class 8 maths chapter 9 deals with algebraic expressions and Identities.
Find NCERT solutions for algebraic expressions and Identities in class 8
Exercises and solutions of algebraic expressions and Identities available here so students can read online view.
Find solutions to exercise 9.1, exercise 9.2 and 9.3 in chapter 9 in class 8 maths.
Ncert Solutions are provided by mensuration.in for all subjects of maths, physics, chemistry, biology for preparation of exams
solutions for all exercises in algebraic expressions class 9
class 8 maths chapter 9 exercise 9.1
class 8 maths chapter 9 exercise 9.1 (1)
1. Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz²– 3zy (ii) 1 + x + x ² (iii) 4x²y²+4x²y²z²+z² (iv) 3 – pq + qr – rp
(v) (ˣ/₂)+(ʸ/₂)-xy (vi) 0.3a – 0.6ab + 0.5b
Solution to class 8 maths chapter 9 exercise 9.1 (1)
(i) Given expression 5xyz²– 3zy
Terms 5xyz² and –3zy
Coefficient in 5xyz² is 5 and in –3zy is -3
(ii) Terms 1,x and x².
Coefficient of x is 1 and Coefficient of x² is 1
(iii) Terms 4x²y², -4x²y²z² and z²
Coefficient in 4x²y² is 4, Coefficient in -4x²y²z² is -4, and Coefficient of z² is 1
(iv) Terms 3, -pq, qr and -rp
Coefficient of -pq is -1, Coefficient of qr is 1, and Coefficient of -rp is -1
(v) Terms ; (ˣ/₂), (ʸ/₂) and -xy
Coefficient of (ˣ/₂) is ½, Coefficient of (ʸ/₂) is ½, and Coefficient of -xy is -1
(vi) Terms 0.3a, -0.6ab and 0.5b
Coefficient of 0.3a is 0.3, Coefficient of -0.6ab is -0.6, and Coefficient of 0.5b is 0.5
class 8 maths chapter 9 exercise 9.1 (2)
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x+x²+x³+x⁴, 7+y+5x, 2y-3y², 2y-3y²+4y³, 5x-4y+3xy, 4z-15z², ab + bc + cd + da, pqr, p²q+pq², 2p+2q
Solution to class 8 maths chapter 9 exercise 9.1 (2)
(i) Since x+y contains two terms. Therefore it is binomial.
(ii) Since 1000 contains one term. Therefore it is monomial.
(iii) Since x+x²+x³+x⁴, contains Four terms. Therefore it is polynomial and it does not fit in above three categories.
(iv) Since 7+y+5x contains three terms. Therefore it is trinomial.
(v) Since 2y-3y² contains two terms. Therefore it is binomial.
(vi) Since 2y-3y²+4y³ contains three terms. Therefore it is trinomial.
(vii) Since 5x-4y+3xy contains three terms. Therefore it is trinomial.
(viii) Since 4x-15z² contains two terms. Therefore it is binomial.
(ix) Since ab+bc+cd+da, contains Four terms. Therefore it is polynomial and it does not fit in above three categories.
(x) Since pqr contains one term. Therefore it is monomial.
(xi) Since p²q+pq² contains two terms. Therefore it is binomial.
(xii) Since 2p+2q contains two terms. Therefore it is binomial.
class 8 maths chapter 9 exercise 9.1 (3)
3. Add the following.
(i) ab-bc, bc-ca, ca-ab (ii)a-b+ab, b-c+bc, c-a+ac
(iii) 2p²q²-3pq+4, 5+7pq-3p²q² (iv) l²+m², m²+n², n²+l², 2lm+2mn+2nl
Solution to class 8 maths chapter 9 exercise 9.1 (3).
(i) ab-bc, bc-ca, ca-ab
ab – bc
+bc – ca
-ab +ca
0 + 0 + 0
(ii) a-b+ab, b-c+bc, c-a+ac
a – b + ab
+b -c + bc
-a +c +ac
0 +0 +ab +0 +bc+ac
(iii) 2p²q²-3pq+4, 5+7pq-3p²q²
2p²q² – 3pq + 4,
-3p²q² + 7pq + 5
-p²q² + 4pq + 9
(iv) l²+m², m²+n², n²+l², 2lm+2mn+2nl
l² + m²
+ m² + n²
+l² + n²
+ 2lm + 2mn + 2nl
2l² + 2m² + 2n² + 2lm + 2mn + 2nl
class 8 maths chapter 9 exercise 9.1 (4)
4. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p ²q – 3pq + 5pq²– 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq²+ 5p²q
Solution to class 8 maths chapter 9 exercise 9.1 (4).
