Karnataka Class 8 Math Solution Chapter Algebraic Expressions and Identities by Experience Math Teacher. Karnataka Board Class 8 Math Part 1 PDF Chapter 6 Exercise 6.1, Exercise 6.2, Exercise 6.3, Exercise 6.4, Exercise 6.5 Solution.
Karnataka Class 8 Math Solution Chapter Algebraic Expressions and Identities
EXERCISE 6.1
1.) Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz 2 – 3zy
Ans: Coefficient of 5xyz 2 – 3zy are 5 and -3.
(ii) 1 + x + x 2
Ans: Coefficient of 1 + x + x 2 are 1, 1 and 1.
(iii) 4x 2y 2 – 4x 2y 2z 2 + z 2
Ans: Coefficient of 4x 2y 2 – 4x 2y 2z 2 + z 2 are 4,-4 and 1.
(iv) 3 – pq + qr – rp
Ans: Coefficient of 3 – pq + qr – rp are 3,-1, 1 and -1.
(v) x/2 + y/2 – xy
Ans: Coefficient of x/2 + y/2 – xy are 1/2, 1/2 and -1.
(vi) 0.3a – 0.6ab + 0.5b
Ans: Coefficient of 0.3a – 0.6ab + 0.5b are 0.3,-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?
x + y , 1000, x + x 2 + x 3 + x 4, 7 + y + 5x, 2y – 3y 2, 2y – 3y 2 + 4y3, 5x – 4y + 3xy, 4z – 15z 2, ab + bc + cd + da, pqr, p²q+pq², 2p + 2q
Ans:
We know,
Expression that contains only one term is called a monomial.
Expression that contains two terms is called a binomial.
Expression containing three terms is a trinomial.
Here we have to classify polynomials as monomials, binomials, trinomials.
Monomials: 1000, pqr
Binomials: x + y, 2y – 3y2, 4z – 15z2, p²q+pq², 2p + 2q
Trinomials: 7 + y + 5x, 2y – 3y 2 + 4y3, 5x – 4y + 3xy
Karnataka class 8 Math Solution paper chapter 6EXERCISE 6.2
1.) Find the product of the following pairs of monomials.
(i) 4, 7p
Ans:
4 x 7p = 4 x 7p = 28p
(ii) – 4p, 7p
Ans:
– 4p x 7p = – 4p x 7p = (- 4 x 7) (p x p) = – 28p2
(iii) – 4p, 7pq
Ans:
– 4p x 7pq = – 4p x 7pq = (- 4 x 7) (p x p x q) = – 28p2q
(iv 4p3, – 3p
Ans:
4p3 x (- 3) p = 4(- 3) x p x p x p = – 12p4
(v) 4p, 0
Ans:
4p x 0 = 4p x 0 = 0
2.) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively. (p. q): (10m, 5n); (20x 2, 5y 2) (4x, 3x 2) (3mn, 4np)
Ans:
We know,
Area of rectangle = Length x Breadth
Area of 1st rectangle = p x q = pq
Area of 2nd rectangle = 10m × 5n = 10 x 5 x mx n = 50 mn
Area of 3rd rectangle = 20x2 x 5y2 = 20 x 5x2y2 = 100x2y2
Area of 4th rectangle = 4x x 3x² = (4 x 3) x3 = 12x3
Area of 5th rectangle = 3mn x 4np = 3 x 4mnnp = 12mn2p
3.) Complete the table of products.
