{"id":779,"date":"2021-03-24T11:55:26","date_gmt":"2021-03-24T06:25:26","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=779"},"modified":"2021-03-24T12:30:44","modified_gmt":"2021-03-24T07:00:44","slug":"samacheer-kalvi-9th-maths-chapter-3-ex-3-4","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-9th-maths-chapter-3-ex-3-4\/","title":{"rendered":"Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4"},"content":{"rendered":"

You can Download Samacheer Kalvi 9th Maths Book Solutions<\/a>\u00a0Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4<\/h2>\n

9th Maths Exercise 3.4 Question 1.<\/strong>
\nExpand the following :
\n(i) (2x + 3y + 4z)2<\/sup>
\n(ii) (-p + 2q + 3r)2<\/sup>
\n(iii) (2p + 3)(2p – 4)(2p – 5)
\n(iv) (3a + 1)(3a – 2)(3a + 4)
\nSolution:
\n(i) (2x + 3y + 4z)2<\/sup>
\n(a + b + c)2<\/sup> \u2261 a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
\n\u2234 (2x + 3y + 4z)2<\/sup> = (2x)2<\/sup> + (3y)2<\/sup> + (4z)2<\/sup> + 2 (2x) (3y) + 2 (3y) (4z) + 2 (4z) (2x)
\n= 4x2<\/sup> + 9y2<\/sup> + 16z2<\/sup> + 12xy + 24yz + 16zx<\/p>\n

(ii) (-p + 2q + 3r)2<\/sup>
\n(a + b + c)2<\/sup> \u2261 a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
\n(-p + 2q + 3r)2<\/sup> = (-p)2<\/sup> + (2q)2<\/sup> + (3r)2<\/sup> + 2(-p) (2q) + 2(2q) (3r) + 2 (-p) (3r)
\n= p2<\/sup> + 4q2<\/sup> + 9r2<\/sup> – 4pq +12qr – 6rp<\/p>\n

(iii) (2p + 3)(2p – 4)(2p – 5)
\n(x + a) (x + b) (x + c) \u2261 x3<\/sup> + (a + b + c) x2<\/sup> + (ab + be + ca) x + abc
\n(2p + 3)(2p – 4)(2p – 5) = (2p)3<\/sup> + (3 – 4 – 5) (2p)2<\/sup> + [(3 \u00d7 – 4)2<\/sup> + (-4 \u00d7 -5) + (-5 \u00d7 3)] 2p + 3 \u00d7 – 4 \u00d7 – 5
\n= 8p3<\/sup> + (-6) (4p2<\/sup>) + [-12 + 20 + (-15)] 2p + 60
\n= 8p3<\/sup> – 24p2<\/sup> + (-7)2p + 60
\n(2p + 3)(2p – 4)(2p – 5) = 8p3<\/sup> – 24p2<\/sup> – 14p + 60<\/p>\n

(iv) (3a + 1)(3a – 2)(3a + 4)
\n(x + a) (x + b) (x + c) \u2261 x3<\/sup> + (a + b + c) x2<\/sup> + (ab + bc + ca) x + abc
\n(3a + 1)(3a – 2)(3a + 4) = (3a)3<\/sup> + (1 – 2 + 4) (3a)2<\/sup> + [1 \u00d7 (- 2) + (-2 \u00d7 4) + 4 \u00d7 1] (3a) + 1 \u00d7 -2 \u00d7 4
\n= 27a3<\/sup> + 3 (9a2<\/sup>) + (-2 – 8 + 4)3a – 8
\n= 27a3<\/sup> + 27a2<\/sup> – 8a – 8<\/p>\n

9th Maths Algebra Exercise 3.4 Question 2.<\/strong>
\nUsing algebraic identity, find the coefficients of x2, x and constant term without actual expansion.
\n(i) (x + 5)(x + 6)(x + 7)
\n(ii) (2x + 3)(2x – 5)(2x – 6)
\nSolution:
\n(i) (x + 5)(x + 6)(x + 7)
\n(x + a) (x + b) (x + c) \u2261 x3<\/sup> + (a + b + c) x2<\/sup> + (ab + bc + ca) x + abc
\nCo-efficient of x2<\/sup> = a + b + c = 5 + 6 + 7 = 18
\nCo-efficient of x2<\/sup> = ab + bc + ca = (5 \u00d7 6) + (6 \u00d7 7) + (7 \u00d7 5)
\n= 30 + 42 + 35 = 107
\nConstant term = abc = 5 \u00d7 6 \u00d7 7
\nCo-efficient of constant term = 210<\/p>\n

(ii) (2x + 3)(2x – 5)(2x \u2014 6)
\n\u2234 Co-efficient of x2<\/sup> = 4 (a + b + c) = 4 (3 + (-5) + (-6))
\n= 4 \u00d7 (-8) = -32
\nCo-efficient of x = 2 (ab + bc + ca)
\n= 2 [3 \u00d7 (-5) + (-5) (-6) + (-6) (3)]
\n= 2[-15 + 30- 18] = 2 \u00d7 (-3) = -6
\nConstant term = abc = 3 \u00d7 (-5) \u00d7 (-6) = 90<\/p>\n

9th Maths 3.4 Question 3.<\/strong>
\nIf (x + a)(x + b)(x + c) = x3<\/sup> + 14x2<\/sup> + 59x + 70, find the value of
\n(i) a + b + c
\n(ii) \\(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\)
\n(iii) a2<\/sup> + b2<\/sup> + c2<\/sup>
\n(iv) \\(\\frac{a}{b c}+\\frac{b}{a c}+\\frac{c}{a b}\\)
\nSolution:
\n(x + a)(x + b)(x + c) = x3<\/sup> + 14x2<\/sup> + 59x + 70 …………….. (1)
\n(x + a) (x + b) (x + c) \u2261 x3<\/sup> + (a + b + c) x2<\/sup> + (ab + bc + ca) x + abc …………. (2)
\n(i) Comparing (1) & (2)
\nWe get, a + b + c = 14
\n(ii) \\(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{b c+a c+a b}{a b c}=\\frac{59}{70}\\)
\n(iii) (a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
\n(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2(ab + bc + ca)
\n= 142<\/sup> – 2 (59) = 196 – 118 = 78
\n(iv) \\(\\frac{a}{b c}+\\frac{b}{a c}+\\frac{c}{a b}=\\frac{a^{2}+b^{2}+c^{2}}{a b c}=\\frac{78}{70}=\\frac{39}{35}\\)<\/p>\n

Exercise 3.4 Class 9 Question 4.<\/strong>
\nExpand
\n(i) (3a – 4b)3<\/sup>
\n(ii) (x + \\(\\frac{1}{y}\\))3<\/sup>
\nSolution:
\n(i) (3a – 4b)3<\/sup> We know that
\n(x – y)3<\/sup> = x3<\/sup> – 3x2<\/sup>y + 3xy2<\/sup> – y3<\/sup>
\n(3a-4b)3<\/sup> = (3a)3<\/sup> – 3 (3a)2<\/sup> (4b) + 3 (3a) (4b)2<\/sup> – (4b)3<\/sup>
\n= 27a3<\/sup> – 108a2<\/sup>b + 144 ab2<\/sup> – 64b3<\/sup><\/p>\n

(ii) (x + \\(\\frac{1}{y}\\))3<\/sup>
\n(x + y)3<\/sup> \u2261 x3<\/sup> + 3x2<\/sup>y + 3xy2<\/sup> + y3<\/sup>
\n\\(\\left(x+\\frac{1}{y}\\right)^{3}=x^{3}+\\frac{3 x^{2}}{y}+\\frac{3 x}{y^{2}}+\\frac{1}{y^{3}}\\)<\/p>\n

9th Maths Exercise 3.4 In Tamil Question 5.<\/strong>
\nEvaluate the following by using identities:
\n(i) 983<\/sup>
\n(ii) 10013<\/sup>
\nSolution:
\n(i) 982<\/sup> = (100 – 2)3<\/sup>
\n(a – b)3<\/sup> \u2261 a3<\/sup> – 3a2<\/sup>b + 3ab2<\/sup> – b3<\/sup>
\n983<\/sup> = (100 – 2)3<\/sup> = 1003<\/sup> – 3 \u00d7 1002<\/sup> \u00d7 + 3 \u00d7 100 \u00d7 22<\/sup> – 23<\/sup>
\n= 1000000 – 3 \u00d7 10000 \u00d7 2 + 300 \u00d7 4 – 8
\n= 1000000 – 60000 + 1200 – 8 = 1001200 – 60008 = 941192<\/p>\n

(ii) 10013<\/sup> = (1000 + 1)3<\/sup>
\n(a + b)3<\/sup> \u2261 a3<\/sup> + 3a2<\/sup>b + 3ab2<\/sup> + b3<\/sup>
\n(1000 + 1)3<\/sup> = 10003<\/sup> + 3(10002<\/sup>) \u00d7 1 + 3 \u00d7 1000 \u00d7 12<\/sup> + 13<\/sup>
\n= 1000,000,000 + 3,000,000 + 3000 + 1 = 1,003,003,001<\/p>\n

Ex 3.4 Class 9 Question 6.<\/strong>
\nIf (x + y + z) = 9 and (xy + yz + zx) = 26 then find the value of x2<\/sup> + y2<\/sup> + z2<\/sup>.
\nSolution:
\n(x + y + z) = 9 and(xy + yz + zx) = 26
\nx2<\/sup> + y2<\/sup> + z2<\/sup> = (x + y + z)2<\/sup> – 2 (xy + yz + zx)
\n= 92<\/sup> – 2 \u00d7 26 = 81 – 52 = 29<\/p>\n

9th Standard Maths Exercise 3.4 Question 7.<\/strong>
\nFind 27a3<\/sup> + 64b3<\/sup>, if 3a + 4b = 10 and ab = 2.
\nSolution:
\n3a + 4b = 10, ab = 2
\n(3a + 4b)3<\/sup> = (3a)3<\/sup> + 3 (3a)2<\/sup> (4b) + 3 (3a) (4b)2<\/sup> + (4b)3<\/sup>
\n(27a3<\/sup> + 64b3<\/sup>) = (3a + 4b)3<\/sup> – 3 (3a) (4b) (3a + 4b)
\n\u2235 x3<\/sup> + y3<\/sup> = (x + y)3<\/sup> – 3xy – (x + y)
\n= 103<\/sup> – 36 ab (10)= 1000 – 36 \u00d7 2 \u00d7 10
\n= 1000 – 720 = 280<\/p>\n

Exercise 3.4 Class 9 Maths Solution Question 8.<\/strong>
\nFind x3<\/sup> – y3<\/sup>, if x – y = 5 and xy = 14.
\nSolution:
\nx – y = 5, xy = 14
\nx3<\/sup> – y3<\/sup> = (x – y)3<\/sup> + 3xy (x – y) = 53<\/sup> + 3 \u00d7 14 \u00d7 5
\n= 125 + 210 = 335<\/p>\n

Class 9 Maths Exercise 3.4 Solutions Question 9.<\/strong>
\nIf a + \\(\\frac{1}{a}\\) = 6, then find the value of a3<\/sup> + \\(\\frac{1}{a^{3}}\\)
\nSolution:
\na3<\/sup> + b3<\/sup> = (a + b)3<\/sup> – 3ab (a + b)
\n\"9th<\/p>\n

9th Math 3.4 Solution Question 10.<\/strong>
\nIf x2<\/sup> + \\(\\frac{1}{x^{2}}\\) = 23, then find the value of x + \\(\\frac{1}{x}\\) and x3<\/sup> + \\(\\frac{1}{x^{3}}\\)
\nSolution:
\n\"9th<\/p>\n

Maths Exercise 3.4 Class 9 Question 11.<\/strong>
\nIf (y – \\(\\frac{1}{y}\\))3<\/sup> = 27, then find the value of y3<\/sup> – \\(\\frac{1}{y^{3}}\\)
\nSolution:
\n\"9th<\/p>\n

9th Class Maths Exercise 3.4 Solution Question 12.<\/strong>
\nSimplify : (i) (2a + 36 + 4c)(4a2<\/sup> + 9b2<\/sup> + 16c2<\/sup> – 6ab – 12bc – 12bc – 8ca)
\n(ii) (x – 2y + 3z) (x2<\/sup> + 4y2<\/sup> + 9z2<\/sup> + 2xy + 6yz – 3xz)
\nSolution:
\n(i) (2a + 36 + 4c)(4a2<\/sup> + 9b2<\/sup> + 16c2<\/sup> – 6ab – 12bc – 12bc – 8ca)
\nWe know that
\n(a + b + c) (a2<\/sup> + b2<\/sup> + c2<\/sup> – ab – bc – ca) = a3<\/sup> + b3<\/sup> + c3<\/sup> \u00d7 3 abc
\n\u2234 (2a + 36 + 4c) (4a2<\/sup> + 9b2<\/sup> + 16 c2<\/sup> – 6 ab – 12 bc – 8 ca)
\n= (2a)3<\/sup> + (3b)3<\/sup> + (4c)3<\/sup> – 3 \u00d7 2a \u00d7 36 \u00d7 4c
\n= 8a3<\/sup> + 27b3<\/sup> + 64c3<\/sup> – 72 abc<\/p>\n

(ii) (x – 2,y + 3z) (x2<\/sup> + 4y2<\/sup> + 9z2<\/sup> + 2xy + 6yz – 3xz)
\n(a + b + c) (a2<\/sup> + b2<\/sup> + c2<\/sup> – ab – bc – ca) = a3<\/sup> + b3<\/sup> + c3<\/sup> – 3abc .
\n\u2234 (x – 2y + 3z) (x2<\/sup> + 4y2<\/sup> + 9z2<\/sup> + 2xy + 6yz – 3xz)
\n= x3<\/sup> + (-2y)3<\/sup> + (3z)3<\/sup> – 3 \u00d7 x \u00d7 (-2y) (3z)
\n= x3<\/sup> – 8y3<\/sup> + 27z3<\/sup> + 18 xyz<\/p>\n

9th Maths Exercise 3.4 Samacheer Kalvi Question 13.<\/strong>
\nBy using identity evaluate the following:
\n(i) 73<\/sup> – 103<\/sup> + 33<\/sup>
\n(ii) 1 + \\(\\frac{1}{8}-\\frac{27}{8}\\)
\nSolution:
\n(i) 73<\/sup> – 103<\/sup> + 33<\/sup>
\n(a + b + c) (a2<\/sup> + b2<\/sup> + c2<\/sup> – ab – bc – ca) = a3<\/sup> + b3<\/sup> + c3<\/sup> – 3 abc
\nIf a + b + c = 0, then a3<\/sup> + b3<\/sup> + c3<\/sup> = 3 abc
\n\u2234 7 – 10 + 3 = 0
\n\u21d2 73<\/sup> – 103<\/sup> + 33<\/sup> = 3 \u00d7 7 \u00d7 – 10 \u00d7 3
\n= 9 \u00d7 -70 = -630
\n\"Exercise<\/p>\n

9th Class Math Exercise 3.4 Solution Question 14.<\/strong>
\nIf 2x – 3y – 4z = 0, then find 8x3<\/sup> – 27y3<\/sup> – 64z3<\/sup>.
\nSolution:
\nIf 2x – 3y – 4z = 0 then 8x3<\/sup> – 27y3<\/sup> – 64z3<\/sup> = ?
\nIf x + y + z = 0 then x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz
\n8x3<\/sup> – 21 y3<\/sup> – 64z3<\/sup> = (2x)3<\/sup> + (-3y)3<\/sup> + (-4z)3<\/sup>
\n= 3 \u00d7 2x \u00d7 -3y \u00d7 – 4z = 72 xyz<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 9th Maths Book Solutions\u00a0Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4 9th Maths Exercise 3.4 Question 1. Expand the following : (i) (2x + 3y + 4z)2 (ii) …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/779"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=779"}],"version-history":[{"count":0,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/779\/revisions"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=779"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=779"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=779"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}