Separate the constants and variables from the following:
and
In the given question
The symbols which has a fixed value are constant such as 15, 3, –3/7 and 7
The symbols which do not have any fixed value, but may be assigned the value (values) according to the requirement are called the variables such as
Where z, x and y are variables which may acquire different values according to the situation.
And a combination of a constant and a variable is a variable.
Separate the monomials, binomials and trinomials from the following:
7xyz, 9 – 4y, 4y2 – xz, x – 2y + 3z, 7x + z2, 8xy, x2y2, 4 + 5y – 6z.
A polynomial which contains only one term is monomials
Such as
A Polynomial which contains two terms is called a Polynomial.
Such as 9–4y, 4y2 – xz, 7x +z2
A polynomial which contains three terms is called as trinomial
Such as x –2y +3z, 4+5y –6z
Classify into like terms:
4x2, x, –8x3, xy, 6x3, 4y, –74x3, 8xy, 7xyz, 3x2.
The terms having same variable with same exponents are called like terms
Here like terms are
(4x2,3x2) with same variable x and exponential power as 2
( has no like term with x as variable and exponential power as 1)
(–8x3, 6x3, –74x3 with variable as x and exponential power as 3 )
7xyz has no like term as no other term has three variables x, y, z
(xy, 8xy has same variables x, y and exponential power of both x and y as 1)
Simplify:
(i) 7x – 9y + 3 – 3x – 5y + 8;
(ii) 3x2 + 5xy – 4y2 + x2 – 8xy – 5y2.
While adding or subtracting, the like term`s numerical coefficient are added or subtracted
(i) 7x – 9y + 3 – 3x – 5y + 8
⇒ (7– 3)x +(–9 –5)y +3+8
⇒ 4x – 14y +11
(ii) 3x2 + 5xy – 4y2 + x2 – 8xy – 5y2
⇒ (3+1) x2 +( –4 –5) y2 +( 5–8) x y
⇒ 4x2 – 9y2 – 3xy
Add:
(i) 5a + 3b, a – 2b and 3a + 5b;
(ii) x3 – x2y + 5xy2 + y, –x3 – 9xy2 + y, and 3x2y + 9xy2.
(i) While adding or subtracting, the like term`s numerical coefficient are added or subtracted
⇒ (5 + 1+ 3)a + (3 – 2 +5) b
⇒ 9 a + 6b
(ii) While adding or subtracting, the like term`s numerical coefficient are added or subtracted
⇒ (1–1) x3 + (–1+3) x2y + (5–9 +9) xy2 + (1+1) y
⇒ 0x3 + 2 x2y +5xy2 +2y
⇒ 2 x2y +5xy2 +2y
Subtract:
(i) –2xy + 3xy2 from 8xy;
(ii) a – b – 2c from 4a + 6b – 2c.
Given problem asks to subtract
(i) –2xy + 3xy2 from 8xy
⇒ 8xy – ( –2xy + 3xy2 )
⇒ 8xy +2xy –3xy2
⇒ 10xy –3xy2
(ii) a – b – 2c from 4a + 6b – 2c
⇒ 4a +6b –2c – (a– b – 2c)
⇒ 4a +6b –2c –a + 2c +b
While adding or subtracting, the like term`s numerical coefficient are added or subtracted
⇒ (4–1) a + (6 +1) b +( –2 +2) c
⇒ 3a +7b
Complete the following table of products of two monomials:
To find out product of monomials
We multiple the numerical coefficient together and variables together
The given table hence can be completed as :
Find the products:
(i) (5x + 8)3x
(ii) (–3pq) (–15p3q2 – q3)
(iii) (3a3 – 3b3)
(iv) –x2(x – 15).
(i) By using the distributive law
(5x +8) 3x
= 5x × 3x + 8× 3x
= 15x2 +24x
(ii) By using the distributive law
= (–3pq)(–15 p3q2) – (–3pq)(q3)
= 45p4q3 + 3pq4
(iii) By using the distributive law
=
(iv) By using the distributive law
= –x2 (x) – (–x2) (15)
= –x3 + 15x2
Simplify the following:
(i) (2xy – xy)(3xy – 5)
(ii) (3xy2 + 1)(4xy – 6xy2)
(iii) (3x2 + 2x)(2x2 + 3)
(iv) (2m3 + 3m)(5m – 1).