Ans
4.) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a 2, 7a 4
Ans:
We know,
Volume = Length × Breadth x Height
Volume = 5a x 3a² × 7a4 = (5x3x7) x (axa2xa4) = 105a7
(ii) 2p, 4q, 8r
Ans:
We know,
Volume = Length × Breadth x Height
Volume = 2p × 4q × 8r = (2×4×8) ×p × q × r = 64pqr
(iii) xy, 2x 2 y, 2x y 2
Ans:
We know,
Volume = Length × Breadth x Height
Volume = xy x 2x²y x 2xy² = (2 x 2) x (xy) x (x²y) x (xy²) = 4x²y4
(iv) a, 2b, 3c
Ans:
We know,
Volume = Length × Breadth x Height
Volume = a × 2b × 3c = 2 × 3 × a x b x c = 6abc
5.) Obtain the product of
(i) xy, yz, zx
Ans:
(xy) x (yz) x (zx) = x2y2z2
(ii) a, – a 2, a 3
Ans:
a x (-a2) x a3 = – a6
(iii) 2, 4y, 8y 2, 16y 3
Ans:
2 x 4y x 8y2 x 16y3 = (2 x 4 x 8 x 16) x (y x y2 x y3) = 1024y6
(iv) a, 2b, 3c, 6abc
Ans:
a x 2b x 3c x 6abc = 2 x 3 x 6 x (abc x abc) = 36a2b2c2
(v) m,– mn, mnp
Ans:
m x (-mn) x mnp = – m3 n2p
EXERCISE 6.3
1.) Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r
Ans:
We have to carry out multiplication.
(4p)(q + r) = (4pq) + (4pr) = 4pq + 4pr
(ii) ab, a – b
Ans:
We have to carry out multiplication.
(ab)(a – b) = (aba) + [ab(- b)] = a2b – ab2
(iii) a + b , 7a 2b 2
Ans:
We have to carry out multiplication.
a + b x 7a 2b 2 = (a x 7a2b2) + (b x7a2b2) =7a3b2 + 7a2b3
(iv) a2 – 9 , 4a
Ans:
We have to carry out multiplication.
(a2 – 9) (4a) = (a2 x 4a) + [(- 9) (4a)] = 4a3 – 36a
(v) pq + qr + rp, 0
Ans:
We have to carry out multiplication.
(pq + qr + rp) x 0 = (pq x 0) + (qr x 0) + (rp x 0) = 0
Q2.) Complete the table
Ans:
3.) Find the product.
(i) (a2)(2a 22)(4a26)
Ans:
a2 x 2a 22 x 4a26
= 2 x 4 x (a50)
(a2)(2a 22)(4a26) = 8a50
(ii) (2/3 xy) ((- 9/10) x 2y 2)
Ans:
(2/3 xy) ((-9/10) x 2y 2)
= (2/3 x – 9/10) x (xy x (x 2y 2))
(2/3 xy) ((- 9/10) x 2y 2) = (-3/5) x 3y 3
(iii) ((- 10/3) pq3) ((6/5) p3q)
Ans:
((- 10/3) pq3) x ((6/5) p3q)
= [(- 10/3) x (6/5)] x pq3 x p3q)
((- 10/3) pq3) ((6/5) p3q) = -4 p4q4
(iv) X x(x 2) x (x 3) x(x 4)
Ans:
X x (x 2) x (x 3) x (x 4) = x 10
4.) (a) Simplify 3x(4x – 5) + 3 and find its values for
(i) x = 3
Ans:
- a) 3x(4x – 5) + 3 = 12x2 – 15x + 3
= 12x2 – 15x + 3
= 12 x (3)2 – 15(3) + 3
=108-45 +3
= 66
For x = 3, 3x (4x – 5) + 3 is 66.
(ii) x = 1/2
Ans:
= 12x2 – 15x + 3
= 12 x (1/2)2 – 15(1/2) + 3
= 12 x 1/4 – 15/2 + 3
= 3 – 15/2 + 3
= 6 – 15/2
= (12 – 15)/2
= – 3/2
For x = 1/2, 3x (4x – 5) + 3 is – 3/2.
(b) Simplify a (a2 + a + 1) + 5 and find its value for
(i) a = 0
Ans:
a (a2 + a + 1) + 5 = a3 + a2 + a + 5
For a = 0,
a3 + a2 + a + 5 =
0 + 0 + 0 + 5
= 5
For a = 0, a (a2 + a + 1) + 5 is 5.