(a)
12a – 9ab + 5b – 3
4a – 7ab + 3b + 12
(-) (+) (-) (-)
8a – 2ab + 2b – 15
(b)
5xy – 2yz – 2zx + 10xyz
3xy + 5yz – 7zx
(-) (-) (+)
2xy – 7yz + 5zx + 10xyz
(c)
5p²q – 2pq² + 5pq – 11q – 3p + 18
4p²q + 5pq² – 3pq + 7q – 8p – 10
(-) (-) (+) (-) (+) (+)
p²q – 7pq² + 8pq – 18q +5p +28
Exercise-9.2
class 8 maths chapter 9 exercise 9.2 (1)
1 carry out the multiplication of the expression in each of the following pairs.
(i) 4p, q+r (ii) ab, a-b (iii) a+b, 7a²b² (iv) a²-9, 4a (v) pq+qr+rp, 0
Solution to class 8 maths chapter 9 exercise 9.2 (1).
(i) 4×7p= 4×7×p=28p
(ii) -4p×7p=(-4×7)×(p×p)=-28p²
(iii) -4p×7pq=(-4×7)(p×pq)=-28p²q
(iv) 4p³×(-3p)=[4×(-3)](p³×p)=-12p⁴
(v) 4p×0=(4×0)(P)=0
class 8 maths chapter 9 exercise 9.2 (2)
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)
Solution to class 8 maths chapter 9 exercise 9.2 (2).
(i) Area of the rectangle = length × bredth
=p×q=pq sq.units
(ii) Area of the rectangle = length × bredth
=10m×5n=(10×5)(m×n)=50mn sq.units
(iii) Area of the rectangle = length × bredth
=20x²×5y²=(20×5)(x²×y²)=100x²y² sq.units
(iv) Area of the rectangle = length × bredth
=4x×3x²=(4×3)(x×x²)=12x³ sq.units
(v) Area of the rectangle = length × bredth
=3mn×4np=(3×4)(mn×np)=12mn²p sq.units
class 8 maths chapter 9 exercise 9.2 (3)
Complete the table of products:
Solution to class 8 maths chapter 9 exercise 9.2 (3).
class 8 maths chapter 9 exercise 9.2 (4)
4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively;
(i) 5a, 3a²7a³a (ii) 2p, 4q, 8r (iii) xy, 2x²y, 2xy² (iv) a, 2b, 3c
Solution to class 8 maths chapter 9 exercise 9.2 (4).
(i) Volume of rectangular box = length × breadth × height
=5a×3a²×7a³a=(5×3×7)(a×a²×a⁴)
= 105a⁷ cubic units
(ii) Volume of rectangular box = length × breadth × height
=2p×4q×8r=(2×4×8)(p×q×r)
= 64pqr cubic units
(iii) Volume of rectangular box = length × breadth × height
=xy×2x²y×2xy²=(1×2×2)(x×x²×x×y×y×y²)
= 4x⁴y⁴ cubic units.
(ii) Volume of rectangular box = length × breadth × height
=a×2b×3c=(1×2×3)(a×b×c)
= 6abc cubic units
class 8 maths chapter 9 exercise 9.2 (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
Solution to class 8 maths chapter 9 exercise 9.2 (5).
(i) xy×yz×zx=x×x×y×y×z×z=x²y²z²
(ii) a×(-a²)×a³=(-1)(a×a²×a³)=-a⁶
(iii) 2×4y×8y²×16y³=(2×4×8×16)(y×y²×y³)=1024y⁶
(iv) a×2b×3c×6abc=(1×2×3×6)(a×b×c×abc)=36a²b²c²
(v) m×(-mn)×mnp=(-1)(m×m×m×n×n×p) =-m³n²p
Exercice-9.3
class 8 maths chapter 9 exercise 9.3 (1)
1 Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q+r (ii) ab, a-b (iii) a+b, 7a²b² (iv) a² – 9, 4a
(v) pq+qr+rp, 0
Solution to class 8 maths chapter 9 exercise 9.3 (1).
