{"id":4706,"date":"2020-09-09T09:01:08","date_gmt":"2020-09-09T03:31:08","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4706"},"modified":"2021-07-02T15:03:36","modified_gmt":"2021-07-02T09:33:36","slug":"kseeb-solutions-class-8-maths-chapter-2-ex-2-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-2-ex-2-4\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.4"},"content":{"rendered":"

Students can Download Maths Chapter 2 Algebraic Expressions Ex 2.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka Board Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.4<\/h2>\n

Question 1.
\nFind the product:<\/p>\n

i. (a + 3) ( a + 5)
\nii. (3t + 1) (3t + 4)
\niii. (a – 8) (a + 2)
\niv. (a – 6) (a – 2)
\nAnswer:
\ni. (a + 3) (a + 5) This is in the form.
\n(x + a)(x + b) = x\u00b2 + (a + b)x + ab
\nHere x = a, a = 3, b = 5
\n(a + 3)(a + 5) = a\u00b2 + (a + 5)a + 3.5
\n= a\u00b2 + 8a + 15<\/p>\n

(ii) (3t + 1)(3t + 4)This is in the form.
\n(x + a)(x + b) = x\u00b2 +(a + b)x + ab
\nHere x =3t,a = 1, b = 4
\n(3t – 1)(3t + 4) = (3t)\u00b2+(1 + 4)3t + 1.4
\n= 9t\u00b2 + 5(3t) + 4 = 9t\u00b2 + 15t + 4<\/p>\n

(iii) (a -8)(a + 2)
\nThis is in the form (x + a)(x + b) = x\u00b2 + (a + b)x + ab
\nHere x = a, a = -8, b = 2
\n(a \u2013 8)(a + 2) = a\u00b2 + (-8 + 2)a + (-8)(2)
\n= a\u00b2 + (-6)a \u2013 16
\n= a\u00b2 \u2013 6a \u2013 16<\/p>\n

(iv) (a – 6)(a – 2)
\nThis is in the form (x + a)(x + b) = x\u00b2 +(a + b)x + ab
\nHerex = a, a = -6,b = -2
\n(a \u2013 6)(a \u2013 2) = a\u00b2 + (- 6 – 2) a + (-6)(-2)
\n=a\u00b2 + (- 8)a + 12
\n= a\u00b2 \u2013 8a + 12<\/p>\n

2. Evaluate using suitable identities<\/p>\n

Question i.
\n53 \u00d7 55
\nAnswer:
\n53 \u00d7 55 = (50 + 3) (50 + 5)
\nThis is in the form (x + a)(x + b) = x\u00b2 + (a + b)x + ab
\nHere x = 50, a = 3, b = 5
\n(50 + 3)(50 + 5) = 50\u00b2 + (3 + 5) 50 + 3 \u00d7 5
\n= 2500 + (8)50 +15
\n= 2500 + 400 + 15
\n\u2234 53 \u00d7 55 = 2915<\/p>\n

Question ii.
\n102 \u00d7 106
\nAnswer:
\n102 \u00d7 106 = (100 + 2) (100 + 6) This is in the form.
\n(x + a)(x + b) = x\u00b2 +(a + b)x + ab
\nHere x = 100, a = 2, b = 6
\n(100 + 2)(100 + 6) = 100\u00b2 +(2 + 6)100 + 2 \u00d7 6
\n= 10000 + (8)100 + 12
\n= 10000 + 800 + 12
\n\u2234102 \u00d7 106 = 10812<\/p>\n

Question iii.
\n34 \u00d7 36
\nAnswer:
\n34 \u00d7 36 = (30 + 4) (30 + 6) This is in the form.
\n(x + a)(x + b) = x\u00b2+ (a + b)x + ab
\nHere x =30, a = 4, b = 6
\n(30 + 4)(30 + 6) = 30\u00b2 + (4 + 6)30 + 4 \u00d7 6
\n= 900 + 300 + 24
\n34 \u00d7 36 = 1224<\/p>\n

