8 Math Algebraic Expressions

Finish Line & Beyond

Algebraic Expressions 1. Expressions are formed from variables and constants. 2. Terms are added to form expressions. Terms themselves are formed as product of factors. 3. Expressions that contain exactly one, two and three terms are called monomials, binomials and trinomials respectively. In general, any expression containing one or more terms with non-zero coefficients (and with variables having non- negative exponents) is called a polynomial. 4. Like terms are formed from the same variables and the powers of these variables are the same, too. Coefficients of like terms need not be the same. 5. While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them; then handle the unlike terms. 6. There are number of situations in which we need to multiply algebraic expressions: for example, in finding area of a rectangle, the sides of which are given as expressions. 7. A monomial multiplied by a monomial always gives a monomial. 8. While multiplying a polynomial by a monomial, we multiply every term in the polynomial by the monomial. 9. In carrying out the multiplication of a polynomial by a binomial (or trinomial), we multiply term by term, i.e., every term of the polynomial is multiplied by every term in the binomial (or trinomial). Note that in such multiplication, we may get terms in the product which are like and have to be combined. 10. An identity is an equality, which is true for all values of the variables in the equality. On the other hand, an equation is true only for certain values of its variables. An equation is not an identity. 11. The following are the standard identities: (a + b)2 = a2 + 2ab + b2 …………………..(I) (a – b) 2 = a2 – 2ab + b2 …………………….(II) (a + b) (a – b) = a2 – b2 …………………….(III) 12. Another useful identity is (x + a) (x + b) = x2 + (a + b) x + ab (IV) 13. The above four identities are useful in carrying out squares and products of algebraic expressions. They also allow easy alternative methods to calculate products of numbers and so on.

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond Exercise 1 1. Identify the terms, their coefficients for each of the following expressions. (i) 5xyz2 – 3zy Answer: Term: xyz2 -- Coefficient = 5 Term: zy – Coefficient = 3 (ii) 1 + x + x2 Answer: Term: x – Coefficient = 1 & Term: x2 – Coefficient = 1 (iii) 4x2y2 – 4x2y2z2 + z2 Answer: 4 is the coefficient for x2y2z2 1 is the term for z2 (iv) 3 – pq + qr – rp Answer: For each term the coefficient is 1

x y + − xy 2 2 1 Answer: is the coefficient for x and y and for xy it is 1 2 (v)

(vi) 0.3a – 0.6ab + 0.5b Answer: 0.3 is the coefficient for a, for b there are two coefficients 0.6 and 0.5 2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories? Answer: x + y: Binomial 1000: Mononomial x + x2 + x3 + x4: Polynomial 7 + y + 5x: Binomial 2y – 3y2: Binomial 2y – 3y2 + 4y3: Trinomial 5x – 4y + 3xy: Trinomial 4z – 15z2: Binomial ab + bc + cd + da: Polynomial pqr: Mononomial p2q + pq2: Binomial 2p + 2q: Binomial

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond 3. Add the following. (i) ab – bc, bc – ca, ca – ab

(ab − bc) + (bc − ca) + (ca − ab) = ab − bc + bc − ca + ca − ab = ab − ab + bc + bc − ca + ca = bc + bc = 2bc

Answer:

(ii) a – b + ab, b – c + bc, c – a + ac

(a − b + ab) + (b − c + bc ) + (c − a + ac) = a − b + ab + b − c + bc + c − a + ac = a − a − b + b − c + c + ab + bc + ac = ab + bc + ca

Answer:

(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2

(2 p ²q² - 3pq + 4) + (5 + 7pq - 3p²q²) 2 p ²q² - 3pq + 4 + 5 + 7pq - 3p²q² 2 p ²q² - 3p²q² - 3pq + 7pq + 4 + 5 − p ²q² + 4pq + 9 4 pq + 9 − p ²q²

Answer:

= = = =

(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl

l ² + m² + m² + n² + n² + l² + 2lm + 2mn + 2nl = 2l ² + 2m² + 2n² + 2lm + 2mn + 2nl = 2(l ² + m² + n² + lm + mn + nl)

Answer:

4. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(12a − 9ab + 5b − 3) − (4a − 7 ab + 3b + 12) = 12a − 9ab + 5b − 3 − 4a + 7 ab − 3b − 12

Answer:

(signs are reversed after –sign once bracket is opened)