(i) 4p×(q+r) = 4p×q + 4p×r
=4pq+4pr
(ii) ab×(a-b) =ab×a – ab×b
=a²b-ab²
(iii) (a+b)×7a²b² =a×7a²b²+b×7a²b²
= 7a³b²+7a²b³
(iv) (a² – 9)×4a =a²×4a-9×4a
=4a³-36a
(v) (pq+qr+rp)×0 = pq×0+qr×0+rp×0
=0+0+0=0
class 8 maths chapter 9 exercise 9.3 (2)
2 Complete the table.
Solution to class 8 maths chapter 9 exercise 9.3 (2).
class 8 maths chapter 9 exercise 9.3 (3)
3 Find the product:
(i) (a²)×(2a²²)×(4a²⁶) (ii)[(⅔)xy]×[(- ⁹/₁₀)x²y²]
(iii) [(- ¹⁰/₃)pq³]×[(⁶/₅)p³q] (iv) x×x²×x³×x⁴
Solution to class 8 maths chapter 9 exercise 9.3 (3).
(i) (a²)×(2a²²)×(4a²⁶)=(2×4)(a²×a²²×a²⁶)
=8×a²⁺²²⁺²⁶=8a⁵⁰
(ii) [(⅔)xy]×[(- ⁹/₁₀)x²y²]=(⅔)×(- ⁹/₁₀)(x×x²×y×y²)
= (- ⅗)x³y³
(iii) [(- ¹⁰/₃)pq³]×[(⁶/₅)p³q]=(- ¹⁰/₃)×(⁶/₅)(p×p³×q³×q)
= – 4p⁴q⁴
(iv) x×x²×x³×x⁴=x¹⁺²⁺³⁺⁴=x¹⁰
class 8 maths chapter 9 exercise 9.3 (4)
4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x = ½
(b) Simplify a(a ² + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a=-1
Solution to class 8 maths chapter 9 exercise 9.3 (4).
(a) 3x (4x – 5) + 3=3x×4x – 3x×5+3=12x² – 15x+3
(i) For x=3, 12x²+15x+3=12(3)²-15×3+3
=12×9-45+3=108-45+3=66
(ii) For x= ½ , 12x²+15x+3=12(½)²-15×(½)+3
=12×(¼)- (¹⁵/₂)+3=6- (¹⁵/₂) =[⁽¹²⁻¹⁵⁾/₂]= – (³/₂)
(b) a(a ²+a+1)+5=a×a²+a×a+a×1+5=a³+a²+a+5
(i) For a=0, a³+a²+a+5=(0)³+(0)²+(0)+5
=0+0+0+5=5
(ii) For a=1, a³+a²+a+5=(1)³+(1)²+(1)+5
=1+1+1+5=8
(iii) For a= -1, a³+a²+a+5=(-1)³+(-1)²+(-1)+5
= -1+1-1+5= -2+6=4
class 8 maths chapter 9 exercise 9.3 (5)
5. (a) Add: p( p – q), q( q – r) and r( r – p)
(b) Add: 2x(z – x – y) and 2y(z – y – x)
(c) Subtract: 3l(l – 4 m + 5 n) from 4l( 10 n – 3 m + 2 l )
(d) Subtract: 3a(a + b + c ) – 2b(a – b + c) from 4c(– a + b + c )
Solution to class 8 maths chapter 9 exercise 9.3 (5).
p( p – q)+q( q – r)+r( r – p)
=p²–pq+q²–qr+r²–rp
=p²+q²+r²–pq–qr–rp
(b) 2x(z – x – y)+2y(z – y – x)
=2xz–2x²–2xy+2yz–2y²–2xy
=2xz–2xy–2xy+2yz–2x²–2y²
= –2x²–2y²–4xy+2yz+2zx
(c) 4l( 10n – 3m+2l )–3l(l – 4 m+5 n)
=40ln-12ml+8l² –3l²+12lm–15ln
=8l²–3l² –12lm+12lm+40ln–15ln
=5l²+25ln
(d)4c(– a+b+c ) –[3a(a +b+c ) – 2b(a – b + c)]
= –4ac+4bc+4c² –[3a²+3ab+3ac–2ab+2b²–2bc]
= –4ac+4bc+4c²–[3a²+2b²+3ab–2bc+3ac–2ab]
=–4ac+4bc+4c²–[3a²+2b²+ab+3ac–2bc]
= –4ac+4bc+4c² –3a²–2b²–ab–3ac+2bc
= –3a² –2b²+4c² –ab+4bc+2bc–4ac–3ac
= –3a² –2b²+4c² –ab+6bc –7ac
Exercise 9.4
class 8 maths chapter 9 exercise 9.4 (1)
1. Multiply the binomials.
(i) (2x + 5) and (4x – 3) (ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m) (iv) (a + 3b) and (x + 5)
(v) (2pq + 3q ²) and (3pq – 2q²)
(vi) [(¾)a²+3b²] and 4[a² –(⅔)b²]
Solution to class 8 maths chapter 9 exercise 9.4 (1).