Question iv.
\n103\u00d7 96
\nAnswer: 103 \u00d7 96 = (100 + 3)(100-4)This is in the form.
\n(x + a)(x + b) = x\u00b2 + (a + b)x + ab
\nHere x = 100,a = 3,b = -4 (100 + 3)(100-4) = 100\u00b2 + (3 – 4)100 + 3 \u00d7 -4
\n= 10000 + (- 1)100 – 12
\n= 10000 – 100 – 12
\n103 \u00d7 96 = 9888<\/p>\n

Question 3.
\nFind the expression for the product (x+a) (x+b) (x+c) using the identity (x+a)(x+b) = x\u00b2 + (a + b)x + ab
\nAnswer:
\n(x + a) (x + b) (x + c)
\nConsider (x + a)(x + b)
\n= x\u00b2 + (a + b)x + ab
\n= x\u00b2 + ax + bx + ab (x + a)(x + b)(x + c)
\n= (x\u00b2 + ax + bx + ab)(x + c)
\n= x(x\u00b2 + ax + bx+ ab) + c
\n(x\u00b2 + ax + bx + ab)
\n= x3<\/sup> + ax\u00b2 + bx\u00b2 + abx + cx\u00b2 + cax +cbx + abc
\n= x3<\/sup> + ax\u00b2 + bx\u00b2 + cx\u00b2 + abx + cax +cbx + abc [by rearranging]
\n= x3<\/sup> + x\u00b2(a + b + c) + x(ab + ca +bc) + abc
\n\u2234 (x + a)(x + b)(x + c)
\n= x3<\/sup> +x\u00b2 (a + b + c) + x(ab + bc + ca) + abc<\/p>\n

4. Using the identify (a+b)2 = a\u00b2 + 2ab+b\u00b2 simplify the following.<\/p>\n

Question i.
\n(a + 6 )\u00b2
\nAnswer:
\nUsing the identity
\n(a + b)\u00b2 = a\u00b2 + 2ab + b\u00b2
\n(a + 6)\u00b2 = a\u00b2 +2.a.6 + 6\u00b2
\n= a\u00b2 +12a + 36<\/p>\n

Question ii.
\n(3x+2y)2
\nAnswer:
\nUsing the identity
\n(a + b)\u00b2 = a\u00b2 +2ab + b\u00b2
\n(3x + 2y)\u00b2 =(3x)\u00b2 + 2.3x.2y + (2y)\u00b2
\n= 9x\u00b2 +12xy + 4y\u00b2<\/p>\n

Question iii.
\n(2p + 3q)\u00b2
\nAnswer:
\nUsing the identity
\n(a + b)\u00b2 =a\u00b2 + 2ab + b\u00b2
\n(2p + 3q)\u00b2 =(2p)\u00b2 +2.2p.3q + (3q)\u00b2
\n= 4p\u00b2 + 12pq + 9q\u00b2<\/p>\n

Question iv.
\n(x\u00b2 + 5)\u00b2
\nAnswer:
\nUsing the identity (a + b)\u00b2 = a\u00b2 +2ab + b\u00b2
\n(x\u00b2 +5)\u00b2 = (x\u00b2)\u00b2 + 2.x\u00b2.5 + 52 = x4<\/sup> + 10x\u00b2 + 25<\/p>\n

5. Evaluate using the identity (a+b)\u00b2 = a\u00b2 + 2ab + b\u00b2<\/p>\n

Question i.
\n(34)\u00b2
\nAnswer:
\n34\u00b2 = (30 + 4)\u00b2
\n(a + b)\u00b2 = a\u00b2 + 2ab+ b\u00b2 Here a = 30, b = 4
\n(30 + 4)\u00b2 = (30)\u00b2 + 2 \u00d7 30 \u00d7 4 + (4)\u00b2
\n= 900 + 240 + 16
\n(34)\u00b2 = 1156<\/p>\n