= 8a − 2ab + 2b − 15

(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

(5 xy − 2 yz − 2 zx + 10 xyz ) − (3 xy + 5 yz − 7 zx ) = 5 xy − 2 yz − 2 zx + 10 xyz − 3xy − 5 yz + 7 zx = 2 xy − 7 yz + 5 zx + 10 xyz

Answer:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond (c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q

(18 − 3 p − 11q + 5 pq − 2 pq ² + 5p²q) - (4p²q - 3pq + 5pq² - 8p + 7 q − 10) = 18 + 3 p − 11q + 5 pq − 2 pq ² + 5p²q - 4p²q + 3pq - 5pq² + 8p - 7q + 10 = 28 + 11 p − 18q + 8 pq 7 pq ² + p²q

Answer:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond Exercise 2 1. Find the product of the following pairs of monomials. (i) 4, 7p Answer: 4 × 7 p (ii) – 4p, 7p Answer:

= 28 p

− 4 p × 7 p = − 28 p ²

(iii) – 4p, 7pq Answer:

− 4 p × 7 pq = − 28 p ²q

(iv) 4p3, – 3p Answer: 4 p ³ × (v) 4p, 0 Answer:

− 3 p = − 12 p 

4p× 0 = 0

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively. (p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np) Answer: Area = Length × breadth (i) p × q = pq (ii) 10m × 5n = 50mn (iii) 20 x ² × 5y² = 100x²y² (iv) 4 x × 3 x ² = 12x ³ (v) 3mn × 4np = 12mn²p

3. Complete the following table of products: First Mononomial Second Mononomial 2x -5y 3x² -4xy 7x²y -9x²y²

2x

-5y

3x²

-4xy

7x²y

-9x²y²

4x² -10xy 6x³ -8x²y 14x³y -18x³y²

-10xy 10y² -15x²y 20xy² -35x²y² 45x²y³

6x³ -15x²y 9x -12x³y 21xy² -27xy²

-8x²y 20xy² -12x³y 16x²y² -28x³y² 36x³y³

14x³y -35x²y² 21xy² -28x³y² 49xy² -63xy³

-18x³y² 45x²y³ -27xy² 36x³y³ -63xy³ 81xy

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond 4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively. (i) 5a, 3a², 7a (ii) 2p, 4q, 8r (iii) xy, 2x²y, 2xy² (iv) a, 2b, 3c Answer: Volume= length × breadth × height (i) 5a × 3a² × 7a=105a7 (ii) 2p × 4q × 8r = 64pqr (iii) xy × 2x²y × 2xy² = 4xy (iv) a × 2b × 3c = 6abc 5. Obtain the product of (i) xy, yz, zx (ii) a, – a², a³ (iii) 2, 4y, 8y², 16y³ (iv) a, 2b, 3c, 6abc (v) m, – mn, mnp Answer: (i) x²y²z² (ii) –a5 (iii) 1024y6 (iv) 36a²b²c² (v) -m³n²p

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond Exercise 3 1. Carry out the multiplication of the expressions in each of the following pairs. (i) 4p, q + r Answer: 4 p (q + (ii) ab, a – b

r ) = 4 pq + 4 pr

Answer: ab(a − b) (iii) a + b, 7a²b² Answer:

= a ²b - ab²

(a + b)7a ²b² = 7a³b²+7a²b³

(iv) a² – 9, 4a Answer:

(a ² - 9)4a = 4a³-36a

(v) pq + qr + rp, 0 Answer:

( pq + qr + rp )0 = 0

2. Find the product. (i) (a²) × (2a22) × (4a26) Answer: As you know an × am × ao = am+n+o So, (a²) × (2a22) × (4a26) = 8a50

2  9  xy ×  − x ²y²  3  10  3 = − x ³y³ 5 10 6 pq ³ × p ³q (iii) − 3 5 (ii)

=-4pq (iv) x × x² × x³ × x =x14

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond 3. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =

1 2

Answer: 3x(4x-5)+3 =12x²-15x+3 (i) putting x=3 in the equation we get 12 × 3²-15 × 3+3 =108-45+3 =63 (ii) putting x=