By using the distributive law
(i) (2xy – xy)(3xy – 5)
= 2xy (3xy – 5) –xy (3xy –5)
= 6x2y2 –10xy – 3x2y2 +5xy
= 3x2y2 – 5xy (after adding and subtracting the like terns)
(ii) (3xy2 + 1)(4xy – 6xy2)
By using the distributive law
3xy2 ((4xy – 6xy2) +1((4xy – 6xy2)
= 3xy2 × 4xy – 3xy2 × 6xy2 + 4xy – 6xy2
= 12x2y3 – 18 x2y4 + 4xy –6xy2
(iii) (3x2 + 2x)(2x2 + 3)
By using the distributive law
3x2 (2x2 + 3) +2x (2x2 + 3)
= 3x2 × 2x2 + 3x2 × 3 + 2x × 2x2 + 2x × 3
= 6x4 + 9x2 + 4x3 +6x
(iv) (2m3 + 3m)(5m – 1).
By using the distributive law
2m3 (5m –1) +3m (5m –1)
= 2m3 × 5m – 1× 2m3 + 3m × 5m – 1× 3m
= 10m4 – 2m3 + 15m2 – 3m
Find the product:
(i) (a + 3)(a + 5)
(ii) (3t + 1)(3t + 4)
(iii) (a – 8)(a + 2)
(iv) (a – 6)(a – 2)
(i) By using the distributive law
= a(a + 5) +3((a+5)
= a2 +5a +3a +15
= a2 +8a +15
(ii) By using the distributive law
= 3t (3t + 4) + 1(3t + 4)
= 9t2 + 12t +3t + 4
= 9t2 + 15t +4
(iii) By using the distributive law
= a( a+2) –8(a+2)
= a2 + 2a – 8a – 16
= a2 – 6a –16
(iv) By using the distributive law
= a (a–2) – 6 ( a – 2)
= a2 –2a – 6a +12
= a2 – 8a +12
Evaluate using suitable identities:
(i) 53 × 55 (ii) 102 × 106
(iii) 34 × 36 (iv) 103 × 96
(i) 53 × 55
We can re–write 53 and 55 as
(50+3)× (50+5)
Using the identity
(x+a) (x+b) = x2 + (a+b) x + ab
(50+3)× (50+5) where x = 50, a = 3 and b = 5
= 502 + ( 3+5) 50 + 3× 5
=2500 + 400 + 15
= 2915
(ii) 102 × 106
= (100 + 2) (100 + 6)
Using the identity
(x+a) (x+b) = x2 + (a+b) x + ab
Here x= 100, a = 2 and b = 6
⇒ (100 + 2) (100 + 6)
= 1002 + ( 2+6) 100 + 2× 6
= 10000 + 800 +12
= 10812
(iii) 34 × 36
= (30+4) + (30+6)
Using the identity
(x+a) (x+b) = x2 + (a+b) x + ab
Here x = 30, a= 4 and b = 6
So, 302 +( 4+6) 30 +(4× 6)
= 900 + 300 +24
= 1224
(iv) 103 × 96
= (90 + 13) (90 +6)
Using the identity
(x+a) (x+b) = x2 + (a+b) x + ab
Here x = 90, a = 13 and b = 6
So, (90 + 13) (90 +6)
= 902 + ( 13+6) 90 +(13× 6)
= 8100 + 1710 +78
= 9888
Find the expression for the product (x + a)(x + b)(x + c) using the identity (x + a)(x + b) = x2 + (a + b)x + ab
first we will expand (x + a)(x + b)
Using the identity
(x+a) (x+b) = x2 + (a+b)x + ab
= x2 + (a+b)x + ab
Now multiplying the expansion with (x+c)
(x2 + (a+b)x + ab) (x+c)
By using the distributive law
x(x2 + (a+b) x + ab) + c(x2 + (a+b) x + ab)
= x3 + (a + b)x2 + abx + cx2 + (a + b)cx + abc
Arranging the like terms
= x3 + (a + b)x2 + cx2 + abx + (a + b)cx + abc= x3 + (a + b + c)x2 + abx + acx + bcx + abc
= x3 + (a + b + c) x2 + x(ab+ ac+ bc) + abc
Using the identity (a +b)2 = a2 + 2ab + b2, simplify the following:
(i) (a + 6)2 (ii) (3x + 2y)2
(iii) (2p + 3q)2 (iv) (x2 + 5)2
Given the identity
(a +b)2 = a2 + 2ab + b2
(i) (a + 6)2
Using the given identity
Here a = a, b = 6
= a2 +2 × a × 6 +62
= a2 +12a +36
(ii) (3x + 2y)2