(i) (2x + 5) and (4x – 3)=2x(4x –3)+5(4x –3)
=(2x× 4x) –(2x ×3)+(5 ×4x) –(5 ×3)
=8x² –6x+20x –15
=8x²+14x –15
(ii)(y –8)(3y –4)= y(3y –4) –8(3y –4)
=(y ×3y) +[y ×(–4 )]–(8 ×3y) –[8 ×(–4)]
=3y² –4y –24y+12
=3y² –28y+12
(iii) (2.5l –0.5m) ×(2.5l+0.5m)
=2.5l ×(2.5l+0.5m) –0.5m ×(2.5l+0.5m)
=2.5l×2.5l+0.5m×2.5l–0.5m×2.5l–0.5m ×0.5m
=6.25l²+1.25ml –1.25ml –0.25m²
=6.25l² –0.25m²
(iv) (a+3b) ×(x+5)=a(x+5) ×3b(x+5)
=a ×x+a ×5+3b ×x+ 3b ×5
=ax+5a+3bx+15b
(v) (2pq+3q²)(3pq –2q²)=2pq ×(3pq –2q²)+3q²(3pq –2q²)
=2pq ×3pq –2pq ×2q²+3q² ×3pq –3q² ×2q²
=6p²q² –4pq³+9pq³ –6q⁴
=6p²q²+5pq³ –6q⁴
(vi) [(¾)a²+3b²]×4[a² –(⅔)b²]
=[(¾)a²+3b²]×[4a² –(⁸/₃)b²]
=(¾)a²×[4a² –(⁸/₃)b²]+3b²×[4a² –(⁸/₃)b²]
=(¾)a² ×4a² –(¾)a²×(⁸/₃)b²+3b² ×4a² –3b² ×(⁸/₃)b²
=3a⁴ –2a²b²+12a²b² –8b⁴
=3a⁴+10a²b² –8b⁴
class 8 maths chapter 9 exercise 9.4 (2)
2 Find the product.
(i) (5 – 2x) (3 + x) (ii) (x + 7y) (7x – y)
(iii) (a²+b)(a+b²) (iv) (p² –q²)(2p+q)
Solution to class 8 maths chapter 9 exercise 9.4 (2).
(i) (5 –2x)(3 +x)=5×(3+x) –2x(3+x)
=5 ×3+5 ×x –2x ×3 –2x ×x
=15+5x –6x –2x²
=15 –x –2x²
(ii) (x+7y)(7x –y) =x(7x –y)+7y(7x –y)
=7x² –xy+49xy –7y²
=7x²+48xy –7y²
(iii) (a²+b)(a+b²)=a²(a+b²)+b(a+b²)
=a² ×a+a² ×b²+b ×a+b ×b²
=a³+a²b²+ab+b³
(iv) (p² –q²)(2p+q)=p²(2p+q) –q²(2p+q)
=p²×2p+p² ×q –q²×2p –q² ×q
=2p³+p²q –2pq² –q³
class 8 maths chapter 9 exercise 9.4 (3)
3 Simplify.
(i) (x² –5)(x + 5)+25 (ii) (a²+5)(b²+3)+5
(iii) (t+s²)(t² –s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
(v) (x + y)(2x + y) + (x + 2y)(x – y)
(vi) (x+y)(x² –xy+y²)
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c)(a + b – c)
Solution to class 8 maths chapter 9 exercise 9.4 (3).