(ii) a = 1
Ans:
a (a2 + a + 1) + 5 = a3 + a2 + a + 5
For a = 1
a3 + a2 + a + 5
= (1)3 + (1)2 + 1 + 5
= 1 + 1 + 1 + 5 = 8
For a = 1, a (a2 + a + 1) + 5 is 8.
(iii) a = – 1
Ans:
a (a2 + a + 1) + 5 = a3 + a2 + a + 5
For a = – 1
a3 + a2 + a + 5
= (- 1)3 + (- 1)2 + (- 1) + 5
= – 1 + 1 – 1 + 5
= 4
For a = -1, a (a2 + a + 1) + 5 is 4.
5.) (a) Add: p(p – q), q(q – r) and r(r – p)
EXERCISE 6.4
1.) Multiply the binomials.
(i) (2x + 5) and (4x – 3)
Ans:
(2x + 5) × (4x − 3)
= 2x × (4x − 3) + 5 × (4x − 3)
= 8x2 − 6x + 20x – 15
Now, we are adding same terms,
(2x + 5) × (4x − 3) = 8x2 + 14x −15
(ii) (y-8) and (3y – 4)
Ans:
(y − 8) × (3y − 4)
= y × (3y − 4) − 8 × (3y − 4)
= 3y2 − 4y − 24y + 32
Now, we are adding same terms,
(y − 8) × (3y − 4) = 3y2 − 28y + 32
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
Ans:
(2.5l − 0.5m) × (2.5l + 0.5m)
= 2.5l × (2.5l + 0.5m) − 0.5m (2.5l + 0.5m)
= 6.25l2 + 1.25lm − 1.25lm − 0.25m2
(2.5l − 0.5m) × (2.5l + 0.5m) = 6.25l2 − 0.25m2
(iv) (a + 3b) and (x + 5)
Ans:
(a + 3b) × (x + 5)
= a × (x + 5) + 3b × (x + 5)
(a + 3b) × (x + 5) = ax + 5a + 3bx + 15b
(v) (2pq + 3q 2) and (3pq – 2q 2)
Ans:
(2pq + 3q2) × (3pq − 2q2)
= 2pq × (3pq − 2q2) + 3q2 × (3pq − 2q2)
= 6p2q2 − 4pq3 + 9pq3 − 6q4
(2pq + 3q2) × (3pq − 2q2) = 6p2q2 + 5pq3 − 6q4
(vi) ((3/4) a 2 + 3b 2) and 4(a 2 – 2/3 b 2)
Ans:
(3/4) a 2 x 4(a 2 – 2/3 b 2) + 3b 2 x 4(a 2 – 2/3 b 2)
= 3a4 -2b2a2 + 12b2a2 -8b4
((3/4) a 2 + 3b 2) x 4(a 2 – 2/3 b 2) = 3a4 +10b2a2 -8b4
2.) Find the product.
(i) (5 – 2x) (3 + x)
Ans:
(5 − 2x) (3 + x)
= 5 (3 + x) − 2x (3 + x)
= 15 + 5x − 6x − 2x2
(5 − 2x) (3 + x) = 15 − x − 2x2
(ii) (x + 7y) (7x – y)
Ans:
(x + 7y) (7x − y)
= x (7x − y) + 7y (7x − y)
= 7x2 − xy + 49xy − 7y2
(x + 7y) (7x − y) = 7x2 + 48xy − 7y2
(iii) (a2 + b) (a + b2)
Ans:
(a2 + b) (a + b2)
= a2 (a + b2) + b (a + b2)
(a2 + b) (a + b2) = a3 + a2b2 + ab + b3
(iv) (p2 – q2) (2p + q)
Ans:
(p2 − q2) (2p + q) = p2 (2p + q) − q2 (2p + q)
(p2 − q2) (2p + q) = 2p3 + p2q − 2pq2 − q3
3.) Simplify.
