{"id":21520,"date":"2020-12-07T10:40:40","date_gmt":"2020-12-07T10:40:40","guid":{"rendered":"https:\/\/tnboardsolutions.com\/?p=21520"},"modified":"2021-07-08T03:19:16","modified_gmt":"2021-07-08T08:49:16","slug":"samacheer-kalvi-11th-maths-guide-chapter-2-ex-2-4","status":"publish","type":"post","link":"https:\/\/tnboardsolutions.com\/samacheer-kalvi-11th-maths-guide-chapter-2-ex-2-4\/","title":{"rendered":"Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.4"},"content":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 11th Maths Guide<\/a> Pdf Chapter 2 Basic Algebra Ex 2.4 Text Book Back Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4<\/h2>\n

Question 1.
\nConstruct a quadratic equation with roots 7 and – 3.
\nAnswer:
\nThe given roots are 7 and -3
\nLet \u03b1 = 7 and \u03b2 = -3
\n\u03b1 + \u03b2 = 7 – 3 = 4
\n\u03b1\u03b2 = (7)(-3) = -21
\nThe quadratic equation with roots \u03b1 and \u03b2 is x2<\/sup> – (\u03b1 + \u03b2) x + \u03b1\u03b2 = 0
\nSo the required quadratic equation is
\nx2<\/sup> – (4) x + (-21) = 0
\n(i.e.,) x2<\/sup> – 4x – 21 = 0<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nA quadratic polynomial has one of its zero 1 + \u221a5 and it satisfies p(1) = 2. Find the quadratic polynomial.
\nAnswer:
\nLet p(x) = ax2<\/sup> + bx + c be the required quadratic polynomial.
\nGiven p (1) = 2 , we have
\na \u00d7 12<\/sup> + b \u00d7 1 + c = 2
\na + b + c = 2 ——— (1)
\nAlso given 1 + \u221a5 is a zero of p(x)
\n\u2234 a(1 + \u221a5)2<\/sup> + b (1 + \u221a5) + c = 0
\na( 1 + 5 + 2\u221a5) + b (1 + \u221a5) + c = 0
\n6a + 2a\u221a5 + b + b\u221a5 + c = 0 ——— (2)
\nIf 1 + \u221a5 is zero then 1 – \u221a5 is also a zero of p (x).
\n\u2234 a(1 – \u221a5)2<\/sup> + b (1 – \u221a5) + c = 0
\na( 1 – 2\u221a5 + 5) + b (1 – \u221a5) + c = 0
\n6a – 2a\u221a5 + b – b\u221a5 + c = 0 ——— (3)
\n\"Samacheer
\nSubstituting the value of a in equation (4)
\n5 \u00d7 – \\(\\frac{2}{5}\\) + 2 \u00d7 – \\(\\frac{2}{5}\\) \u00d7 \u221a5 + b\u221a5 = – 2
\n– 2 – \\(\\frac{4}{5}\\)\u221a5 + b\u221a5 = – 2
\nb\u221a5 = – 2 + 2 + \\(\\frac{4}{5}\\) . \u221a5
\nb\u221a5 = \\(\\frac{4}{5}\\) . \u221a5
\nb = \\(\\frac{4}{5}\\)
\nSubstituting the value of a and b in equation (1), we have<\/p>\n