1 in the equation we get 2

1 1 -15 × +3 4 2 15 =3+3 2 6 − 15 + 6 15 1 = = = 7 2 2 2 12 ×

(b) Simplify a (a² + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = – 1. Answer: a(a²+a+1) =a³+a²+a (i) putting a= 0 in the equation we get 0³+0²+0=0 (ii) putting a=1 in the equation we get 1³+1²+1=1+1+1=3 (iii) putting a=-1 in the equation we get -1³+1²+1=-1+1+1=1 5. (a) Add: p ( p – q), q ( q – r) and r ( r – p)

p ( p − q ) + q (q − r ) + r (r − p ) = p ² - pq + q² - qr + r² - rp = p ² + q² + r² - pq - qr - pr

Answer:

(b) Add: 2x (z – x – y) and 2y (z – y – x)

2 x( z − x − y ) + 2 y ( z − y − x ) = 2 xz − 2 x ² - 2xy + 2yz - 2y² - 2xy = 2 xz − 2 xy + 2 yz − 2 x ² - 2y²

Answer:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond (c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )

4l (10n − 3m + 2l ) − 3l (l − 4m + 5n) = 40 ln − 12lm + 8l ² - 3l² + 12lm - 15ln = 25lm + 5l ²

Answer:

(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c) from 4c ( – a + b + c )

4c(− a + b + c) − ( 3a (a + b + c) − 2b(a − b + c) ) − 4ac + 4bc + 4c ² - (3a² + 3ab + 3ac - 2ab + 2b² - 2bc) − 4ac + 4bc + 4c ² - 3a² - 3ab - 3ac + 2ab - 2b² + 2bc − 7 ac + 6bc + 4c ² - 3a² - ab - b² 4c ² - 3a² - 2b² - ab + 6bc - 7ac

Answer:

= = = =

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond Exercise 4 1. Multiply the binomials. (i) (2x + 5) and (4x – 3)

(2 x + 5)(4 x − 3) = 2x × 4x − 2x × 3 + 5 × 4x − 5 × 3 = 8 x ² - 6x + 20x - 15 = 8 x ² + 14x - 15

Answer:

(ii) (y – 8) and (3y – 4)

( y − 8)(3 y − 4) = y × 3y − y × 4 − 8 × 3y + 8 × 4 = 3 y ² - 4y - 24y + 32 = 3 y ² - 28y + 32

Answer:

(iii) (2.5l – 0.5m) and (2.5l + 0.5m) Answer: (2.5l − 0.5m)(2.5l + 0.5m) This can be solved using following identity: (a − b)(a + b) = a ² - b² ; Assume 2.5l = a and Resulting Equation

=6.25l²-0m

0.5m = b

= 2.5²l² - 0.5²m²

(iv) (a + 3b) and (x + 5)

(a+b3)x5 = a × x + a × 5 + 3b × x + 3b × 5 = ax + 5a + 3bx + 15b

Answer:

(v) (2pq + 3q²) and (3pq – 2q²)

(2 pq + 3q ²)(3pq - 2q²) = 2 pq × 3 pq − 2 pq × 2q ² + 3q² × 3pq - 3q² × 3q² = 6 p ²q² - 4pq³ + 9pq³ - 9q  =6p²q²+5pq³-9q Answer:

2  3    a ² + 3b²  × 4 a ² - b ²  3  4   8  3  =  a ² + 3b²   4a ² + b ²  3  4 

(vi)

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond =

3 3 8 8 a ² × 4a² - a ² × b² + 3b² × 4a² - 3b² × b² 4 4 3 3

= 3a-2a²b²+12a²b²-8b

=3a+10a²b²-8b

2. Find the product. (i) (5 – 2x) (3 + x)

(5 − 2 x )(3 + x) = 5 × 3 + 5 × x − 2x × 3 − 2x × x = 15 + 5 x − 6 x − 2 x ² = 15 − x − 2 x ²

Answer:

(ii) (x + 7y) (7x – y)

( x + 7 y )(7 x − y ) = x × 7x − x × y + 7 y × 7x − 7 y × y = 7 x ² - xy + 49xy - 7y² = 7 x ² + 48xy - 7y²

Answer:

(iii) (a² + b) (a + b²)

(a ² + b)(a + b²) = a ² × a + a² × b² + b × a + b × b² = a ³ + a ²b² + ab + b³ = a ³ + b³ + a²b² + ab

Answer:

(iv) (p² – q²) (2p + q)

( p ² - q²)(2p + q) = p ² × 2p + p²q - q² × 2p - q² × q = 2 p ³ + p ²q - 2pq² - q³