(i) (x 2 – 5) (x + 5) + 25
Ans:
We have to simplify (x 2 – 5) (x + 5) + 25
= x2 (x + 5) − 5 (x + 5) + 25
= x3 + 5x2 − 5x − 25 + 25
= x3 + 5x2 − 5x
(ii) (a2 + 5) (b 3 + 3) + 5
Ans:
We have to simplify (a 2 + 5) (b 3 + 3) + 5
= a2 (b3 + 3) + 5 (b3 + 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 3a2 + 5b3 + 20
(iii) (t + s 2) (t 2 – s)
Ans:
We have to simplify (t + s 2) (t 2 – s)
= t (t2 − s) + s2 (t2 − s)
= t3 − st + s2t2 − s3
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
Ans:
We have to simplify (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
= a (c − d) + b (c − d) + a (c + d) − b (c + d) + 2 (ac + bd)
= ac − ad + bc − bd + ac + ad − bc − bd + 2ac + 2bd
= (ac + ac + 2ac) + (ad − ad) + (bc − bc) + (2bd − bd − bd)
By cancelling we get,
= 4ac
(v) (x + y) (2x + y) + (x + 2y)(x – y)
Ans:
We have to simplify (x + y) (2x + y) + (x + 2y)(x – y)
= x (2x + y) + y (2x + y) + x (x − y) + 2y (x − y)
= 2x2 + xy + 2xy + y2 + x2 − xy + 2xy − 2y2
= (2x2 + x2) + (y2 − 2y2) + (xy + 2xy − xy + 2xy)
= 3x2 − y2 + 4xy
(vi) (x + y)(x 2 – xy + y 2)
Ans:
We have to simplify(x + y) (x 2 – xy + y 2)
= x (x2 − xy + y2) + y (x2 − xy + y2)
= x3 − x2y + xy2 + x2y − xy2 + y3
= x3 + y3 + (xy2 − xy2) + (x2y − x2y)
By cancelling, we get,
= x3 + y3
(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
Ans:
We have to simplify (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x (1.5x + 4y + 3) − 4y (1.5x + 4y + 3) − 4.5x + 12y
= 2.25 x2 + 6xy + 4.5x − 6xy − 16y2 − 12y − 4.5x + 12y
= 2.25 x2 + (6xy − 6xy) + (4.5x − 4.5x) − 16y2 + (12y − 12y)
By cancelling, we get,
= 2.25x2 − 16y2
(viii) (a + b + c) (a + b – c)
Ans:
We have to simplify (a + b + c) (a + b – c)
= a (a + b − c) + b (a + b − c) + c (a + b − c)
= a2 + ab − ac + ab + b2 − bc + ca + bc − c2
= a2 + b2 − c2 + (ab + ab) + (bc − bc) + (ca − ca)
By cancelling, we get,
= a2 + b2 − c2 + 2ab
EXERCISE 6.5
1.) Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3)
Ans:
(x + 3) (x + 3) = (x + 3)2
Here, we use [(a + b)2 = a2 + 2ab + b2]
= (x)2 + 2(x) (3) + (3)2
= x2 + 6x + 9
(ii) (2y + 5) (2y + 5)
Ans:
(2y + 5) (2y + 5) = (2y + 5)2
Here, we use [(a + b)2 = a2 + 2ab + b2]
= (2y)2 + 2(2y) (5) + (5)2
= 4y2 + 20y + 25
(iii) (2a – 7) (2a – 7)
Ans:
(2a − 7) (2a − 7) = (2a − 7)2
Here, we use [(a − b)2 = a2 − 2ab + b2]
= (2a)2− 2(2a) (7) + (7)2
= 4a2 − 28a + 49
(iv) (3a – 1/2) (3a – 1/2)
Ans:
3a – 1/2) (3a – 1/2) = 3a – 1/2)2
Here, we use [(a − b)2 = a2 − 2ab + b2]
= (3a)2− 2(3a) (1/2) + (1/2)2