\"Samacheer
\n\u2234 The required quadratic polynomial is
\np(x) = \\(-\\frac{2}{5}\\)x2<\/sup> + \\(\\frac{4}{5}\\)x + \\(\\frac{8}{5}\\)
\np(x) = \\(-\\frac{2}{5}\\)(x2<\/sup> – 2x – 4)<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nIf \u03b1 and \u03b2 are the roots of the quadratic equation x2<\/sup> + \u221a2x + 3 = 0 form a quadratic polynomial with zeros \\(\\frac{1}{\\alpha}, \\frac{1}{\\beta}\\).
\nAnswer:
\nGiven \u03b1 and \u03b2 are the roots of the quadratic polynomial
\nx2<\/sup> + \u221a2x + 3 = 0 ——— (1)
\n\"Samacheer
\n\u2234 The required quadratic equation whose roots are \\(\\frac{1}{\\alpha}, \\frac{1}{\\beta}\\) is
\nx2<\/sup> – (sum of the roots)x + product of the roots = 0
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nIf one root of k (x – 1)2<\/sup> = 5x – 7 is double the other root, show that k = 2 or – 25
\nAnswer:
\nThe given quadratic equation is
\nk(x – 1)2<\/sup> = 5x – 7
\nk(x2<\/sup> – 2x + 1) – 5x + 7 = 0
\nkx2<\/sup> – 2kx + k – 5x + 7 = 0
\nkx2<\/sup> – (2k + 5)x + k + 7 = 0 ———- (1)
\nLet the roots be \u03b1 and 2\u03b1
\nSum of the roots \u03b1 + 2\u03b1 = –\\(\\frac{b}{a}\\)
\n\"Samacheer
\nProduct of te roots \u03b1(2\u03b1) = \\(\\frac{c}{a}\\)
\n\"Samacheer
\nUsing equation (2) and (3) we have
\n\"Samacheer
\n2(4k2<\/sup> + 20k + 25) = 9k(k + 7)
\n8k2<\/sup> + 40k + 50 = 9k2<\/sup> + 63k
\n9k2<\/sup> + 63k – 8k2<\/sup> – 40k – 50 = 0
\nk2<\/sup> + 23k – 50 = 0
\nk2<\/sup> + 25k – 2k – 50 = 0
\nk(k + 25) – 2(k + 25) = 0
\n(k – 2) (k + 25) = 0
\nk – 2 = 0 or k + 25 = 0
\nk = 2 or k = – 25<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nIf the difference of the roots of the equation 2x2<\/sup> – (a + 1)x + a – 1 = 0 is equal to their product then prove that a = 2.
\nAnswer:
\nThe given quadratic equation is
\n2x2<\/sup> – (a + 1) x + a – 1 = 0 ———– (1)
\nLet \u03b1 and \u03b2 be the roots of the given equation
\nGiven that \u03b1 – \u03b2 = \u03b1\u03b2 —— (2)
\nSum of the roots \u03b1 + \u03b2 = – \\(\\frac{b}{a}\\)
\n\"Samacheer
\n\"Samacheer
\nSubstituting the values of \u03b1 and \u03b2 in equation (2)
\n\"Samacheer
\n2(a – 1) = a
\n2a – 2 – a = 0
\na – 2 = 0
\n\u21d2 a = 2<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nFind the condition that one of the roots of ax2<\/sup> + bx + c may be
\n(a) negative of the other
\n(b) thrice the other
\n(c) reciprocal of the other.
\nAnswer:
\nThe given quadratic equation is
\nax2<\/sup> + bx + c = 0 ——- (1)
\nLet \u03b1 and \u03b2 be the roots of the equation (1) then
\nSum of the roots \u03b1 + \u03b2 = ——- (2)
\nProduct of the roots \u03b1\u03b2 = ——- (3)<\/p>\n

(a) Given one root is the negative of the other
\n\u03b2 = – \u03b1
\n(2) \u21d2 \u03b1 + (-\u03b1) = – \\(\\frac{b}{a}\\)
\n0 = – \\(\\frac{b}{a}\\)
\n\u21d2 b = 0
\n(3) \u21d2 \u03b1(-\u03b1) = \\(\\frac{c}{a}\\)
\n– \u03b12<\/sup> = \\(\\frac{c}{a}\\)
\nHence the required condition is b = 0<\/p>\n

(b) Given that one root is thrice the other
\n\u03b2 = 3\u03b1
\n\"Samacheer
\nWhen is the required condition?<\/p>\n