(i) (x² –5)(x+5)+25=x²(x+5) –5(x+5)+25
=x² ×x+x² ×5 –5 ×x –5 ×5+25
=x³+5x² –5x –25+25
=x³+5x² –5x
(ii) (a²+5)(b³+3)+5=a²(b³+3)+5(b³+3)+5
=(a² ×b³)+(a² ×3)+(5 ×b³)+(5 ×3)+5
=a²b³+3a²+5b³+15+5
=a²b³+3a²+5b³+20
(iii) (t+s²)(t² –s) =t(t² –s)+s²(t² –s)
=(t ×t²) –(t ×s)+(s² ×t²) –(s²×s)
=t³ –st+s²t² –s³
(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
=a(c –d)+b(c –d)+a(c+d) –b(c+d)+2ac+2bd
=ac –ad+bc –bd+ac+ad –bc –bd+2ac+2bd
=ac+ac –ad+ad+bc–bc–bd–bd+2ac+2bd
=2ac –2bd+2ac+2bd
=4ac
(v) (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+xy –xy+2xy+2xy+y² –2y²
=3x²+4xy –y²
(vi) (x+y)(x² –xy+y²)=x(x² –xy+y²)+y(x² –xy+y²)
=x³ –x²y+xy²+x²y –xy²+y³
=x³ –x²y+x²y+xy² –xy²+y³
=x³+y³
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
=1.5x(1.5x+4y+3) –4y(1.5x+4y+3) –4.5x+12y
=2.25x²+6.0xy+4.5x–6.0xy–16y²–12y–4.5x+12y
=2.25x²+6.0xy –6.0xy+4.5x–4.5x–16y²–12y+12y
=2.25x² –16y²
(viii) (a+b+c)(a+b –c)
=a(a+b –c)+b(a+b –c)+c(a+b –c)
=a²+ab –ac+ab+b² –bc+ac+bc –c²
=a²+b²–c²+ab+ab –bc+bc –ac+ac
=a²+b²–c²+2ab
Exercise 9.5
class 8 maths chapter 9 exercise 9.5 (1)
1. Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5) (iii) (2a – 7)(2a – 7)
(iv) (3a – ½) (3a – ½) (v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a²+b²)( –a²+b²) (vii) (6x – 7) (6x + 7) (viii) (– a + c)(– a + c)
(ix) (ˣ/₂ + ³ʸ/₄)(ˣ/₂ + ³ʸ/₄) (x) (7a – 9b) (7a – 9b)
Solution to class 8 maths chapter 9 exercise 9.5 (1).
(i) (x+3)(x+3) = (x+3)²=(x)² +(2 ×x×3)+ (3)² [ ∵Using identity (a+b)² =a² + 2ab + b² ]
=x²+6x+9
(ii) (2y + 5)(2y + 5) = (2y + 5)²
= (2y)²+ 2×2y×5+(5)² [ ∵Using identity (a+b)² = a² +2ab+b²]
= 4y²+20y+25
(ii) (2a –7)(2a –7) = (2a –7)²
= (2a)² – 2×2a×7+(7)² [ ∵Using identity (a–b)² = a² –2ab+b²]
= 4a² –28a+49
(iv) [3a – (½)][3a –(½)] = [3a –(½)]²
= (3a)² – 2×3a×(½)+(½)² [ ∵Using identity (a–b)² = a² –2ab+b²]
= 9a² –3a+(¼)
(v) (1.1m –0.4)(1.1m+0.4)=(1.1m)² –(0.4)²
[ ∵ (a –b)(a+b)=a² –b²]
=1.21m² –0.16
(vi) (a²+b²)( –a²+b²)=(b²+a²)(b² –a²)
=(b²)² –(a²)² [ ∵ (a –b)(a+b)=a² –b²]
=b⁴ –a⁴
(vii) (6x –7)(6x+7)=(6x)² –(7)² [ ∵ (a –b)(a+b)=a² –b²]
=36x² –49
(viii) ( –a+c)( –a+c)=(c –a)(c –a)=(c –a)²
=(c)² –2 × c ×a+(a)²
[∵(a–b)² = a² –2ab+b²]
=c² –2ca+a²
(ix) (ˣ/₂ + ³ʸ/₄)(ˣ/₂ + ³ʸ/₄)= (ˣ/₂ + ³ʸ/₄)²
=(ˣ/₂)²+2 (ˣ/₂)(³ʸ/₄) +(³ʸ/₄)²
{∵(a+b)²=a²+2ab+b²}
= x²(¼)+(¾)xy+(⁹/₁₆)y²
(x) (7a-9b)(7a-9b)=(7a-9b)²
=(7a)² –2 ×7a×9b+(9b)² {∵(a –b)²=a²–2ab+b²}
=49a² –126ab+81b²
class 8 maths chapter 9 exercise 9.5 (2)
2. Use the identity (x + a) (x + b) = x ² + (a + b)x + ab to find the following products.
(i) (x + 3) (x + 7) (ii) (4x + 5) (4x + 1)
(iii) (4x – 5) (4x – 1) (iv) (4x + 5) (4x – 1)
(v) (2x + 5y) (2x + 3y) (vi) (2a ² + 9) (2a² + 5)
(vii) (xyz – 4) (xyz – 2)
Solution to class 8 maths chapter 9 exercise 9.5 (2).