= 9a2 − 3a + 1/4
(v) (1.1m – 0.4) (1.1m + 0.4)
Ans:
(1.1m − 0.4) (1.1 m + 0.4)
Here, we use [(a + b) (a − b) = a2 − b2]
= (1.1m)2 − (0.4)2
= 1.21m2 − 0.16
(vi) (a2 + b 2) (- a 2 + b 2)
Ans:
(a2 + b2) (− a2 + b2) = (b2 + a2) (b2 − a2)
Here, we use [(a + b) (a − b) = a2 − b2]
= (b2)2 − (a2)2
= b4− a4
(vii) (6x – 7) (6x + 7)
Ans:
(6x − 7) (6x + 7)
Here we use, [(a + b) (a − b) = a2 − b2]
= (6x)2− (7)2
= 36x2 − 49
(viii) (- a + c) (- a + c)
Ans:
(− a + c) (− a + c) = (− a + c)2
Here, we use [(a + b)2 = a2 + 2ab + b2]
= (− a)2+ 2(− a) (c) + (c)2
= a2 − 2ac + c2
(ix) (x/2 + (3y)/4) (x/2 + (3y)/4)
Ans:
(X/2 + (3y)/4) (x/2 + (3y)/4) = (x/2 + (3y)/4)2
Here, we use [(a + b)2 = a2 + 2ab + b2]
(X/2)2 + 2(x/2 + (3y)/4) + (3y)/4)2
On solving we get,
= X2/4 + 3xy/4 + 9y2/16
(x) (7a – 9b) (7a – 9b)
Ans:
(7a − 9b) (7a − 9b) = (7a − 9b)2
Here, we use [(a − b)2 = a2 − 2ab + b2]
= (7a)2 − 2(7a)(9b) + (9b)2
= 49a2 − 126ab + 81b2
2.) Use the identity (x + a)(x + b) = x 2 + (a + b) x + a to find the following products.
(1) (x + 3) (x + 7)
Ans:
(x + 3) (x + 7)
= x2 + (3 + 7) x + (3) (7)
= x2 + 10x + 21
(ii) (4x + 5)(4x + 1)
Ans:
(4x + 5) (4x + 1)
= (4x)2 + (5 + 1) (4x) + (5) (1)
= 16x2 + 24x + 5
(iii) (4x – 5)(4x – 1)
Ans:
(4x – 5) (4x – 1)
= (4x)2 + (-5 + -1) (4x) + (-5) (-1)
= 16x2 – 24x + 5
(iv) (4x + 5)(4x – 1)
Ans:
(4x + 5)(4x – 1)
= (4x)2 + (+5 + -1) (4x) + (+5) (-1)
= 16x2 + 16x − 5
(v) (2x + 5y) (2x + 3y)
Ans:
(2x +5y) (2x + 3y)
= (2x)2 + (5y + 3y) (2x) + (5y) (3y)
= 4x2 + 16xy + 15y2
(vi) (2a 2 + 9)(2a 2 + 5)
Ans:
(2a2 +9) (2a2 + 5) =
(2a2)2 + (9 + 5) (2a2) + (9) (5)
= 4a4 + 28a2 + 45
(vii) (xyz – 4) (xyz – 2)
Ans:
(xyz − 4) (xyz − 2)
= (xyz)2+ (-4 -2) xyz+ (-4 -2)
= x2y2z2 − 6xyz + 8
3.) Find the following squares by using the identities.
(i) (b – 7) 2
Ans:
(b − 7)2 =
Here we use,[(a − b)2 = a2 − 2ab + b2]
(b)2 − 2(b) (7) + (7)2
= b2 − 14b + 49
(ii) (xy + 3z) 2
Ans:
(xy + 3z)2
Here we use, [(a + b)2 = a2 + 2ab + b2]
= (xy)2 + 2(xy) (3z) + (3z)2
= x2y2 + 6xyz + 9z2
(iii) (6x2 – 5y) 2
Ans:
(6×2 − 5y)2
Here we use, [(a − b)2 = a2 − 2ab + b2]
= (6x2)2 − 2(6x2) (5y) + (5y)2
= 36x4 − 60x2y + 25y2
(iv) (2/3m + 3/2n) 2
Ans:
(2/3m + 3/2n) 2
Here we use, [(a + b)2 = a2 + 2ab + b2]
= 4/9 m2 + 2mn + 9/4 n2
(v) (0.4p − 0.5q)2
Here we use,[(a − b)2 = a2 − 2ab + b2]
= (0.4p)2 − 2 (0.4p) (0.5q) + (0.5q)2
= 0.16p2 − 0.4pq + 0.25q2
(vi) (2xy + 5y)2
Here, we use [(a + b)2 = a2 + 2ab + b2]
= (2xy)2 + 2(2xy) (5y) + (5y)2
= 4x2y2 + 20xy2 + 25y2
4.) Simplify.