(c) One root is reciprocal of the other
\n\"Samacheer
\nWhen is the required condition?<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nIf the equations x2<\/sup> – ax + b = 0 and x2<\/sup> – ex + f = 0 have one root in common and if the second equation has equal roots then prove that ae = 2(b + f).
\nAnswer:
\nThe given quadratic equations are
\nx2<\/sup> – ax + b = 0 ———- (1)
\nx2<\/sup> – ex + f = 0 ——— (2)
\nLet \u03b1 be the common root of the given quadratic equations (1) and (2)
\nLet \u03b1, \u03b2 be the roots of x2<\/sup> – ax + b = 0
\nSum of the roots \u03b1 + \u03b2 = \\(-\\left(-\\frac{a}{1}\\right)\\)
\n\u03b1 + \u03b2 = a ———- (3)
\nProduct of the roots \u03b1\u03b2 = \\(\\frac{b}{1}\\)
\n\u03b1\u03b2 = b ——– (4)
\nGiven that the second equation has equal roots.
\n\u2234 The roots of the second equation are a, a
\nSum of the roots \u03b1 + \u03b1 = \\(-\\left(-\\frac{e}{1}\\right)\\)
\n2\u03b1 = e ——— (5)
\nProduct of the roots \u03b1.\u03b1 = \\(\\frac{f}{1}\\)
\n\u03b12<\/sup> = f ———- (6)
\nae = (\u03b1 + \u03b2)2\u03b1 (Multiplying equations (3) and (5))
\nae = 2\u03b12 + 2\u03b1\u03b2
\nae= 2 (f) + 2b From equations (4) and (6)
\nae= 2(f + b) Hence proved.<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nDiscuss the nature of roots of
\n(i) – x2<\/sup> + 3x + 1 = 0
\n(ii) 4x2<\/sup> – x – 2 = 0
\n(iii) 9x2<\/sup> + 5x = 0.
\nAnswer:
\n(i) -x2<\/sup> + 3x + 1 = 0
\n\u21d2 comparing with ax2<\/sup> + bx + c = 0
\n\u2206 = b2<\/sup> – 4ac = (3)2<\/sup> – 4(1)(-1) = 9 + 4 = 13 > 0
\n\u21d2 The roots are real and distinct<\/p>\n

(ii) 4x2<\/sup> – x – 2 = 0
\na = 4, b = -1, c = -2
\n\u2206 = b2<\/sup> – 4ac = (-1)2<\/sup> – 4(4)(-2) = 1 + 32 = 33 >0
\n\u21d2 The roots are real and distinct<\/p>\n

(iii) 9x2<\/sup> + 5x = 0
\na = 9, b = 5, c = 0
\n\u2206 = b2<\/sup> – 4ac = 52<\/sup> – 4(9)(0) = 25 > 0
\n\u21d2 The roots are real and distinct<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nWithout sketching the graphs, find whether the graphs of the following functions will intersect the x-axis and if so in how many points.
\n(i) y = x2<\/sup> + x + 2
\n(ii) y = x2<\/sup> – 3x – 7
\n(iii) y = x2<\/sup> + 6x + 9
\nAnswer:
\n(i) y = x2<\/sup> + x + 2
\ny = x2<\/sup> + x + 2 ——— (1)
\nCompare this equation with the equation
\nax2<\/sup> + bx + c = 0
\nwe have a = 1 , b = 1, c = 2
\nb2<\/sup> – 4ac = 12<\/sup> – 4 \u00d7 1 \u00d7 2 = 1 – 8
\nb2<\/sup> – 4ac = – 7 < 0
\nSince the discriminant is negative the quadratic equation has no real roots and therfore the graph does not meet x-axis.<\/p>\n

(ii) y = x2<\/sup> – 3x – 7
\ny = x2<\/sup> – 3x – 7 ——— (2)
\nCompare this equation with the equation ax2<\/sup> + bx + c = 0
\nwe have a = 1 , b = – 3 , c = – 1
\nb2<\/sup> – 4ac = (-3)2<\/sup> – 4(1)(-1)
\n= 9 + 4
\nb2<\/sup> – 4ac = 13 > 0
\nSince the discriminant is positive the quadratic equation has real and distinct roots and therefore the graph intersect the x – axis at two different points,<\/p>\n

(iii) y = x2<\/sup> + 6x + 9
\ny = x2<\/sup> + 6x + 9 ——— (3)
\nCompare this equation with the equation
\nax2<\/sup> + bx + c = 0
\nwe have a = 1 , b = 6, c = 9
\nb2<\/sup> – 4ac = 62<\/sup> – 4 \u00d7 1 \u00d7 9
\n= 36 – 36 =0
\nSince the discriminant is zero the quadratic equation has real and equal roots and therefore the graph touches the x-axis at one point.<\/p>\n

\"Samacheer<\/p>\n

Question 10.
\nWrite f(x) = x2<\/sup> + 5x + 4 in completed square form.
\nAnswer:
\nThe given quadratic equation is
\nf(x) = x2<\/sup> + 5x + 4
\nLet y = x2<\/sup> + 5x + 4
\ny – 4 = x2<\/sup> + 5x
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.4 Text Book Back Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 Question 1. Construct a quadratic equation with roots 7 and – 3. Answer: The given roots are 7 and -3 …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/21520"}],"collection":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/comments?post=21520"}],"version-history":[{"count":0,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/21520\/revisions"}],"wp:attachment":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/media?parent=21520"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/categories?post=21520"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/tags?post=21520"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}