Find Solutions to class 8 maths chapter 9 exercise 9.5
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