(i) (a2 – b 2) 2
Ans:
(a2 − b2)2
Here we use,[(a − b)2 = a2 − 2ab + b2]
= (a2)2 − 2(a2) (b2) + (b2)2
= a4 − 2a2b2 + b4
(ii) (2x +5)2 − (2x − 5)2
Here we use,[(a − b)2 = a2 − 2ab + b2] and [(a + b)2 = a2 + 2ab + b2]
= (2x)2 + 2(2x) (5) + (5)2 − [(2x)2 − 2(2x) (5) + (5)2]
= 4x2 + 20x + 25 − [4x2 − 20x + 25]
= 4x2 + 20x + 25 − 4x2 + 20x − 25
= 40x
(iii) (7m – 8n) 2 + (7m + 8n) 2
Ans:
(7m − 8n)2 + (7m + 8n)2
Here we use,[(a − b)2 = a2 − 2ab + b2 and (a + b)2 = a2 + 2ab + b2]
= (7m)2 − 2(7m) (8n) + (8n)2 + (7m)2 + 2(7m) (8n) + (8n)2
= 49m2 − 112mn + 64n2 + 49m2 + 112mn + 64n2
= 98m2 + 128n2
5.) Show that.
(i) (3x + 7)2 – 84x = (3x – 7)2
Ans:
Here L.H.S = (3x + 7)2 − 84x
= (3x)2 + 2(3x)(7) + (7)2 − 84x
= 9x2 + 42x + 49 − 84x
= 9x2 − 42x + 49
Now we solve RHS.
R.H.S = (3x − 7)2 = (3x)2 − 2(3x)(7) +(7)2
= 9x2 − 42x + 49
L.H.S = R.H.S
(3x + 7)2 – 84x = (3x – 7)2
- ii) (9p – 5q)2+ 180pq = (9p + 5q)2
Ans:
Here L.H.S = (9p − 5q)2 + 180pq
= (9p)2 − 2(9p)(5q) + (5q)2 − 180pq
= 81p2 − 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
Now we solve RHS.
R.H.S = (9p + 5q)2
= (9p)2+ 2(9p)(5q) + (5q)2
= 81p2 + 90pq + 25q2
L.H.S = R.H.S
(9p – 5q)2+ 180pq = (9p + 5q)2
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
Ans:
L.H.S = (4pq + 3q)2 − (4pq − 3q)2
= (4pq)2 + 2(4pq)(3q) + (3q)2 − [(4pq)2 − 2(4pq) (3q) + (3q)2]
= 16p2q2 + 24pq2 + 9q2 − [16p2q2 − 24pq2 + 9q2]
= 16p2q2 + 24pq2 + 9q2 −16p2q2 + 24pq2 − 9q2
= 48pq2 = R.H.S
(4pq + 3q) 2 – (4pq – 3q) 2 = 48pq2
(v) (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = 0
Ans:
Here, L.H.S = (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a)
= (a2 − b2) + (b2 − c2) + (c2 − a2)
= 0 = R.H.S.
(a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
6.) Using identities, evaluate.