Answer:

3. Simplify. (i) (x² – 5) (x + 5) + 25

( x ² - 5)(x + 5) + 25 = ( x ² × x + 5 × x - 5 × x - 5 × 5) + 25 = x ³ + 5 x − 5 x − 25 + 25 = x³

Answer:

(ii) (a² + 5) (b³ + 3) + 5

(a ² + 5)(b³ + 3) + 5 = a ² × b³ + a² × 3 + 5 × b³ + 5 × 3 = a ²b ³ + 3a ² + 5b ³ + 15

Answer:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond (iii) (t + s²) (t² – s)

(t + s ²)(t² - s) = t × t ² - t × s + s² × t² - s² × s = t ³ − ts + s ²t² - s³

Answer:

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

(a + b)(c − d ) + (a − b)(c + d ) + 2(ac + bd ) = ( ac − ad + bc − bd ) + (ac + ad − bc − bd ) + 2ac + 2bd = ac − ad + bc − bd + ac + ad − bc − bd + 2ac + 2bd = 4ac

Answer:

(v) (x + y)(2x + y) + (x + 2y)(x – y)

( x + y )(2 x + y ) + ( x + 2 y )( x − y ) = ( 2 x ² + xy + 2xy + y²) + ( x² - xy + 2xy - 2y²) = 2 x ² + 3xy + y² + x² + xy - 2y² = 3x ² + 4xy - y²

Answer:

(vi) (x + y)(x² – xy + y²)

( x + y )( x ² - xy + y²) = x ³ − x ²y + xy² + x²y - xy² + y³ = x ³ + y³

Answer:

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

(1.5 x − 4 y )(1.5 x + 4 y + 3) − 4.5 x + 12 y = ( 2.25 x ² + 6xy + 4.5x - 6xy - 16y² - 12y) - 4.5x + 12y = 2.25 x ² + 4.5x - 16y² - 12y - 4.5x + 12y = 2.25 x ² - 16y²

Answer:

(viii) (a + b + c)(a + b – c)

(a + b + c)(a + b − c) = a ² + ab - ac + ab + b² - bc + ac + bc - c² = a ² + b² - c² + 2ab

Answer:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond Exercise 5 1. Use a suitable identity to get each of the following products. (i) (x + 3) (x + 3) Answer: Using (a+b) ²=a²+2ab+b², we get the following equation:

x ² + 6x + 9

(ii) (2y + 5) (2y + 5) Answer:

4y² + 20y + 25 (same as previous question)

(iii) (2a – 7) (2a – 7) Answer: As you know So,

(a − b)² = a² - 2ab + b²

(a2−)7²=4-8+9 1 1  3a −   3a −  2 2 

(iv) 

Answer:

9a ² - 3a +

1 (same as previous question) 4

(v) (1.1m – 0.4) (1.1m + 0.4) Answer: Use 1.21m²-0.16

(a − b)(a + b) = a ² - b²

(vi) (a² + b²) (– a² + b²) Answer: (a²+b²)(-a²+b²) =(b²+a²)(b²-a²) =b-a (same as previous question) (vii) (6x – 7) (6x + 7) Answer: 36x²-49 (same as previous question) (viii) (– a + c) (– a + c) Answer: (-a+c)(-a+c) =(c-a)(c-a) =c²+a²-2ac

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond  x 3y   x 3y  +  +   2 4  2 4 

(ix) 

Answer: Use (a+b) ² =

x ² 9y² 3xy + + 4 16 4

(x) (7a – 9b) (7a – 9b) Answer: Using (a-b) ²

= 49a ² + 81b² - 126ab

2. Use the identity (x + a) (x + b) = x² + (a + b) x + ab to find the following products. (i) (x + 3) (x + 7)

( x + 3)( x + 7) = x ² + (3 + 7)x + 21 = x ² + 10x + 21

Answer:

(ii) (4x + 5) (4x + 1)

(4 x + 5)(4 x + 1) = 16 x ² + (5 + 1)4x + 5 = 16 x ² + 24x + 5

Answer:

(iii) (4x – 5) (4x – 1)

(4 x − 5)(4 x − 1) = 16 x ² + (-5 + -1)4x + (-5) = 16x ² - 24x - 5

Answer:

(iv) (4x + 5) (4x – 1)

(4 x + 5)(4 x − 1) = 16 x ² + (5 - 1)4x - 5 = 16 x ² + 4x - 5

Answer:

(v) (2x + 5y) (2x + 3y)

(2 x + 5 y )(2 x + 3 y ) = 4 x ² + (5y + 3y)2x + 15y² = 4 x ² + 8xy + 15y²

Answer:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond (vi) (2a² + 9) (2a² + 5) Answer: (2a ² + 9)(2a² + =2a+(9+5)2a²+45 =2a+28a²+45

5)

(vii) (xyz – 4) (xyz – 2) Answer: ( xyz −

4)( xyz − 2) = x ²y²z² + (-4 - 2)xyz + 8 = x ²y²z² - 6xyz + 8

3. Find the following squares by using the identities. (i) (b – 7) ² Answer: Use (a-b) ²=a²+b²-2ab

(b − 7)² = b² + 49 - 14b

(ii) (xy + 3z) ²

(a + b) ² = a² + b² + 2ab ( xy + 3 z )² = x²y² + 9z² + 6xyz

Answer: Use

(iii) (6x² – 5y) ² Answer: Use

(a − b)² = a² + b² - 2ab

(6 x ² - 5y)² =36x+25y²-60x²y (iv)

2 3 ( m + n) ² 3 2

(a + b) ² = a² + b² + 2ab 2 3 4 9 2 3 ( m + n)² = m² + n² + 2 × × × mn 3 2 9 4 3 2 4 9 = m² + n ² + 2mn 9 4

Answer: Use

(v) (0.4p – 0.5q) ²

(a − b)² = a² + b² + 2ab (0.4 p − 0.5q )² = 0.16p² + 0.25q² - 0.4pq

Answer: Use

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond (vi) (2xy + 5y) ²

(a + b) ² = a² + b² + 2ab (2 xy + 5 y )² = 4x²y² + 25y² + 20xy²

Answer: Use

4. Simplify. (i) (a² – b²)² Answer: Use (a − b)² = (a²-b²)²=a+b-2a²b²

a² + b² + 2ab

(ii) (2x + 5) ² – (2x – 5) ²

(2 x + 5)² - (2x - 5)² = ( 4 x ² + 25 + 20x) - (4x² + 25 - 20x) = 4 x ² + 25 + 20x - 4x² - 25 + 20x = 20 x

Answer:

(iii) (7m – 8n) ² + (7m + 8n) ²

(7 m − 8n)² + (7m + 8n)² = ( 49m² + 64n² - 112mn) + (49m² + 64n² - 112mn) = 49m² + 64n² + 112mn + 49m² + 64n² - 112mn = 98m² + 128n²

Answer:

(iv) (4m + 5n) ² + (5m + 4n) ² Answer: (4m +

5n)² + (5m + 4n)² = 16m² + 25n² + 40mn + 25m² + 16n² + 40mn

= 41m ² + 41n² + 8

(v) (2.5p – 1.5q) ² – (1.5p – 2.5q) ²

(2.5 p − 1.5q )² - (1.5p - 2.5q)² = (6.25 p ² + 2.25q² - 0.75pq) - (2.25p² + 6.25q² - 0.75pq) = 6.25 p ² + 2.25q² - 0.75pq - 2.25p² - 6.25q² + 0.75pq = 4 p ² - 4q²

Answer:

(vi) (ab + bc) ² – 2ab²c

(ab + bc)² - 2ab²c = ( a ²b² + b²c² + 2ab²c) - 2ab²c = a ²b² + b²c²

Answer:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond (vii) (m² – n²m) ² + 2m³n² Answer: (m ² - n²m)² + 2m³n² =m+nm²-2mn²+2m³n² =m+nm² 5. Show that. (i) (3x + 7)² – 84x = (3x – 7) ²

(3x + 7)² - 84x = 9 x ² + 49 + 42x - 84x = 9 x ² + 49 - 42x ………………………..(1)

Answer: LHS:

(3x − 7)² = 9 x ² + 49 - 42x ………………………………………(2)

RHS

Equations (1) & (2) prove that LHS =RHS

(ii) (9p – 5q) ² + 180pq = (9p + 5q) ²

(9 p − 5q )² + 180pq = 81 p ² + 25q² - 90pq + 180pq = 81 p ² + 25q² + 90pq …………………….(1)

Answer: LHS:

(9 p + 5q)² = 81 p ² + 25q² + 90pq ……………………..(2)

RHS:

Equations (1) & (2) prove that LHS = RHS (iii) (

4 3 16 9 m − n)² + 2mn = m² + n² 3 4 9 16

Answer: LHS:

16 m² + 9 16 = m² + 9 =

4 3 ( m − n)² + 2mn 3 4

9 n² - 2mn + 2mn 16 9 n² = RHS Proved 16

(iv) (4pq + 3q) ² – (4pq – 3q) ² = 48pq²

(4 pq + 3q)² - (4pq - 3q)² = (16 p ²q² + 9q² + 24pq²) - (16p²q² + 9q² - 24pq²) = 16 p ²q² + 9q² + 24pq² - 16p²q² - 9q² + 24pq² = 48 pq ² = RHS Proved

Answer: LHS:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond (v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

(a − b)(a + b) + (b − c)(b + c) + (c − a)(c + a ) = a ² - b² + b² - c² + c² - a² = 0 Proved

Answer:

6. Using identities, evaluate. (i) 71² Answer: 71² Can be written as follows:

(70 + 1)² which can be solved using (a + b)² = a² + b² + 2ab = 70² + 1² + 2 × 70 × 1 = 4900 + 1 + 140 = 5041

(ii) 99² Answer: 99² can be written as follows: (100-1) ² and can be solved using (a − b) ²

= a² + b² - 2ab

= 100² + 1² - 2 × 100 × 1 = 10000 + 1 − 200 = 99801 (iii) 102²

102² = (100 + 2)² = 100² + 2² + 2 × 100 × 2 = 10000 + 4 + 400 = 10404

Answer:

(iv) 998²

998² = (1000 - 2)² = 1000² + 2² - 2 × 1000 × 2 = 1000000 + 4 − 4000 = 996004

Answer:

(v) 5.2²

5.2² = (5 + 0.2)² = 5² + 0.2² + 2 × 5 × 0.2 = 25 + 0.04 + 2 = 27.04

Answer:

(vi) 297 × 303

(297 × 303) = (300 − 3)(300 + 3) Using (a − b)(a + b) = a ² - b² we get Answer:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond 300² - 3² = 90000 − 9 = 89991 (vii) 78 × 82 Answer:

78 × 82 = (80 − 2)(80 + 2)

= 80² - 2² = 6400 − 4 = 6396

(viii) 8.9²

8.9² = (9 - 0.1)² = 9² + 0.1² - 2 × 9 × 0.1 = 81 + 0.01 − 1.8 = 79.21

Answer:

(ix) 1.05 × 9.5 Answer: 10.5 ×

9.5 = (10 + 0.5)(10 − 0.5)

= 10² - 0.5² = 100 − 0.25 = 99.75

7. Using a² – b² = (a + b) (a – b), find (i) 51² – 49²

51² - 49² = (51 + 49)(51 − 49) = 100 × 2 = 200

Answer:

(ii) (1.02) ² – (0.98) ² Answer: (1.02)² - (0.98)²

= (1.02 + 0.98)(1.02 − 0.98) = 2 × 0.04 = 0.08 (iii) 153² – 147² Answer: 153² - 147²

= (153 + 147)(153 − 147) = 300 × 6 = 1800

(iv) 12.1² – 7.9²

12.1² - 7.9² = (12.1 + 7.9)(12.1 − 7.9) = 20 × 4.2 = 84

Answer:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond 8. Using (x + a) (x + b) = x² + (a + b) x + ab, find (i) 103 × 104

103× 104 (100 + 3)(100 + 4) 100² + (3 + 4)100 + 4 × 3 10000 + 700 + 12 10712

Answer:

= = = =

(ii) 5.1 × 5.2

5.1 × 5.2 (5 + 0.1)(5 + 0.2) 5² + (0.1 + 0.2)5 + 0.1 × 0.2 25 + 1.5 + 0.02 26.52

Answer:

= = = =

(iii) 103 × 98 Answer: 103×

98 = (90 + 13)(90 + 8) = 90² + (13 + 8)90 + 13 × 8 = 8100 + 1890 + 104 = 10094

(iv) 9.7 × 9.8

9 .7 × 9 .8 (9 + 0.7)(9 + 0.8) 9² + (0.7 + 0.8)9 + 0.7 × 0.8 81 + 13.5 + 0.56 95.05

Answer:

= = = =

www.excellup.com ©2009 send your queries to [email protected]