(i) (x+3)(x+7)=x²+(3+7)x+3×7 {∵(x+a)(x+b)=x²+(a+b)x+ab}
=x²+10x+21
(ii) (4x+5)(4x+1)=(4x)²+(5+1)4x+5×1
{∵(x+a)(x+b)=x²+(a+b)x+ab}
=16x²+6×4x+5=16x²+24x+5
(iii) (4x-5)(4x-1) = (4x)²+{(–5)+(-1)}×4x+(-5)(-1)
{∵ (x+a)(x+b)= x +(a+b)x+ab}
= 16x²+(-6)×4x+ 5= 16x² -24x +5
(iv)(4x+5)(4x – 1) = (4x)²+[5+(-1)]×4x+5×(-1)
[∵ (x + a)(x+b)=x²+(a+b)x+ ab]
= 16x² +(5-1)×4x-5
= 16x² +4×4×x–5
= 16x²+16x-5
(v) (2x+5y) (2x+3y) = (2x)²+(5y+3y) ×2x+(5y×3y)
[∵ (x+a)(x+b)= x +(a+b)x+ab]
= 4x²+8y×2x+15y²= 4x²+16xy +15y²
(iv) (2a²+9)(2a²+5)=(2a²)²+(9+5)×2a²+9×5
[∵ (x+a)(x+b)= x +(a+b)x+ab]
=4a²+14×2a²+45
=4a²+28a²+45
(vii) (xyz–4)(xyz–2) =(xyz)²+(–4–2)×xyz+(–4) ×(–2)
[∵ (x+a)(x+b)= x +(a+b)x+ab]
=x²y²z² –6xyz+8
class 8 maths chapter 9 exercise 9.5 (3)
3. Find the following squares by using the identities.
(i) (b – 7)² (ii) (xy + 3z)² (iii) (6x ² – 5y)²
(iv) [(⅔)m+(³/₂)n]² (v) (0.4p – 0.5q)²
(vi) (2xy + 5y)²
Solution to class 8 maths chapter 9 exercise 9.5 (3).
(i) (b–7)²=b²–2(b)(7)+7² {∵(a –b)²=a²–2ab+b²}
=b² –14b+49
(ii) (xy + 3z)²=(xy)²+2(xy)(3z)+(3z)² {∵(a+b)²=a²+2ab+b²}
=x²y²+6xyz+9z²
(iii)(6x ² – 5y)² =(6x²)² –2(6x²)(5y)+(5y)² {∵(a –b)²=a²–2ab+b²}
=36x⁴ –60x²y+25y²
(iv) [(⅔)m+(³/₂)n]²=[(⅔)m]²+2×(⅔)m×(³/₂)n+[(³/₂)n]²
{∵(a+b)²=a²+2ab+b²}
=(⁴/₃)m²+2mn+(⁹/₄)n²
(v) (0.4p – 0.5q)²=(0.4p)²–2×(0.4p)×(0.5q)+(0.5q)²
{∵(a –b)²=a²–2ab+b²}
=0.16p²–0.40pq+0.25q²
(vi) (2xy + 5y)²=(2xy)²+2×(2xy)×(5y)+(5y)²
{∵(a+b)²=a²+2ab+b²}
=4x²y²+20xy²+25y²
class 8 maths chapter 9 exercise 9.5 (4)
4 Simplify:
(i) (a² –b²)² (ii)(2x+5)² –(2x –5)²
(iii)(7m–8n)² +(7m+8n)²
(iv)(4m+5n)²+(5m+4n)²
(v) (2.5p–1.5q)²–(1.5p–2.5q)²
(vi) (ab+bc)² –2ab²c
(vii)(m² –n²m)²+2m³n²
Solution to class 8 maths chapter 9 exercise 9.5 (4).