Using the given identity
Here a = 3x, b = 2y
= (3x)2 + 2 × 3x × 2y +(2y)2
= 9x2 + 12xy +4y2
(iii) (2p + 3q)2
Using the given identity
Here a = 2p, b = 3q
= (2p)2 +2× 2p × 3q + (3q)2
= 4p2 + 12pq +9q2
(iv) (x2 + 5)2
Using the given identity
Here a = x2, b = 5
= (x2)2 +2 × x2 × 5 +52
= x4 + 10x2 +25
Evaluate using the identity (a + b)2 = a2 + 2ab + b2
(i) (34)2 (ii) (10.2)2
(iii) (53)2 (iv) (41)2
Given identity (a + b)2 = a2 + 2ab + b2
(i) (34)2
= (30+4)2
Here a = 30 and b = 4
= 302 +2× 30× 4 + 42
= 900 + 240 +16
= 1156
(ii)(10.2)2
= (10 +0.2)2
Using the given identity
Here a = 10, b = 0.2
= 102 + 2× 10 × 0.2 +(0.2)2
= 100 + 4 + 0.04
= 104.04
(iii) (53)2
= (50+3)2
Using the given identity
Here a = 50, b = 3
= (50)2 +2× 50 × 3 + 32
= 2500 + 300 +9
= 2809
(iv) (41)2
= (40+1) 2
Using the given identity
Here a = 40, b = 1
= 402 +2 × 40 × 1 + 12
= 1600 + 80 +1
= 1681
Use the identity (a – b)2 = a2 – 2ab + b2 to compute:
(i) (x – 6)2 (ii) (3x – 5y)2
(iii) (5a – 4b)2 (iv) (p2 + q2)2
Given identity
(a – b)2 = a2 – 2ab + b2
(i) (x – 6)2
Using the given identity
Here a = x, b = 6
= x2 – 2× x × 6 +62
= x2 –12x +36
(ii)(3x – 5y)2
Using the given identity
Here a = 3x, b = 5y
= (3x)2 – 2× 3x × 5y + (5y)2
= 9x2 – 30xy +25y2
(iii)(5a – 4b)2
Using the given identity
Here a = 5a, b = 4b
= (5a)2 –2× 5a × 4b + (4b)2
= 25a2 – 40 ab + 16b2
(iv) (p2 – q2)2
Using the given identity
Here a = p2, b = q2
= (p2)2 – 2 × p2 × q2 + (q2)2
= p4 – 2 p2 q2 + q4
Evaluate using the identity (a – b)2 = a2 – 2ab + b2
(i) (49)2 (ii) (9.8)2
(iii) (59)2 (iv) (198)2
The given identity is (a – b)2 = a2 – 2ab + b2
(i) (49)2
= (50 –1)2
Using the given identity
Here a = 50, b = 1
= 502 – 2× 50 × 1 +12
= 2500 –100 +1
= 2401
(ii) (9.8)2
= (10 – 0.2)2
Using the given identity
Here a = 10, b = 0.2
= 102 – 2× 10 × 0.2 + (0.2)2
= 100 – 4 + 0.04
= 96.04
(iii) (59)2
= (60 – 1)2
Using the given identity
Here a = 60, b = 1
= 602 – 2× 60× 1 + 12
= 3600 – 120 +1
= 3481
(iv) (198)2
= (200 –2)2
Using the given identity
Here a = 200, b = 2
= 2002 – 2× 200 × 2 + 22
= 40000 – 800 +4
= 39204
Use the identity (a + b)(a – b) = a2 – b2 to find the products:
(i) (x – 6) (x + 6)
(ii) (3x + 5)(3x + 5)
(iii) (2a + 4b)(2a – 4b)
(iv)
The given identity is (a+b) (a–b) = a2 –b2
(i) (x – 6) (x + 6)
Using the given identity
Here a = x and b = 6
= x2 – 62
= x2 – 36
(ii) (3x + 5)(3x + 5)
Using the given identity
Here a = 3x and b = 5
= (3x)2 – 52
= 9x2 – 25
(iii) (2a + 4b)(2a – 4b)
Using the given identity
Here a = 2a and b = 4b
= (2a)2 – (4b)2
= 4a2 – 16b2
(iv)
Using the given identity
Here
=
Evaluate these using identity:
(i) 55 × 45 (ii) 33 × 27
(iii) 8.5 × 9.5 (iv) 102 × 98
(i) 55 × 45
We can split 55 as (50+5)
And 45 as (50–5)
Now 55 × 45
= (50+5) (50–5)
Using the identity (a +b) (a–b) = a2 – b2
Here a = 50 and b = 5