Question ii.
\n(10.2)\u00b2
\nAnswer:
\n(10.2)\u00b2 = (10 + 0.2)\u00b2
\n(a + b)\u00b2 = a\u00b2 + 2ab + b\u00b2 Here a = 10, b=0.2
\n[(10 + (0.2))]\u00b2 = 10\u00b2 +2.(10).(0.2) + (0.2)\u00b2 ]
\n= 100 + 4 + 0.04
\n(10.2)\u00b2 = 104.04<\/p>\n

Question iii.
\n53\u00b2
\nAnswer:
\n53\u00b2 = (50 + 3)\u00b2
\n(a + b)\u00b2 =a\u00b2 +2ab + b\u00b2
\nHere a = 50, b = 3 (50 + 3)\u00b2 =50\u00b2 + 2 \u00d7 50 \u00d7 3 + (3)\u00b2
\n= 2500 + 300 + 9
\n53\u00b2 = 2809<\/p>\n

Question iv.
\n(41 )\u00b2
\nAnswer:
\n41\u00b2 = (40 + 1)\u00b2
\n(a + b)\u00b2 = a\u00b2 + 2ab + b\u00b2
\nHere a = 40,b = 1
\n(40 + 1)\u00b2 = 40\u00b2 + 2 \u00d7 40 \u00d7 1 + (1)\u00b2
\n= 1600 + 80 + 1
\n41\u00b2 = 1681<\/p>\n

6. Use the identify (a -b)\u00b2 = a\u00b2 – 2ab + b\u00b2 to compute.<\/p>\n

Question i.
\n(x – 6)\u00b2
\nAnswer:
\n(a – b)\u00b2 = a\u00b2 – 2ab + b\u00b2
\nHerea = x, b=6
\n(x – 6)\u00b2 = x\u00b2 – (2)(x)(6) + 6\u00b2
\n= x\u00b2 – 12x+ 36<\/p>\n

Question ii.
\n(3x – 5y)\u00b2
\nAnswer:
\n(a – b)\u00b2 = a\u00b2 – 2ab + b\u00b2 Here a = 3x, b = 5y
\n(3x – 5y)\u00b2 = (3x)\u00b2 -(2)(3x)(5y) + (5y)\u00b2 = 9x\u00b2 – 30xy + 25y\u00b2<\/p>\n

Question iii.
\n(5a – 4b)\u00b2
\nAnswer:
\n(a – b)\u00b2 = a\u00b2 – 2ab + b\u00b2
\nHere a = 5a, b = 4b
\n(5a – 4b)\u00b2 = (5a)\u00b2 – (2)(5a)(4b) + (4b)\u00b2
\n= 25a\u00b2 – 40ab + 16b\u00b2<\/p>\n

Question iv.
\n(p\u00b2 – q\u00b2)\u00b2
\nAnswer:
\n(a – b)\u00b2 = a\u00b2 – 2ab + b\u00b2 Here a = p\u00b2 , b = q2
\n(p\u00b2 – q\u00b2)\u00b2 = (p\u00b2)\u00b2 – (2)(p\u00b2)(q\u00b2) + (q\u00b2)\u00b2
\n= p4<\/sup> – 2p\u00b2q\u00b2 + q4<\/sup><\/p>\n

7. Evaluate using the identify (a – b)\u00b2 = a\u00b2 – 2ab + b\u00b2.<\/p>\n

Question i.
\n(49)\u00b2
\nAnswer:
\n49\u00b2 = (50 – 1)\u00b2
\n(a – b)\u00b2 = a\u00b2 – 2ab + b\u00b2 Here a = 50, b = 1
\n(50 – 1)\u00b2 = 50\u00b2 – (2)(50)(1) + 1\u00b2 = 2500 – 100 + 1
\n(49)\u00b2 = 2401<\/p>\n