(i) (a²–b²)²=(a²)²–2(a²)(b²)+(b²)² {∵(a –b)²=a²–2ab+b²}
=a⁴–2a²b²+b⁴
(ii)(2x+5)² –(2x –5)²
=(2x)²+2×(2x)×(5)+(5)²–[(2x)²+2×(2x)×(5)+(5)²]
{∵(a+b)²=a²+2ab+b² and (a–b)²=a²–2ab+b²}
=4x²+20x+25 –[4x²–20x+25]
=4x²+20x+25 –4x²+20x–25
=40x
(iii) (7m–8n)² +(7m+8n)²
=(7m)²+2×(7m)×(8n)+(8n)²+[(7m)²+2×(7m)×(8n)+(8n)²]
{∵(a+b)²=a²+2ab+b² and (a–b)²=a²–2ab+b²}
=49m²–112mn+64n² +[49m²+112mn+64n²]
=49m²–112mn+64n²+49m²+112mn+64n²
=98m²+128n²
(iv) (4m+5n)²+(5m+4n)²
=(4m)²+2×(4m)×(5n)+(5n)²+[(5m)²+2×(5m)×(4n)+(4n)²]
{∵(a+b)²=a²+2ab+b² }
=16m²+40mn+25n² +[25m²+40mn+16n²]
=16m²+40mn+25n² +25m²+40mn+16n²
=41m²+80mn+41n²
(v) (2.5p–1.5q)²–(1.5p–2.5q)²
=(2.5p)² –2×(2.5p)×(1.5q)+(1.5q)² –[(1.5p)² –2×(1.5p)×(2.5q)+(2.5q)²]
{∵(a –b)²=a² –2ab+b² }
=6.25p² –7.5pq+2.25q² –[2.25p² –7.5pq+6.25q²]
=6.25p² –7.5pq+2.25q² –2.25p²+7.5pq –6.25q²
=4p² –4q²
(vi) (ab+bc)² –2ab²c
=(ab)²+2×(ab)×(bc)+(bc)²–2ab²c
{∵(a +b)²=a²+2ab+b²}
=a²b²+2ab²c+b²c²–2ab²c
=a²b²+b²c²
(vii)(m² –n²m)²+2m³n²
=(m²)²+2×(m²)×(n²m)+(n²m)²–2m³n²
{∵(a +b)²=a²+2ab+b²}
=m⁴+2m³n²+n⁴m²–2m³n²
=m⁴+n⁴m²
class 8 maths chapter 9 exercise 9.5 (5)
5 Show that
(i) (3x+7)² –84x=(3x –7)²
(ii) (9p – 5q) ²+180pq = (9p + 5q) ²
(iii) [(⁴/₃)m –(¾)n]²+2mn=(¹⁶/₉)m²+(⁹/₁₆)n²
(iv) (4pq + 3q) ²–(4pq – 3q)²= 48pq²
(v) (a–b)(a+b)+(b–c)(b+c)+(c–a)(c+a)=0
Solution to class 8 maths chapter 9 exercise 9.5 (5).
L.H.S
(i)(3x+7)² –84x =[(3x)²+2×(3x)×(7)+(7)²]–84x
{∵(a+b)²=a²+2ab+b²}
=9x²+42x+49–84x
=9x² –42x+49
=(3x–7)² {∵(a –b)²=a²–2ab+b²}
=R.H.S
(ii)L.H.S
=(9p–5q)²+180pq
=[(9p)²–2×(9p)×(5q)+(5q)²]–180pq
{∵(a+b)²=a²+2ab+b²}
=81p² –90pq+25q²–180pq
=81p²+90pq+25q²
=(9p+5q)² {∵(a+b)²=a²+2ab+b²}
=R.H.S
(iii)L.H.S=
[(⁴/₃)m –(¾)n]²+2mn
={[(⁴/₃)m]²–2×[(⁴/₃)m]×[(¾)n]+[(¾)n]²}–2mn
{∵(a–b)²=a²–2ab+b²}
=(¹⁶/₉)m²–2mn+(⁹/₁₆)n²+2mn
=(¹⁶/₉)m²+(⁹/₁₆)n²
=R.H.S
(iv) L.H.S
=(4pq + 3q)²– (4pq – 3q)²
=(4pq)²+2×(4pq)×(3q)+(3q)²–[(4pq)² –2×(4pq)×(3q)+(3q)²]
{∵(a+b)²=a²+2ab+b² and (a+b)²=a²+2ab+b² }
=16p²q²+24pq²+9q² –[16p²q²–24pq²+9q²]
=16p²q²+24pq²+9q² –16p²q²+24pq²–9q²
=48pq²
=R.H.S
(v) L.H.S
=(a–b)(a+b)+(b–c)(b+c)+(c–a)(c+a)
=a²–b²+b²–c²+c²–a² {∵(a+b)(a–b)=a²–b²}
=0
=R.H.S
class 8 maths chapter 9 exercise 9.5 (6)
6 Using identities Evaluate:
(i) 71² (ii) 99² (iii) 102² (iv)998²
(v) 5.2²(vi) 297 ×303 (vii) 78 ×82
(viii) 8.9² (ix)1.05×9.5
Solution to class 8 maths chapter 9 exercise 9.5 (6).