= 502 – 52
= 2500 –25
= 2475
(ii) 33 × 27
= (30+3) (30–3)
Using the identity (a +b) (a–b) = a2 – b2
Here a = 30 and b = 3
= (30)2 – 32
= 900 – 9
= 891
(iii) 8.5 × 9.5
= (9 – 0.5) (9 + 0.5)
Using the identity (a +b) (a–b) = a2 – b2
Here a = 9 and b = 0.5
= 92 – (0.5)2
= 81 – 0.25
= 80.75
(iv) 102 × 98
= (100 + 2) (100 – 2)
Using the identity (a +b) (a–b) = a2 – b2
Here a = 100 and b = 2
= (100)2 – 22
= 10000 – 4
= 9996
Find the product:
(x – 3)(x + 3)(x2 + 9)
First solving (x – 3)(x + 3)
Using the identity (a +b) (a–b) = a2 – b2
Here a = x and b = 3
= x2 –32
= x2 – 9
Now (x2 – 9) (x2 + 9)
Again, using the identity (a +b) (a–b) = a2 – b2
Here a = x2 and b = 9
= (x2)2 – 92
= x4 –81 ( ∵ (ax)m = axm)
Find the product:
(2a + 3)(2a – 3)(4a2 + 9)
First solving (2a + 3)(2a – 3)
Using the identity (a +b) (a–b) = a2 – b2
Here a = 2a and b = 3
= (2a)2 –32
= 4a2 – 9
Now (4a2 – 9)(4a2 + 9)
Using the identity (a +b) (a – b) = a2 – b2
Here a = 4a and b = 9
= (4a)2 – 9 2
= 16a4 – 81
Find the product:
(p + 2) (p – 2)(p2 + 4)
First solving (p+2) (p–2)
Using the identity (a +b) (a–b) = a2 – b2
Here a = p and b = 2
= p2 – 22
= p2 – 4
Now solving (p2 + 4) (p2 – 4)
Again Using the identity (a +b) (a–b) = a2 – b2
Here a = p2 and b = 4
= (p2)2 – 42
= p4 – 16 (∵ (ax)m = a xm)
Find the product:
First solving
Using the identity (a +b) (a–b) = a2 – b2
Here
Now solving
Again, Using the identity (a +b) (a–b) = a2 – b2
Here
=
Find the product:
(2x – y)(2x + y)(4x2 + y2)
First solving (2x – y) (2x+y)
Using the identity (a +b) (a–b) = a2 – b2
Here a = 2x and b = y
= (2x)2 – y2
= 4x2 – y2
Now solving (4x2 – y2 ) (4x2 + y2)
Again using the identity (a +b) (a–b) = a2 – b2
Here a = 4x2 and b = y2
= (4x2)2 – (y2)2
= 16 x4 – y4
Find the product:
(2x – 3y)(2x + 3y)(4x2 + 9y2)
First solving for (2x – 3y)(2x + 3y)
Using the identity (a +b) (a–b) = a2 – b2
Here a = 2x and b = 3y
= (2x)2 – (3y)2
= 4x2 – 9y2
Now solving for ( 4x2 – 9y2 ) (4x2 + 9y2 )
Again Using the identity (a +b) (a–b) = a2 – b2
Here a = 4x2 and b = 9y2
= (4x2 )2 – (9y2)2
= 16 x4 – 81 y4
Terms having the same literal factors with same exponents are called.
A. exponents
B. like terms
C. factors
D. unlike terms
Terms having the same literal factors with same exponents are called like terms
So B is the correct option
The coefficient of ab in 2ab is:
A. ab
B. 2
C. 2a
D. 2b
The coefficient is the term excluding ab in 2ab.
So B is the correct option
The exponential form of a × a × a is:
A. 3a
B. 3 + a
C. a3
D. 3 – a
The exponential form of a × a × a is a3.
So C is the correct option.
Sum of two negative integers is:
A. negative
B. positive
C. zero
D. infinite
Sum of two negative integers is always negative
So A is the correct option.
What should be added to a2 + 2ab to make it a complete square?
A. b2
B. 2ab
C. ab
D. 2a
(a + b)2 = a2 + 2ab + b2
In the above expression we have to add b2 to make it (a + b)2
So A is the correct option.
What is the product of (x + 2)(x–3)?
A. 2x – 6
B. 3x – 2
C. x2 – x – 6
D. x2 – 6x
(x + 2)(x-3)
⇒ (x + 2)(x-3) = x(x-3) + 2(x-3)
⇒ (x + 2)(x-3) = x2-3x + 2x-6
⇒ (x + 2)(x-3) = x2 – x – 6
So C is the correct option.
The value of (7.2)2 is (use an identity to expand)
A. 49.4
B. 14.4
C. 51.84
D. 49.04
(7.2)2 = (7 + 0.2)2
⇒ (7.2)2 = 72 + 0.22 + 2× 7× 0.2
⇒ (7.2)2 = 49 + 0.04 + 2.8
⇒ (7.2)2 = 51.84
So C is the correct option.
The expansion of (2x – 3y)2 is:
A. 2x2 + 3y2 + 6xy
B. 4x2 + 9y2 – 12xy
C. 2x2 + 3y2 – 6xy
D. 4x2 + 9y2 + 12xy
(a + b)2 = a2 + 2ab + b2
⇒ (2x – 3y)2 = 4x2 + 9y2 – 12xy
So B is the correct option.
The product 58 × 62 is
A. 4596
B. 2596
C. 3596
D. 6596
58× 62 = (60-2)× (60 + 2)
We use the form (a + b)(a-b) = a2 - b2
⇒ 58× 62 = 602-22
⇒ 58× 62 = 3600-4 = 3596
So C is the correct option.
Take away 8x – 7y – 8p + 10q from 10x + 10y – 7p + 9q.
We need to subtract the two expressions
10x + 10y – 7p + 9q-8x + 7y + 8p - 10q = 2x-17y + p-q
Expand:
(i) (4x + 3)2
(ii) (x + 2y)2
(iii)
(iv)
(i) We use (a + b)2 = a2 + 2ab + b2
⇒ (4x + 3)2 = (4x)2 + 2× 4x × 3 + 32
⇒ (4x + 3)2 = 16x2 + 24x + 9
(ii) We use (a + b)2 = a2 + 2ab + b2
⇒(x + 2y)2 = x2 + 4xy + 4y2
(iii) We use (a + b)2 = a2 + 2ab + b2
(iv) We use (a - b)2 = a2- 2ab + b2
Expand:
(i) (2t + 5)(2t – 5);
(ii) (xy + 8)(xy – 8);
(iii) (2x + 3y)(2x – 3y).
(i) We use the form (a + b)(a-b) = a2 - b2
⇒ (2t + 5)(2t – 5) = 4t2-25
(ii) We use the form (a + b)(a-b) = a2 - b2
⇒(xy + 8)(xy – 8) = x2y2-64
(iii) We use the form (a + b)(a-b) = a2 - b2
⇒(2x + 3y)(2x – 3y) = 4x2-9y2
Expand:
(n – 1)(n + 1)(n2 + 1)
Applying the formula (a + b)(a - b) = a2 - b2 on first two terms
(n – 1)(n + 1)(n2 + 1) = (n2 - 1) (n2 + 1)
Applying the formula (a + b)(a - b) = a2 - b2 again
⇒ (n – 1)(n + 1)(n2 + 1) = (n4 - 1)
Expand:
Applying the formula (a + b)(a - b) = a2 - b2 on first two terms
Applying the formula (a + b)(a - b) = a2 - b2 again
Expand:
(x – 1)(x + 1)(x2 + 1)(x4 + 1)
Applying the formula (a + b)(a - b) = a2 - b2 on first two terms
(x – 1)(x + 1)(x2 + 1)(x4 + 1) = (x2 - 1) (x2 + 1)(x4 + 1)
Applying the formula (a + b)(a - b) = a2 - b2 again
⇒ (x – 1)(x + 1)(x2 + 1)(x4 + 1) = (x4 - 1)(x4 + 1)
Applying the formula (a + b)(a - b) = a2 - b2 again
⇒ (x – 1)(x + 1)(x2 + 1)(x4 + 1) = (x8 - 1)
Expand:
(2x – y)(2x + y)(4x2 + y2)
Applying the formula (a + b)(a - b) = a2 - b2 on first two terms
(2x – y)(2x + y)(4x2 + y2) = (4x2 - y2) (4x2 + y2)
Applying the formula (a + b)(a - b) = a2 - b2 again
⇒ (2x – y)(2x + y)(4x2 + y2) = (16x4 – y4)
Use appropriate formulae and compute:
(103)2
(103)2 = (100 + 3)2
We use (a + b)2 = a2 + 2ab + b2
⇒ (103)2 = 1002 + 2× 100× 3 + 32
⇒ (103)2 = 10000 + 600 + 9 = 10609
Use appropriate formulae and compute:
(96)2
(96)2 = (100-4)2
We use (a - b)2 = a2- 2ab + b2
⇒ (96)2 = 1002-2× 100× 4 + 42
⇒ (96)2 = 10000-800 + 16 = 9216
Use appropriate formulae and compute:
107 × 93
107 × 93 = (100 + 7)(100 – 7)
We use (a + b)(a - b) = a2 - b2
⇒ 107 × 93 = 1002-72
⇒ 107 × 93 = 10000-49
⇒ 107 × 93 = 9951
Use appropriate formulae and compute:
1008×992
1008 × 992 = (1000 + 8)(1000 – 8)
We use (a + b)(a - b) = a2 - b2
⇒ 1008 × 992 = 10002-82
⇒ 1008 × 992 = 1000000-64
⇒ 1008 × 992 = 999936
Use appropriate formulae and compute:
1852 – 1152
1852 – 1152= (150 + 35)2-(150-35)2
(a + b)2-(a-b)2 = 4ab
⇒ 1852 – 1152= 4× 150× 35
⇒ 1852 – 1152 = 21000
If x + y = 7 and xy = 12, find x2 + y.
x + y = 7
⇒ y = 7-x …Equation (i)
xy = 12
Putting the value of y from above Equation (i) we get
x(7-x) = 12
⇒ 7x-x2 = 12
⇒x2-7x + 12 = 0
Solving the above equation by the method of factorization we get
⇒ x2-4x-3x + 12 = 0
⇒x(x-4)-3(x-4) = 0
⇒(x-4)(x-3) = 0
x = 4,3
When x = 4, y = 3
⇒ x2 + y = 19
When x = 3, y = 4
⇒ x2 + y = 13
If x + y = 12 and xy = 32, find x2 + y.
x + y = 12
⇒ y = 12-x …Equation (i)
xy = 32
Putting the value of y from above Equation (i) we get
x(12-x) = 32
⇒ 12x-x2 = 32
⇒x2-12x + 32 = 0
Solving the above equation by the method of factorization we get
⇒ x2-8x-4x + 12 = 0
⇒x(x-8)-4(x-8) = 0
⇒(x-4)(x-8) = 0
x = 4,8
When x = 4, y = 8
⇒ x2 + y = 24
When x = 8, y = 4
⇒ x2 + y = 68
If 4x2 + y2 = 40 and xy = 6, find 2x + y.
4x2 + y2 = 40 …Equation (i)
xy = 6
⇒ 4xy = 24 …Equation (ii)
Adding Equation (i) and (ii)
4x2 + y2 + 2xy = 64
⇒ (2x + y)2 = 82
⇒ 2x + y = ±8
If x – y = 3 and xy = 10, find x2 + y.
x - y = 3
⇒ y = x-3 …Equation (i)
xy = 10
Putting the value of y from above Equation (i) we get
x(x-3) = 10
⇒ x2-3x = 10
⇒x2-3x-10 = 0
Solving the above equation by the method of factorization we get
⇒ x2-5x + 2x + 10 = 0
⇒x(x-5) + 2(x-5) = 0
⇒(x + 2)(x-5) = 0
x = -2,5
Neglecting the negative value
When x = 5, y = 2
⇒ x2 + y = 27
…Equation (i)
Squaring both sides of the equation we get
Cubing both sides of the equation (i) we get
If x + = 6, find x2 + and x4 +
Squaring both sides of the equation we get
Again Squaring both sides of the above equation we get
Simplify:
(i) (x + y)2 + (x – y)2;
(ii) (x + y)2 × (x – y)2.