Question ii.
\n(9.8)\u00b2
\nAnswer:
\n(9.8)\u00b2 = (10 – 0.2)\u00b2
\n(a – b)\u00b2 =a\u00b2 – 2ab + b\u00b2 Here a = 10, b=0.2
\n(10 – 0.2)\u00b2 = 10\u00b2 – (2)(10)(0.2) + (0.2)\u00b2
\n= 100 – 4 + 0.04
\n(9.8)\u00b2 =96.04<\/p>\n

Question iii.
\n59\u00b2
\nAnswer:
\n59\u00b2 = (60 – 1)\u00b2
\n(a – b)\u00b2 = a\u00b2 – 2ab + b\u00b2 Here a = 60, b = 1
\n(60 – 1)\u00b2 = 60\u00b2 – (2)(60)(1) +12 = 3600 – 120 + 1
\n59\u00b2 =3481<\/p>\n

Question iv.
\n(198)\u00b2
\nAnswer:
\n(198)\u00b2 = (200 – 2)\u00b2
\n(a – b)\u00b2 = a\u00b2 – 2ab + b\u00b2 Here a = 200, b = 2
\n(200 – 2)\u00b2 = 200\u00b2 – (2)(200)(2) + 2\u00b2
\n= 40000 – 800 + 4
\n(198)\u00b2 = 39204<\/p>\n

8. Use the identify (a + b) (a – b) = a\u00b2 – b\u00b2 to find the product.<\/p>\n

Question i.
\n(x + 6) (x – 6)
\nAnswer:
\n(a + b) (a – b) = a2 – b2
\nHere a = x,b = 6
\n(x + 6)(x – 6) = x\u00b2 – 6\u00b2 = x\u00b2 – 36<\/p>\n

Question ii.
\n(3x + 5) (3x – 5)
\nAnswer:
\n(a + b) (a – b) = a\u00b2 – b\u00b2
\nHere a = 3x, b = 5
\n(3x + 5)(3x – 5) = (3x)\u00b2 – (5)\u00b2
\n= 9x\u00b2 -25<\/p>\n

Question iii.
\n(2a + 4b) (2a – 4b)
\nAnswer:
\n(a + b) (a – b) = a\u00b2 – b\u00b2
\nHere a = 2a,b = 4b
\n(2a + 4b)(2a-4b) = (2a)\u00b2 – (4b)\u00b2
\n= 4a\u00b2 – 16b\u00b2<\/p>\n

Question iv.
\n\\(\\left(\\frac{2 x}{3}+1\\right)\\left(\\frac{2 x}{3}-1\\right)\\)
\nAnswer:
\n(a + b) (a – b) = a\u00b2 – b\u00b2
\nHere a = \\(\\frac{2 x}{3}\\) ,b = 1
\n\\(\\left(\\frac{2 x}{3}+1\\right)\\left(\\frac{2 x}{3}-1\\right)\\) = \\(\\left(\\frac{2 x}{3}\\right)^{2}-1^{2}\\)
\n= \\(\\frac{4 x^{2}}{9}-1\\)<\/p>\n

9. Evaluate these using identity :<\/p>\n

Question i.
\n55 \u00d7 45
\nAnswer:
\n55 \u00d7 45 = (50 + 5) (50 – 5)
\nIdentity (a + b)(a – b) = a\u00b2 – b\u00b2
\nHere a = 50,b=5
\n(50 + 5)(50 – 5) = (50)\u00b2 – 5\u00b2
\n= 2500 – 25
\n55 \u00d7 45 = 2475<\/p>\n

Question ii.
\n33 \u00d7 27
\nAnswer:
\n33 \u00d7 27 = (30 + 3) (30 – 3)
\nidentity (a + b)(a – b) = a\u00b2 – b\u00b2
\nHere a = 30,b = 3
\n(30 + 3)(30 – 3) = 30\u00b2 – 3\u00b2 = 900 – 9
\n33 \u00d7 27 = 891<\/p>\n

