3 ( − 299 ) 3
Solution:
( − 299 ) 3 = − ( 299 3 ) (-299)^3=-\left(299^3\right) ( − 299 ) 3 = − ( 29 9 3 )
( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 (a-b)^3=a^3-3 a^2 b+3 a b^2-b^3 ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 299 = 300 − 1 299=300-1 299 = 300 − 1
299 3 = ( 300 − 1 ) 3 299^3=(300-1)^3 29 9 3 = ( 300 − 1 ) 3 = 300 3 − 3 ( 300 2 ) ( 1 ) + 3 ( 300 ) ( 1 2 ) − 1 3 =300^3-3\left(300^2\right)(1)+3(300)\left(1^2\right)-1^3 = 30 0 3 − 3 ( 30 0 2 ) ( 1 ) + 3 ( 300 ) ( 1 2 ) − 1 3 = 27000000 − 270000 + 900 − 1 =27000000-270000+900-1 = 27000000 − 270000 + 900 − 1 = 26730900 − 1 = 26730899 =26730900-1=26730899 = 26730900 − 1 = 26730899
( − 299 ) 3 = − 26730899 (-299)^3=-26730899 ( − 299 ) 3 = − 26730899
Q.86:
Factor the algebraic expression 4 y 2 + 1 + 1 16 y 2 4 y^2+1+\frac{1}{16 y^2} 4 y 2 + 1 + 16 y 2 1
Solution:
4 y 2 + 1 + 1 16 y 2 4 y^2+1+\frac{1}{16 y^2} 4 y 2 + 1 + 16 y 2 1 = ( 2 y ) 2 + 2 ( 2 y ) ( 1 4 y ) + ( 1 4 y ) 2 = ( 2 y + 1 4 y ) 2 =(2 y)^2+2(2 y)\left(\frac{1}{4 y}\right)+\left(\frac{1}{4 y}\right)^2=\left(2 y+\frac{1}{4 y}\right)^2 = ( 2 y ) 2 + 2 ( 2 y ) ( 4 y 1 ) + ( 4 y 1 ) 2 = ( 2 y + 4 y 1 ) 2
Q.87:
Factor the algebraic expression 9 m 2 − 1 25 n 2 9 m^2-\frac{1}{25 n^2} 9 m 2 − 25 n 2 1
Solution:
9 m 2 − 1 25 m 2 9 m^2-\frac{1}{25 m^2} 9 m 2 − 25 m 2 1 = ( 3 m ) 2 − ( 1 5 n ) 2 = ( 3 m − 1 5 n ) ( 3 m + 1 5 n ) =(3 m)^2-\left(\frac{1}{5 n}\right)^2=\left(3 m-\frac{1}{5 n}\right)\left(3 m+\frac{1}{5 n}\right) = ( 3 m ) 2 − ( 5 n 1 ) 2 = ( 3 m − 5 n 1 ) ( 3 m + 5 n 1 )
Q.88:
Factor the algebraic expression 27 b 3 − 1 64 b 3 27 b^3-\frac{1}{64 b^3} 27 b 3 − 64 b 3 1
Solution:
27 b 3 − 1 64 b 3 27 b^3-\frac{1}{64 b^3} 27 b 3 − 64 b 3 1 = ( 3 b ) 3 − ( 1 4 b ) 3 = ( 3 b − 1 4 b ) ( 9 b 2 + 3 4 + 1 16 b 2 ) =(3 b)^3-\left(\frac{1}{4 b}\right)^3=\left(3 b-\frac{1}{4 b}\right)\left(9 b^2+\frac{3}{4}+\frac{1}{16 b^2}\right) = ( 3 b ) 3 − ( 4 b 1 ) 3 = ( 3 b − 4 b 1 ) ( 9 b 2 + 4 3 + 16 b 2 1 )
Q.89:
Factor the algebraic expression x 2 + 5 x 6 + 1 6 x^2+\frac{5 x}{6}+\frac{1}{6} x 2 + 6 5 x + 6 1
Solution:
x 2 + 5 x 6 + 1 6 x^2+\frac{5 x}{6}+\frac{1}{6} x 2 + 6 5 x + 6 1 = x 2 + x + ( − 1 6 x + 1 6 ) = ( x + 1 ) ( x + 1 6 ) =x^2+x+\left(-\frac{1}{6} x+\frac{1}{6}\right)=(x+1)\left(x+\frac{1}{6}\right) = x 2 + x + ( − 6 1 x + 6 1 ) = ( x + 1 ) ( x + 6 1 )
Q.90:
Factor the algebraic expression 27 u 3 − 1 125 − 27 u 2 5 + 9 u 25 27 u^3-\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25} 27 u 3 − 125 1 − 5 27 u 2 + 25 9 u
Solution:
27 u 3 − 1 125 − 27 u 2 5 + 9 u 25 27 u^3-\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25} 27 u 3 − 125 1 − 5 27 u 2 + 25 9 u = ( 3 u ) 3 − ( 1 5 ) 3 − 3 ( 3 u ) 2 ( 1 5 ) + 3 ( 3 u ) ( 1 5 ) 2 =(3 u)^3-\left(\frac{1}{5}\right)^3-3(3 u)^2\left(\frac{1}{5}\right)+3(3 u)\left(\frac{1}{5}\right)^2 = ( 3 u ) 3 − ( 5 1 ) 3 − 3 ( 3 u ) 2 ( 5 1 ) + 3 ( 3 u ) ( 5 1 ) 2 = ( 3 u − 1 5 ) 3 =\left(3 u-\frac{1}{5}\right)^3 = ( 3 u − 5 1 ) 3
Q.91:
Factor the algebraic expression 64 y 3 + 1 125 z 3 64 y^3+\frac{1}{125} z^3 64 y 3 + 125 1 z 3
Solution:
64 y 3 + 1 125 z 3 64 y^3+\frac{1}{125} z^3 64 y 3 + 125 1 z 3 = ( 4 y ) 3 + ( z 5 ) 3 = ( 4 y + z 5 ) ( 16 y 2 − 4 y z 5 + z 2 25 ) =(4 y)^3+\left(\frac{z}{5}\right)^3=\left(4 y+\frac{z}{5}\right)\left(16 y^2-\frac{4 y z}{5}+\frac{z^2}{25}\right) = ( 4 y ) 3 + ( 5 z ) 3 = ( 4 y + 5 z ) ( 16 y 2 − 5 4 y z + 25 z 2 )
Q.92:
Factor the algebraic expression p 3 + 27 q 3 + r 3 − 9 p q r p^3+27 q^3+r^3-9 p q r p 3 + 27 q 3 + r 3 − 9 pq r
Solution:
p 3 + 27 q 3 + r 3 − 9 p q r p^3+27 q^3+r^3-9 p q r p 3 + 27 q 3 + r 3 − 9 pq r
Using identity:
a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) a 3 + b 3 + c 3 − 3 ab c = ( a + b + c ) ( a 2 + b 2 + c 2 − ab − b c − c a ) = ( p + 3 q + r ) ( p 2 + 9 q 2 + r 2 − 3 p q − 3 q r − p r ) =(p+3 q+r)\left(p^2+9 q^2+r^2-3 p q-3 q r-p r\right) = ( p + 3 q + r ) ( p 2 + 9 q 2 + r 2 − 3 pq − 3 q r − p r )
Q.93:
Factor the algebraic expression 9 m 2 − 12 m + 4 9 m^2-12 m+4 9 m 2 − 12 m + 4
Solution:
9 m 2 − 12 m + 4 9 m^2-12 m+4 9 m 2 − 12 m + 4 = ( 3 m ) 2 − 2 ( 3 m ) ( 2 ) + 2 2 = ( 3 m − 2 ) 2 =(3 m)^2-2(3 m)(2)+2^2=(3 m-2)^2 = ( 3 m ) 2 − 2 ( 3 m ) ( 2 ) + 2 2 = ( 3 m − 2 ) 2
Q.94:
Factor the algebraic expression 9 x 3 − 8 3 y 3 + z 3 3 + 6 x y z 9 x^3-\frac{8}{3} y^3+\frac{z^3}{3}+6 x y z 9 x 3 − 3 8 y 3 + 3 z 3 + 6 x y z
Solution:
9 x 3 − 8 3 y 3 + z 3 3 + 6 x y z 9 x^3-\frac{8}{3} y^3+\frac{z^3}{3}+6 x y z 9 x 3 − 3 8 y 3 + 3 z 3 + 6 x y z
Take 1 3 \frac{1}{3} 3 1
= 1 3 ( 27 x 3 − 8 y 3 + z 3 + 18 x y z ) =\frac{1}{3}\left(27 x^3-8 y^3+z^3+18 x y z\right) = 3 1 ( 27 x 3 − 8 y 3 + z 3 + 18 x y z ) = 1 3 ( 3 x − 2 y + z ) 3 =\frac{1}{3}(3 x-2 y+z)^3 = 3 1 ( 3 x − 2 y + z ) 3
Q.95:
Factor the algebraic expression 4 x 2 + 9 y 2 + 36 z 2 + 12 x z + 36 y z + 24 x y 4 x^2+9 y^2+36 z^2+12 x z+36 y z+24 x y 4 x 2 + 9 y 2 + 36 z 2 + 12 x z + 36 y z + 24 x y
Solution:
4 x 2 + 9 y 2 + 36 z 2 + 12 x z + 36 y z + 24 x y 4 x^2+9 y^2+36 z^2+12 x z+36 y z+24 x y 4 x 2 + 9 y 2 + 36 z 2 + 12 x z + 36 y z + 24 x y = ( 2 x ) 2 + ( 3 y ) 2 + ( 6 z ) 2 + 2 ( 2 x ) ( 3 y ) =(2 x)^2+(3 y)^2+(6 z)^2+2(2 x)(3 y) = ( 2 x ) 2 + ( 3 y ) 2 + ( 6 z ) 2 + 2 ( 2 x ) ( 3 y ) + 2 ( 3 y ) ( 6 z ) + 2 ( 2 x ) ( 6 z ) +2(3 y)(6 z)+2(2 x)(6 z) + 2 ( 3 y ) ( 6 z ) + 2 ( 2 x ) ( 6 z ) = ( 2 x + 3 y + 6 z ) 2 =(2 x+3 y+6 z)^2 = ( 2 x + 3 y + 6 z ) 2
Q.96:
Factor the algebraic expression 27 u 3 − 1 216 − 9 u 2 2 + u 4 27 u^3-\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4} 27 u 3 − 216 1 − 2 9 u 2 + 4 u
Solution:
27 u 3 − 1 216 − 9 u 2 2 + u 4 27 u^3-\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4} 27 u 3 − 216 1 − 2 9 u 2 + 4 u = ( 3 u ) 3 − ( 1 6 ) 3 − 3 ( 3 u ) 2 ( 1 6 ) + 3 ( 3 u ) ( 1 6 ) 2 =(3 u)^3-\left(\frac{1}{6}\right)^3-3(3 u)^2\left(\frac{1}{6}\right)+3(3 u)\left(\frac{1}{6}\right)^2 = ( 3 u ) 3 − ( 6 1 ) 3 − 3 ( 3 u ) 2 ( 6 1 ) + 3 ( 3 u ) ( 6 1 ) 2 = ( 3 u − 1 6 ) 3 =\left(3 u-\frac{1}{6}\right)^3 = ( 3 u − 6 1 ) 3
Q.97:
Simplify: 4 x 2 + 4 x + 1 4 x 2 − 1 \frac{4 x^2+4 x+1}{4 x^2-1} 4 x 2 − 1 4 x 2 + 4 x + 1
Solution:
4 x 2 + 4 x + 1 4 x 2 − 1 \frac{4 x^2+4 x+1}{4 x^2-1} 4 x 2 − 1 4 x 2 + 4 x + 1 4 x 2 + 4 x + 1 = ( 2 x + 1 ) 2 , 4 x 2 − 1 4 x^2+4 x+1 =(2 x+1)^2, 4 x^2-1 4 x 2 + 4 x + 1 = ( 2 x + 1 ) 2 , 4 x 2 − 1 = ( 2 x − 1 ) ( 2 x + 1 ) =(2 x-1)(2 x+1) = ( 2 x − 1 ) ( 2 x + 1 ) = ( 2 x + 1 ) 2 ( 2 x − 1 ) ( 2 x + 1 ) = 2 x + 1 2 x − 1 =\frac{(2 x+1)^2}{(2 x-1)(2 x+1)}=\frac{2 x+1}{2 x-1} = ( 2 x − 1 ) ( 2 x + 1 ) ( 2 x + 1 ) 2 = 2 x − 1 2 x + 1
Q.98:
Simplify: 9 ( 3 a 3 − 24 b 3 ) 9 a 2 − 36 b 2 \frac{9\left(3 a^3-24 b^3\right)}{9 a^2-36 b^2} 9 a 2 − 36 b 2 9 ( 3 a 3 − 24 b 3 )
Solution:
9 ( 3 a 3 − 24 b 3 ) 9 a 2 − 36 b 2 \frac{9\left(3 a^3-24 b^3\right)}{9 a^2-36 b^2} 9 a 2 − 36 b 2 9 ( 3 a 3 − 24 b 3 ) = 9 ⋅ 3 ( a 3 − 8 b 3 ) 9 ( a 2 − 4 b 2 ) = 27 ( a − 2 b ) ( a 2 + 2 a b + 4 b 2 ) 9 ( a − 2 b ) ( a + 2 b ) =\frac{9 \cdot 3\left(a^3-8 b^3\right)}{9\left(a^2-4 b^2\right)}=\frac{27(a-2 b)\left(a^2+2 a b+4 b^2\right)}{9(a-2 b)(a+2 b)} = 9 ( a 2 − 4 b 2 ) 9 ⋅ 3 ( a 3 − 8 b 3 ) = 9 ( a − 2 b ) ( a + 2 b ) 27 ( a − 2 b ) ( a 2 + 2 ab + 4 b 2 )
= 3 ( a 2 + 2 a b + 4 b 2 ) a + 2 b =\frac{3\left(a^2+2 a b+4 b^2\right)}{a+2 b} = a + 2 b 3 ( a 2 + 2 ab + 4 b 2 )
Q.99:
Simplify: s 3 + 125 t 3 s 2 − 2 s t − 35 t 2 \frac{s^3+125 t^3}{s^2-2 s t-35 t^2} s 2 − 2 s t − 35 t 2 s 3 + 125 t 3
Solution:
s 3 + 125 t 3 s 2 − 2 s t − 35 t 2 \frac{s^3+125 t^3}{s^2-2 s t-35 t^2} s 2 − 2 s t − 35 t 2 s 3 + 125 t 3
s 3 + 125 t 3 = ( s + 5 t ) ( s 2 − 5 s t + 25 t 2 ) s^3+125 t^3=(s+5 t)\left(s^2-5 s t+25 t^2\right) s 3 + 125 t 3 = ( s + 5 t ) ( s 2 − 5 s t + 25 t 2 ) s 2 − 2 s t − 35 t 2 = ( s − 7 t ) ( s + 5 t ) s^2-2 s t-35 t^2=(s-7 t)(s+5 t) s 2 − 2 s t − 35 t 2 = ( s − 7 t ) ( s + 5 t ) ( s + 5 t ) (s+5 t) ( s + 5 t )
= s 2 − 5 s t + 25 t 2 s − 7 t =\frac{s^2-5 s t+25 t^2}{s-7 t} = s − 7 t s 2 − 5 s t + 25 t 2
Q.100:
Find possible expression for the length and breadth of the rectangle whose area is given by the following expression in square units.25 a 2 − 30 a b + 9 b 2 25 a^2-30 a b+9 b^2 25 a 2 − 30 ab + 9 b 2
Solution:
25 a 2 − 30 a b + 9 b 2 25 a^2-30 a b+9 b^2 25 a 2 − 30 ab + 9 b 2
Recognise identity: ( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2=a^2-2 a b+b^2 ( a − b ) 2 = a 2 − 2 ab + b 2
= ( 5 a ) 2 − 2 ( 5 a ) ( 3 b ) + ( 3 b ) 2 = ( 5 a − 3 b ) 2 =(5 a)^2-2(5 a)(3 b)+(3 b)^2=(5 a-3 b)^2 = ( 5 a ) 2 − 2 ( 5 a ) ( 3 b ) + ( 3 b ) 2 = ( 5 a − 3 b ) 2 = 5 a − 3 b =5 a-3 b = 5 a − 3 b = 5 a − 3 b =5 a-3 b = 5 a − 3 b
Q.101:
Find possible expression for the length and breadth of the rectangle whose area is given by the following expression in square units.36 s 2 − 49 t 2 36 s^2-49 t^2 36 s 2 − 49 t 2
Solution:
36 s 2 − 49 t 2 36 s^2-49 t^2 36 s 2 − 49 t 2
Recognise identity: a 2 − b 2 = ( a − b ) ( a + b ) a^2-b^2=(a-b)(a+b) a 2 − b 2 = ( a − b ) ( a + b )
= ( 6 s ) 2 − ( 7 t ) 2 = ( 6 s − 7 t ) ( 6 s + 7 t ) =(6 s)^2-(7 t)^2=(6 s-7 t)(6 s+7 t) = ( 6 s ) 2 − ( 7 t ) 2 = ( 6 s − 7 t ) ( 6 s + 7 t ) = 6 s − 7 t =6 s-7 t = 6 s − 7 t = 6 s + 7 t =6 s+7 t = 6 s + 7 t
Q.102:
Find possible expressions for the length, breadth, and height of the following cuboid whose volume is given by the following expressions in cubic units. 6 a 2 − 24 b 2 6 a^2-24 b^2 6 a 2 − 24 b 2
Solution:
6 a 2 − 24 b 2 6 a^2-24 b^2 6 a 2 − 24 b 2
Take common factor:
= 6 ( a 2 − 4 b 2 ) =6\left(a^2-4 b^2\right) = 6 ( a 2 − 4 b 2 ) a 2 − b 2 = ( a − b ) ( a + b ) a^2-b^2=(a-b)(a+b) a 2 − b 2 = ( a − b ) ( a + b )
= 6 ( a − 2 b ) ( a + 2 b ) =6(a-2 b)(a+2 b) = 6 ( a − 2 b ) ( a + 2 b )
Length = 6 , Breadth = a − 2 b , Height = a + 2 b \text { Length }=6, \text { Breadth }=a-2 b, \text { Height }=a+2 b Length = 6 , Breadth = a − 2 b , Height = a + 2 b
Q.103:
Find possible expressions for the length, breadth, and height of the following cuboid whose volume is given by the following expressions in cubic units. 3 p s 2 − 15 p s + 12 p 3 p s^2-15 p s+12 p 3 p s 2 − 15 p s + 12 p
Solution:
3 p s 2 − 15 p s + 12 p 3 p s^2-15 p s+12 p 3 p s 2 − 15 p s + 12 p
Take common factor:
= 3 p ( s 2 − 5 s + 4 ) =3 p\left(s^2-5 s+4\right) = 3 p ( s 2 − 5 s + 4 )
= 3 p ( s − 1 ) ( s − 4 ) =3 p(s-1)(s-4) = 3 p ( s − 1 ) ( s − 4 )
Length = 3 p , Breadth = s − 1 , Height = s − 4 \text { Length }=3 p, \text { Breadth }=s-1, \text { Height }=s-4 Length = 3 p , Breadth = s − 1 , Height = s − 4
Q.104:
The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Solution:
Side of square playground = 40 m =40 {~m} = 40 m
Path of width s s s
= 40 + 2 s =40+2 s = 40 + 2 s
( 40 + 2 s ) 2 (40+2 s)^2 ( 40 + 2 s ) 2
40 2 = 1600 40^2=1600 4 0 2 = 1600
= ( 40 + 2 s ) 2 − 1600 =(40+2 s)^2-1600 = ( 40 + 2 s ) 2 − 1600 = ( 1600 + 160 s + 4 s 2 ) − 1600 =\left(1600+160 s+4 s^2\right)-1600 = ( 1600 + 160 s + 4 s 2 ) − 1600 = 160 s + 4 s 2 =160 s+4 s^2 = 160 s + 4 s 2
Q.105:
If a number plus its reciprocal equals 10 3 \frac{10}{3} 3 10
Solution:
Let the number be x x x
x + 1 x = 10 3 x+\frac{1}{x}=\frac{10}{3} x + x 1 = 3 10 3 x 3 x 3 x
3 x 2 + 3 = 10 x 3 x^2+3=10 x 3 x 2 + 3 = 10 x
3 x 2 − 10 x + 3 = 0 3 x^2-10 x+3=0 3 x 2 − 10 x + 3 = 0
( 3 x − 1 ) ( x − 3 ) = 0 (3 x-1)(x-3)=0 ( 3 x − 1 ) ( x − 3 ) = 0
x = 1 3 or x = 3 x=\frac{1}{3} \text { or } x=3 x = 3 1 or x = 3
Q.106:
A rectangular pool has area 2x2 + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
Solution:
Area of rectangle = 2 x 2 + 7 x + 3 =2 x^2+7 x+3 = 2 x 2 + 7 x + 3 = 2 x + 1 =2 x+1 = 2 x + 1
Area = Length × Width \text { Area }=\text { Length × Width } Area = Length × Width
Length = 2 x 2 + 7 x + 3 2 x + 1 \text { Length }=\frac{2 x^2+7 x+3}{2 x+1} Length = 2 x + 1 2 x 2 + 7 x + 3
2 x 2 + 7 x + 3 = ( 2 x + 1 ) ( x + 3 ) 2 x^2+7 x+3=(2 x+1)(x+3) 2 x 2 + 7 x + 3 = ( 2 x + 1 ) ( x + 3 )
Length = ( 2 x + 1 ) ( x + 3 ) 2 x + 1 = x + 3 \text { Length }=\frac{(2 x+1)(x+3)}{2 x+1}=x+3 Length = 2 x + 1 ( 2 x + 1 ) ( x + 3 ) = x + 3
Q.107:
If both x − 2 x-2 x − 2 x − 1 2 x-\frac{1}{2} x − 2 1 p x 2 + 5 x + r p x^2+5 x+r p x 2 + 5 x + r p = r p=r p = r
Solution:
Given polynomial:
p x 2 + 5 x + r p x^2+5 x+r p x 2 + 5 x + r x − 2 x-2 x − 2 x − 1 2 x-\frac{1}{2} x − 2 1
Put x = 2 x=2 x = 2
p ( 2 ) 2 + 5 ( 2 ) + r = 0 ⇒ 4 p + 10 + r = 0 p(2)^2+5(2)+r=0 \Rightarrow 4 p+10+r=0 p ( 2 ) 2 + 5 ( 2 ) + r = 0 ⇒ 4 p + 10 + r = 0 x = 1 2 x=\frac{1}{2} x = 2 1
p ( 1 2 ) 2 + 5 ( 1 2 ) + r = 0 p\left(\frac{1}{2}\right)^2+5\left(\frac{1}{2}\right)+r=0 p ( 2 1 ) 2 + 5 ( 2 1 ) + r = 0 ⇒ p 4 + 5 2 + r = 0 \Rightarrow \frac{p}{4}+\frac{5}{2}+r=0 ⇒ 4 p + 2 5 + r = 0
p + 10 + 4 r = 0 p+10+4 r=0 p + 10 + 4 r = 0
4 p + r = − 10 4 p+r=-10 4 p + r = − 10
p + 4 r = − 10 p+4 r=-10 p + 4 r = − 10
4 p + 16 r = − 40 4 p+16 r=-40 4 p + 16 r = − 40
( 4 p + 16 r ) − ( 4 p + r ) = − 40 − ( − 10 ) (4 p+16 r)-(4 p+r)=-40-(-10) ( 4 p + 16 r ) − ( 4 p + r ) = − 40 − ( − 10 ) 15 r = − 30 ⇒ r = − 2 15 r=-30 \Rightarrow r=-2 15 r = − 30 ⇒ r = − 2
Q.108:
If a + b + c = 5 a+b+c=5 a + b + c = 5 a b + b c + c a = 10 a b+b c+c a=10 ab + b c + c a = 10 a 3 + b 3 + c 3 − 3 a b c = − 25 a^3+b^3+c^3-3 a b c=-25 a 3 + b 3 + c 3 − 3 ab c = − 25
Solution:
Given:
a + b + c = 5 , a b + b c + c a = 10 a+b+c=5, a b+b c+c a=10 a + b + c = 5 , ab + b c + c a = 10
a 3 + b 3 + c 3 − 3 a b c = a^3+b^3+c^3-3 a b c= a 3 + b 3 + c 3 − 3 ab c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) (a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) ( a + b + c ) ( a 2 + b 2 + c 2 − ab − b c − c a ) a 2 + b 2 + c 2 a^2+b^2+c^2 a 2 + b 2 + c 2
a 2 + b 2 + c 2 = ( a + b + c ) 2 − 2 ( a b + b c + c a ) a^2+b^2+c^2= (a+b+c)^2-2(a b+b c+c a) a 2 + b 2 + c 2 = ( a + b + c ) 2 − 2 ( ab + b c + c a ) = 5 2 − 2 ( 10 ) =5^2-2(10) = 5 2 − 2 ( 10 ) = 25 − 20 = 5 =25-20=5 = 25 − 20 = 5
a 3 + b 3 + c 3 − 3 a b c = a^3+b^3+c^3-3 a b c= a 3 + b 3 + c 3 − 3 ab c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) (a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) ( a + b + c ) ( a 2 + b 2 + c 2 − ab − b c − c a ) = 5 ( 5 − 10 ) =5(5-10) = 5 ( 5 − 10 ) = 5 × ( − 5 ) =5 \times(-5) = 5 × ( − 5 ) = − 25 =-25 = − 25
a 3 + b 3 + c 3 − 3 a b c = − 25 a^3+b^3+c^3-3 a b c=-25 a 3 + b 3 + c 3 − 3 ab c = − 25
Q.109:
By factoring the expression, check that n3 – n is always divisible by 6 for all natural numbers n. Give reasons.
Solution:
We need to check whether n 3 − n n^3-n n 3 − n
Step 1: Factorise
n 3 − n = n ( n 2 − 1 ) n^3-n=n\left(n^2-1\right) n 3 − n = n ( n 2 − 1 ) = n ( n − 1 ) ( n + 1 ) =n(n-1)(n+1) = n ( n − 1 ) ( n + 1 )
n ( n − 1 ) ( n + 1 ) n(n-1)(n+1) n ( n − 1 ) ( n + 1 )
are three consecutive natural numbers.
Step 3: Use number properties
One number is always divisible by 2 (even number) One number is always divisible by 3 Step 4: Conclusion
n ( n − 1 ) ( n + 1 ) n(n-1)(n+1) n ( n − 1 ) ( n + 1 )
is divisible by:
2 × 3 = 6 2 \times 3=6 2 × 3 = 6 n 3 − n n^3-n n 3 − n n n n
Q.110:
Find the value of x3 + y3 – 12xy + 64, when x + y = - 4
Solution:
Given x + y = − 4 x+y=-4 x + y = − 4
Expression:
x 3 + y 3 − 12 x y + 64 x^3+y^3-12 x y+64 x 3 + y 3 − 12 x y + 64
x 3 + y 3 = ( x + y ) 3 − 3 x y ( x + y ) x^3+y^3=(x+y)^3-3 x y(x+y) x 3 + y 3 = ( x + y ) 3 − 3 x y ( x + y )
x 3 + y 3 − 12 x y + 64 = x^3+y^3-12 x y+64= x 3 + y 3 − 12 x y + 64 = ( x + y ) 3 − 3 x y ( x + y ) − 12 x y + 64 (x+y)^3-3 x y(x+y)-12 x y+64 ( x + y ) 3 − 3 x y ( x + y ) − 12 x y + 64 x + y = − 4 x+y=-4 x + y = − 4
= ( − 4 ) 3 − 3 x y ( − 4 ) − 12 x y + 64 =(-4)^3-3 x y(-4)-12 x y+64 = ( − 4 ) 3 − 3 x y ( − 4 ) − 12 x y + 64 = − 64 + 12 x y − 12 x y + 64 =-64+12 x y-12 x y+64 = − 64 + 12 x y − 12 x y + 64 = 0 =0 = 0
Q.111:
Find the value of x3 – 8y3 – 36xy – 216, when x = 2y + 6.
Solution:
Given x = 2 y + 6 x=2 y+6 x = 2 y + 6
Expression:
x 3 − 8 y 3 − 36 x y − 216 x^3-8 y^3-36 x y-216 x 3 − 8 y 3 − 36 x y − 216
x 3 − ( 2 y ) 3 = ( x − 2 y ) ( x 2 + 2 x y + 4 y 2 ) x^3-(2 y)^3=(x-2 y)\left(x^2+2 x y+4 y^2\right) x 3 − ( 2 y ) 3 = ( x − 2 y ) ( x 2 + 2 x y + 4 y 2 ) = ( x − 2 y ) ( x 2 + 2 x y + 4 y 2 ) − 36 x y − 216 = (x-2 y)\left(x^2+2 x y+4 y^2\right)-36 x y-216 = ( x − 2 y ) ( x 2 + 2 x y + 4 y 2 ) − 36 x y − 216 x = 2 y + 6 x=2 y+6 x = 2 y + 6
x − 2 y = 6 x-2 y=6 x − 2 y = 6 = 6 ( x 2 + 2 x y + 4 y 2 ) − 36 x y − 216 =6\left(x^2+2 x y+4 y^2\right)-36 x y-216 = 6 ( x 2 + 2 x y + 4 y 2 ) − 36 x y − 216 = 6 x 2 + 12 x y + 24 y 2 − 36 x y − 216 =6 x^2+12 x y+24 y^2-36 x y-216 = 6 x 2 + 12 x y + 24 y 2 − 36 x y − 216 = 6 x 2 − 24 x y + 24 y 2 − 216 =6 x^2-24 x y+24 y^2-216 = 6 x 2 − 24 x y + 24 y 2 − 216 = 6 ( x 2 − 4 x y + 4 y 2 ) − 216 =6\left(x^2-4 x y+4 y^2\right)-216 = 6 ( x 2 − 4 x y + 4 y 2 ) − 216 = 6 ( x − 2 y ) 2 − 216 =6(x-2 y)^2-216 = 6 ( x − 2 y ) 2 − 216 x − 2 y = 6 x-2 y=6 x − 2 y = 6
= 6 ( 6 2 ) − 216 = 216 − 216 = 0 =6\left(6^2\right)-216=216-216=0 = 6 ( 6 2 ) − 216 = 216 − 216 = 0