(i) 71²=(70+1)²=(70)²+2×(70)×(1)+1²
{∵(a +b)²=a²+2ab+b²}
=4900 +140+1=5041
(ii) 99²=(100–1)²=(100)²–2×(100)×(1)+1²
{∵(a –b)²=a² –2ab+b²}
=10000 –200+1=9801
(iii) 102²=(100+2)²
=(100)²+2×(100)×(2)+2²
{∵(a +b)²=a²+2ab+b²}
=10000 +400+4=10404
(iv) 998²=(1000–2)²=(1000)²–2×(1000)×(2)+2²
{∵(a –b)²=a² –2ab+b²}
=1000000 –4000+4=996004
(v) 5.2²=(5+0.2)²=(5)²+2×(5)×(0.2)+(0.2)²
{∵(a +b)²=a²+2ab+b²}
=25+2+0.04=27.04
(vi) 297×303
=(300 –3)×(300+3) {∵(a+b)(a–b)=a²–b²}
=(300)² –(3)²
=90000 –9
=89991
(vii)78×82
=(80 –2)×(80+2) {∵(a+b)(a–b)=a²–b²}
=(80)² –(2)²
=6400 –4
=6396
(viii) 8.9²=(8+0.9)²
=(8)²+2×(8)×(0.9)+(0.9)²
{∵(a +b)²=a²+2ab+b²}
=64 +14.4+0.81=79.21
(ix) 10.5 ×9.5
=(10 +0.5)×(10 –0.5) {∵(a+b)(a–b)=a²–b²}
=(10)² –(0.5)²
=100 –0.25
=99.75
class 8 maths chapter 9 exercise 9.5 (7)
7 Using a²–b²=(a+b)(a–b), find
(i) 51²–49² (ii) (1.02)² –(0.98)² (iii)153² –147² (iv) (12.1)² –(7.9)²
Solution to class 8 maths chapter 9 exercise 9.5 (7).
(i) 51² –49²=(51+49)(51–49) {∵a²–b²=(a+b)(a–b)}
=100 ×2=200
(ii)(1.02)² –(0.98)²
=(1.02+0.98)(1.02–0.98) {∵a²–b²=(a+b)(a–b)}
=2.00×0.04=0.08
(iii) 153² –147²
=(153+147)(153–147) {∵a²–b²=(a+b)(a–b)}
=300 ×6=1800
(iv) (12.1)² –(7.9)²
=(12.1+7.9)(12.1–7.9) {∵a²–b²=(a+b)(a–b)}
=20.0×4.2=84.0=84
class 8 maths chapter 9 exercise 9.5 (8)
8 Using (x+a)(x+b)= x²+(a+b)x+ab, find
(1) 103 ×104 (ii) 5.1 ×5.2
(iii) 103 ×98 (iv) 9.7 ×9.8
Solution to class 8 maths chapter 9 exercise 9.5 (8).
(1) 103 × 104 = (100+3)×(100+4)
=(100)²+(3+4)×100+3×4 [∵ (x+a)(x+b)= x²+(a+b)x+ab]
=10000+700+12=10712.
(ii) 5.1×5.2=5.1×5.2=(5+0.1)×(5 + 0.2)
= (5)²+(0.1+0.2)×5+0.1×0.2
[∵ (x+a)(x+b)=x²+(a+b)x+ab]
=25+0.3×5+0.02
= 25+1.5+0.02= 26.52
(iii)103×98 = (100+3)×(100 – 2)
=(100)²+(3-2)×100+3×(-2)
[∵ (x+a)(x+b)=x²+(a+b)x+ab]
= 10000+[(3 – 2)×100]-6
=10000+100 -6 = 10094
(iv)9.7×9.8 =(10 -0.3)×(10 – 0.2)
= (10)² +{(-0.3) +(-0.2)}×10+(-0.3)×(-0.2)
[∵(x + a)(x+b)=x²+ (a+b)x+ ab]
= 100 +{-0.3-0.2}x10+0.06
= 100-0.5×10 +0.06 =100-5+0.06=95.06
By
Sudheer Kumar B