(i) We use the formula (a + b)2 + (a - b)2 = 2(a2 + b2)
(x + y)2 + (x – y)2 = 2(x2 + y2)
(ii) Applying the formula (a + b)2(a - b)2 = (a2 - b2)2
(x + y)2 × (x – y)2 = (x2 - y2)2
⇒ (x + y)2 × (x – y)2 = x4-2x2y2 + y4
Express the following as difference of two squares:
(i) (x + 2z)(2x + z);
(ii) 4(x + 2y)(2x + y);
(iii) (x + 98)(x + 102);
(iv) 505 × 495.
(i) We use
(x + 2z)(2x + z) =
⇒ (x + 2z)(2x + z) =
(ii) 4(x + 2y)(2x + y) = (2x + 4y)(4x + 2y)
We use
⇒ 4(x + 2y)(2x + y) =
⇒ 4(x + 2y)(2x + y) = ( 3x + 3y)2-(2y – 2x)2
(iii) We use
(x + 98)(x + 102) =
⇒ (x + 98)(x + 102) = (x + 100)2-(x-100)2
(iv) 505 × 495 = (500 + 5)(500-5)
We use (a + b)(a - b) = a2 - b2
⇒ 505 × 495 = 5002-52
If a = 3x – 5y, b = 6x + 3y and c = 2y – 4x, find
(i) a + b – c;
(ii) 2a–3b + 4c.
(i) a + b – c = 3x – 5y + 6x + 3y-2y + 4x
⇒ a + b – c = 13x - 4y
(ii) 2a–3b + 4c = 2(3x – 5y) + 3(6x + 3y)-4(2y – 4x)
⇒ 2a–3b + 4c = 6x-10y + 18x + 9y-8y + 16x
⇒ 2a–3b + 4c = 40x-9y
The perimeter of a triangle is 15x2 – 23x + 9 and two of its sides are 5x2 + 8x – 1 and 6x2 – 9x + 4. Find the third side.
Perimeter = 15x2 – 23x + 9
First side = 5x2 + 8x – 1
Second side = 6x2 – 9x + 4
Sum of first two side = 11x2-x + 3
Third side = (Perimeter- Sum of first two side)
⇒ Third side = 4x2-22x + 6
The two adjacent sides of a rectangle are 2x2 – 5xy + 3z2 and 4xy – x2 – z.
Two adjacent sides are 2x2 – 5xy + 3z2 and 4xy – x2 – z.
Area of rectangle = (2x2 – 5xy + 3z2)× (4xy – x2 – z)
⇒ Area of rectangle = 2x2(4xy – x2 – z)-5xy(4xy – x2 – z) + 3z2(4xy – x2 – z)
⇒ Area of rectangle = 8x3y-2x4-2x2z-20x2y2 + 5x3y + 5xyz + 12xyz2-3x2z2-3z3
Perimeter of rectangle = 2× (2x2 – 5xy + 3z2 + 4xy – x2 – z)
⇒ Perimeter of rectangle = 2× (x2-xy + 3z2-z)
⇒ Perimeter of rectangle = (2x2-2xy + 6z2-2z)
The base and the altitude of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.
Base of triangle = (3x – 4y)
Altitude of triangle = (6x + 5y)
Area of triangle
⇒ Area of triangle
⇒ Area of triangle
⇒ Area of triangle
The sides of a rectangle are 2x + 3y and 3x + 2y. From this a square of side length x + y is removed. What is the area of the remaining region?
Length of rectangle = 2x + 3y
Breadth of rectangle = 3x + 2y
Area of rectangle = (Length× breadth) = (2x + 3y)× (3x + 2y)
⇒ Area of rectangle = 6x2 + 13xy + 6y2
Side of square = x + y
Area of Square = (Side× Side) = (x + y)2
⇒ Area of Square = x2 + y2 + 2xy
Area of remaining region = ( Area of rectangle- Area of Square) = 5x2 + 5y2 + 11xy
If a, b are rational numbers such that a2 + b2 + c2 – ab – bc – ca = 0, prove that a = b = c.
a2 + b2 + c2 – ab – bc – ca = 0
Multiplying both sides by 2 we get
2 (a2 + b2 + c2 – ab – bc – ca ) = 0
⇒ (a2 + b2 - 2ab) + ( b2 + c2 - 2bc) + (c2 + a2 -2ac) = 0
The individual terms inside the brackets can be expressed as a whole square
⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0
Since a, b, c are rational and none of the term is equal to zero so each of the terms inside the bracket must individually be equal to zero
⇒ a – b = 0
⇒ a = b
⇒ b – c = 0
⇒ b = c
⇒ c – a = 0
⇒ c = a
So together we can say that
a = b = c