Question iii.
\n8.5 \u00d7 9.5
\nAnswer:
\n8.5 \u00d7 9.5 = (9 – 0.5) (9 + 0.5)
\nHere a = 9, b = 0.5
\n(9 – 0.5)(9 + 0.5) = (9\u00b2) – (0.5)\u00b2 = 81 – 0.25
\n= 80.75<\/p>\n

Question iv.
\n102 \u00d7 98
\nAnswer:
\n102 \u00d7 98 = (100+ 2) (100 – 2)
\nIdentity (a + b)(a – b) = a2<\/sup> – b2<\/sup>
\nHere a = 100, b= 2
\n(100 + 2)(100 – 2) = 100\u00b2 – 2\u00b2 =10000 – 4
\n102 \u00d7 98 = 9996<\/p>\n

10. Find the product.<\/p>\n

Question i.
\n(x – 3)(x + 3) (x2 + 9)
\nAnswer:
\n(x – 3)(x + 3) (x2 + 9)
\nIdentity(a + b)(a-b) = a\u00b2 – b\u00b2
\n= (x\u00b2 – 3\u00b2)(x\u00b2 + 9)
\n= (x\u00b2 – 9)(x\u00b2 + 9)
\n= (x\u00b2)\u00b2 – 9\u00b2
\n= x4<\/sup> – 81<\/p>\n

Question ii.
\n(2a + 3)(2a – 3) (4a\u00b2 + 9)
\nAnswer:
\n(2a + 3)(2a – 3) (4a\u00b2 + 9)
\nIdentity (a + b)(a-b) = a\u00b2 – b\u00b2
\n= [((2a)\u00b2 – 3\u00b2)(4a\u00b2+9)\u00b2
\n= (4a\u00b2 – 9)(4a\u00b2 + 9)
\n= (4a\u00b2)\u00b2 – 9\u00b2 = 16a4<\/sup> – 81<\/p>\n

Question iii.
\n(P + 2) (P – 2) (P\u00b2 + 4)
\nAnswer:
\n(P + 2) (P – 2) (P\u00b2 + 4)
\nIdentity (a + b)(a – b) = a\u00b2 – b\u00b2
\n= (P\u00b2 – 2\u00b2)(P\u00b2+ 4)
\n= (P\u00b2 – 4)(P\u00b2 + 4)
\n= (P\u00b2)\u00b2 – 4\u00b2 = P4<\/sup> – 16<\/p>\n

Question iv.
\n\\(\\left(\\frac{1}{2} m-\\frac{1}{3}\\right)\\left(\\frac{1}{2} m+\\frac{1}{3}\\right)\\left(\\frac{1}{4} m^{2}+\\frac{1}{9}\\right)\\)
\nAnswer:
\n\"KSEEB<\/p>\n

Question v.
\n(2x – y)(2x + y) (4x\u00b2 + y\u00b2)
\nAnswer:
\n(2x – y)(2x + y) (4x\u00b2 + y\u00b2)
\nIdentity (a + b)(a – b) = a\u00b2 – b\u00b2
\n[(2x)\u00b2 – y\u00b2](4x\u00b2 + y\u00b2)
\n= (4x\u00b2 – y\u00b2)(4x\u00b2 + y\u00b2)
\n= I6x4<\/sup> – y4<\/sup><\/p>\n

Question vi.
\n(2x – 3y)(2x + 3y) (4x\u00b2 + 9y\u00b2)
\nAnswer:
\n(2x – 3y)(2x + 3y) (4x\u00b2 + 9y\u00b2)
\nIdentity (a + b)(a-b) = a\u00b2 -b\u00b2
\n= [(2x)\u00b2 – (3y)\u00b2](4x\u00b2+ 9y\u00b2)
\n= (4x\u00b2 – 9y\u00b2)(4x\u00b2 + 9y\u00b2)
\n= (4x\u00b2)\u00b2 – (9y\u00b2)\u00b2
\n= 16x4<\/sup> – 81y4<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 2 Algebraic Expressions Ex 2.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka Board Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.4 Question 1. Find the product: i. …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4706"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=4706"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4706\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4706"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4706"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4706"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}