(i) 712
Ans:
712 = (70 + 1)2
Here we use,[(a + b)2 = a2 + 2ab + b2]
= (70)2 + 2(70) (1) + (1)2
= 4900 + 140 + 1
712 = 5041
(ii) 992
Ans:
992 = (100 − 1)2
Here we use,[( [(a − b)2 = a2 − 2ab + b2 ]
= (100)2 − 2(100) (1) + (1)2
= 10000 − 200 + 1
992= 9801
(ii) 1022
Ans:
1022 = (100 + 2)2
Here we use [(a + b)2 = a2 + 2ab + b2]
= (100)2 + 2(100)(2) + (2)2
= 10000 + 400 + 4
1022 = 10404
(iv) 9982
Ans:
9982 = (1000 − 2)2
Here we use [(a − b)2 = a2 − 2ab + b2]
= (1000)2 − 2(1000)(2) + (2)2
= 1000000 − 4000 + 4
9982 = 996004
(v) 5.22
Ans:
(5.2)2 = (5.0 + 0.2)2
Here we use [(a + b)2 = a2 + 2ab + b2]
= (5.0)2 + 2(5.0) (0.2) + (0.2)2
= 25 + 2 + 0.04
(5.2)2 = 27.04
(vi) 297 x 303
Ans:
297 × 303 = (300 − 3) × (300 + 3)
Here we use [(a + b) (a − b) = a2 − b2]
= (300)2 − (3)2
= 90000 − 9
297 × 303 = 89991
(vii) 78 x 82
Ans:
78 × 82 = (80 − 2) (80 + 2)
Here we use [(a + b) (a − b) = a2 − b2]
= (80)2 − (2)2
= 6400 − 4
78 × 82 = 6396
(viii) 8.92
Ans:
8.92 = (9.0 − 0.1)2
Here we use [(a − b)2 = a2 − 2ab + b2]
= (9.0)2 − 2(9.0) (0.1) + (0.1)2
= 81 − 1.8 + 0.01
8.92 = 79.21
(ix) 10.5 x 9.5
Ans:
1.05 × 9.5 = 1.05 × (0.95 × 10)
= (1 + 0.05) (1− 0.05) × 10
Here we use [(a + b) (a − b) = a2 − b2]
= [(1)2 − (0.05)2] × 10
= [1 − 0.0025] × 10
= 0.9975 × 10
1.05 × 9.5 = 9.975
- Using
a2 – b2 = (a + b) (a – b) find
(i) 512 – 492
Ans:
512 − 492 = (51 + 49) (51 − 49)
= (100) x (2)
= 200
(ii) (1.02)2 – (0.98)2
Ans:
(1.02)2 − (0.98)2
= (1.02 + 0.98) (1.02 − 0.98)
= (2) x (0.04)
= 0.08
(iii) 1532 – 1472
Ans:
1532 − 1472
= (153 + 147) (153 − 147)
= (300) x (6)
= 1800
(iv) 12.12 – 7.92
Ans:
12.12 − 7.92
= (12.1 + 7.9) (12.1 − 7.9)
= (20.0) x (4.2)
= 84
8.) Using (x + a) (x + b) = x2 + (a + b) x + ab
Find
(i) 103 x 104
Ans:
103 × 104 = (100 + 3) (100 + 4)
= (100)2 + (3 + 4) (100) + (3) x (4)
= 10000 + 700 + 12
= 10712
(ii) 5.1 x 5.2
Ans:
5.1 × 5.2
= (5 + 0.1) (5 + 0.2)
= (5)2 + (0.1 + 0.2) (5) + (0.1) x (0.2)
= 25 + 1.5 + 0.02
= 26.52
(iii) 103 x 98
Ans:
103 × 98 = (100 + 3) (100 − 2)
= (100)2 + [3 + (− 2)] (100) + (3) x (− 2)
= 10000 + 100 − 6
= 10094
(iv) 9.7 x 9.8
Ans:
9.7 × 9.8 = (10 − 0.3) (10 − 0.2)
= (10)2 + [(− 0.3) + (− 0.2)] (10) + (− 0.3) x (− 0.2)
= 100 + (− 0.5) x 10 + 0.06
= 100.06 − 5
= 95.06
Also See: