{"id":168032,"date":"2024-03-01T17:50:57","date_gmt":"2024-03-01T12:20:57","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=168032"},"modified":"2024-03-01T17:50:57","modified_gmt":"2024-03-01T12:20:57","slug":"op-malhotra-class-11-maths-solutions-chapter-10-ex-10b","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/op-malhotra-class-11-maths-solutions-chapter-10-ex-10b\/","title":{"rendered":"OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Ex 10(b)"},"content":{"rendered":"

The availability of step-by-step OP Malhotra Maths Class 11 Solutions<\/a> Chapter 10 Quadratic Equations Ex 10(b) can make challenging problems more manageable.<\/p>\n

S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(b)<\/h2>\n

Solve the following equations :<\/p>\n

Question 1.
\nx4<\/sup> – 5x\u00b2 + 6 = 0.
\nSolution:
\nGiven eqn. be, x4<\/sup> – 5x\u00b2 + 6 = 0 … (1)
\nput x\u00b2 = t in eqn. (1) ; we have
\nt\u00b2 – 5t + 6 = 0
\n\u21d2 (t – 2) (t – 3) = 0
\n\u21d2 t – 2 = 0 or t – 3 = 0
\n\u21d2 x\u00b2 = 2 or x\u00b2 = 3
\n\u21d2 x = \u00b1 \\(\\sqrt{2}\\) or x = \u00b1 \\(\\sqrt{3}\\)
\n\u2234 x = \u00b1 \\(\\sqrt{2}\\), \u00b1 \\(\\sqrt{3}\\)<\/p>\n

Question 2.
\nx5<\/sup> + 242 = \\(\\frac{243}{x^5}\\).
\nSolution:
\nGiven eqn. be,
\nx5<\/sup> + 242 = \\(\\frac{243}{x^5}\\) … (1)
\nputting x5<\/sup> = y in eqn. (1) ; we have
\ny + 242 = \\(\\frac { 243 }{ y }\\) \u21d2 y\u00b2 + 242y – 243 = 0
\n\u21d2 y\u00b2 – y + 243y – 243 = 0
\n\u21d2 y(y – 1) + 243 (y – 1) = 0
\n\u21d2 (y – 1) (y + 243) = 0
\n\u21d2 y = 1 = 0 or y + 243 = 0
\n\u21d2 y = 1 or y = – 243
\n\u21d2 x5<\/sup> = 1 or x5<\/sup> = (- 3)5<\/sup>
\n\u21d2 x = 1 or x = – 3
\nThus, s = – 1, – 3<\/p>\n

Question 3.
\n10x-2<\/sup> – 9 – x-4<\/sup> = 0.
\nSolution:
\nGiven eqn. be, 10x-2<\/sup> – 9 – x-4<\/sup> = 0 … (1)
\nputting x-2<\/sup> = t in eqn. (1); we have
\n10t – 9 – t\u00b2 = 0
\n\u21d2 t\u00b2 – 10t + 9 = 0
\n\u21d2 t\u00b2 – t – 9t + 9 = 0
\n\u21d2 t (t – 1) – 9 (t – 1) = 0
\n\u21d2 (t – 9)(t – 1) = 0
\neither t = 9 = 0 or t – 1 = 0
\n\u21d2 x-2<\/sup> = 9 or x-2<\/sup> – 1 = 0
\n\u21d2 \\(\\frac{1}{x^2}\\) = 9 or \\(\\frac{1}{x^2}\\) = 1
\n\u21d2 x = \u00b1 \\(\\frac { 1 }{ 3 }\\) or x = \u00b1 1
\nThus, x = \u00b1 \\(\\frac { 1 }{ 3 }\\), \u00b1 1<\/p>\n

\"OP<\/p>\n

Question 4.
\n32x<\/sup> – 10 x 3x<\/sup> + 9 = 0.
\nSolution:
\nGiven eqn. be,
\n32x<\/sup> – 10 x 3x<\/sup> + 9 = 0 …(1)
\nputting 3x<\/sup> = t in eqn. (1) ; we have
\nt2<\/sup> – 10t + 9 = 0
\n\u21d2 t\u00b2 – 9t – t + 9 = 0
\n\u21d2 t(t – 9) – 1 (t – 9) = 0
\n\u21d2 (t – 9) (t – 1) = 0
\neither t – 9 = 0 or t – 1 = 0
\n\u21d2 t = 9 or t = 1
\n\u21d2 3x<\/sup> = 3x<\/sup> or 3x<\/sup> = 3\u00b0
\n\u21d2 x = 2 or x = 0
\nThus, x = 0, 2<\/p>\n

Question 5.
\n22x-1<\/sup> – 9 x 2x-2<\/sup> + 1 = 0.
\nSolution:
\nGiven equation be,
\n22x-1<\/sup> – 9 x 2x-2<\/sup> + 1 = 0
\n\u21d2 (2x<\/sup>)\u00b2 x 2-1<\/sup> – 9 x 2x<\/sup> x 2-2<\/sup> + 1 = 0 …(1)
\nputting 2x<\/sup> = t in eqn. (1); we have
\nt\u00b2 x \\(\\frac { 1 }{ 2 }\\) – 9 x t x \\(\\frac { 1 }{ 4 }\\) + 1 = 0
\n\u21d2 2t\u00b2 – 9t + 4 = 0
\n\u21d2 2t\u00b2 – 8t – t + 4 = 0
\n\u21d2 2t (t – 4) – 1 (t – 4) = 0
\n\u21d2 (2t – 1) (t – 4) = 0
\neither 2t – 1 = 0 or t – 4 = 0
\n\u21d2 t = \\(\\frac { 1 }{ 2 }\\) or t = 4
\n\u21d2 2x<\/sup> = \\(\\frac { 1 }{ 2 }\\) = 2-1<\/sup> or 2x<\/sup> = 2\u00b2
\n\u21d2 x = – 1 or x = 2
\n\u2234 x = – 1 or 2<\/p>\n

Question 6.
\n32x+1<\/sup> + 3\u00b2 = 3x+3<\/sup> + 3x<\/sup>.
\nSolution:
\nGiven eqn. be,
\n32x+1<\/sup> + 3\u00b2 = 3x+3<\/sup> + 3x<\/sup>
\n\u21d2 (3x<\/sup>)\u00b2 x 3 + 9 = 3x<\/sup> (3\u00b3 + 1) …(1)
\nputting 3x<\/sup> = t in eqn. (1) ; we have
\n3t\u00b2 + 9 = 28t \u21d2 3t\u00b2 – 28t + 9 = 0
\n\u21d2 3t\u00b2 – 27t – t + 9 = 0
\n\u21d2 3t(t – 9) – 1 (t – 9) = 0
\n\u21d2 (t – 9) (3t – 1) = 0
\neither t – 9 = 0 or 3t – 1 = 0
\n\u21d2 t = 9 or t = \\(\\frac { 1 }{ 3 }\\)
\n\u21d2 3x<\/sup> = 3\u00b2 or 3x<\/sup> = 3-1<\/sup>
\n\u21d2 x = 2 or x = – 1
\nThus, x = 2, – 1<\/p>\n

\"OP<\/p>\n

Question 7.
\n\\(\\sqrt{x^2-3 x}=4 x^2-12 x-3\\)
\nSolution:
\nGiven equation be,
\n\\(\\sqrt{x^2-3 x}=4 x^2-12 x-3\\) – 3 …(1)
\nputting x\u00b2 – 3x = t in eqn. (1); we have
\n\\(\\sqrt{t}\\) = 4t – 3 ;
\non squaring both sides ; we have
\n\u21d2 t = (4t – 3)\u00b2 \u21d2 t = 16t\u00b2 – 24t + 9 \u21d2 16t\u00b2 – 25t + 9 = 0
\n\"OP<\/p>\n

Question 8.
\n\\(\\sqrt{\\frac{x^2+2}{x^2-2}}+6 \\sqrt{\\frac{x^2-2}{x^2+2}}\\) = 5.
\nSolution:
\nGiven equation be
\n\\(\\sqrt{\\frac{x^2+2}{x^2-2}}+6 \\sqrt{\\frac{x^2-2}{x^2+2}}\\) = 5
\nputting \\(\\sqrt{\\frac{x^2+2}{x^2-2}}\\) = y in eqn. (1) ; we have
\ny + \\(\\frac { 6 }{ y }\\) = 5
\n\u21d2 y\u00b2 – 5y + 6 = 0
\n\u21d2 (y – 2)(y – 3) = 0
\n\u21d2 y – 2 = 0 or y – 3 = 0
\n\u21d2 y = 2 or y = 3
\nCase-I: When y = 2 \u21d2 \\(\\sqrt{\\frac{x^2+2}{x^2-2}}\\) = 2
\nOn squaring both sides ; we have
\n\\(\\frac{x^2+2}{x^2-2}\\) = 4
\n\u21d2 x\u00b2 + 2 = 4x\u00b2 – 8
\n\u21d2 3x\u00b2 = 10
\n\u21d2 x\u00b2 = \\(\\frac { 10 }{ 3 }\\)
\n\u21d2 x = \u00b1 \\(\\sqrt{\\frac{10}{3}}\\)<\/p>\n

Case-II: When y = 3 \u21d2 \\(\\sqrt{\\frac{x^2+2}{x^2-2}}\\) = 3
\nOn squaring both sides ; we have
\n\\(\\frac{x^2+2}{x^2-2}\\) = 9 \u21d2 x\u00b2 + 2 = 9x\u00b2 – 18
\n\u21d2 8x\u00b2 = 20
\n\u21d2 x\u00b2 = \\(\\frac { 5 }{ 2 }\\)
\n\u21d2 x = \u00b1 \\(\\sqrt{\\frac{5}{2}}\\)
\nHence x = \u00b1 \\(\\sqrt{\\frac{10}{3}}, \\pm \\sqrt{\\frac{5}{2}}\\)<\/p>\n

Question 9.
\n\\(\\sqrt{\\frac{2 x^2+1}{x^2-1}}+6 \\sqrt{\\frac{x^2-1}{2 x^2+1}}\\) = 5.
\nSolution:
\nGiven equation be
\n\\(\\sqrt{\\frac{2 x^2+1}{x^2-1}}+6 \\sqrt{\\frac{x^2-1}{2 x^2+1}}\\) = 5 … (1)
\nputting \\(\\sqrt{\\frac{2 x^2+1}{x^2-1}}\\) = t in eqn. (1); we have
\nt + \\(\\frac { 6 }{ t }\\) = 5
\n\u21d2 t\u00b2 – 5t + 6 = 0
\n\u21d2 (t – 2)(t – 3) = 0
\neither t – 2 = 0 or t – 3 = 0
\n\u21d2 t = 2 or t = 3
\nCase-I: When t = 2
\n\u21d2 \\(\\sqrt{\\frac{2 x^2+1}{x^2-1}}\\) = 2
\nOn squaring both sides ; we have 2*2 + 1
\n\\(\\frac{2 x^2+1}{x^2-1}\\) = 4
\n\u21d2 2x\u00b2 + 1 = 4x\u00b2 – 4
\n\u21d2 2x\u00b2 = 5
\n\u21d2 x = \\(\\pm \\sqrt{\\frac{5}{2}}\\)<\/p>\n

Case-II: When t = 3
\n\"OP<\/p>\n

Question 10.
\nx (x – 1) (x + 2) (x- 3) + 8 = 0.
\nSolution:
\nGiven equation be
\nx (x – 1) (x + 2) (x – 3) + 8 = 0
\n\u21d2 (x\u00b2 – x) [x\u00b2 – x – 6] + 8 = 0 … (1)
\nputting x\u00b2 – x = t in eqn. (1) ; we have
\nt(t – 6) + 8 = 0 \u21d2 t\u00b2 – 6t + 8 = 0
\n\u21d2 t\u00b2 – 4t – 2t + 8 = 0
\n\u21d2 t(t – 4) – 2(t – 4) = 0
\n\u21d2 (t – 4)(t – 2) = 0
\neither t – 4 = 0 or t – 2 = 0
\n\u21d2 t = 4 or t = 2
\nCase-I: When t – 4
\n\u21d2 x\u00b2 – x – 4 = 0
\n\u21d2 x = \\(\\frac{-b \\pm \\sqrt{b^2-4 a c}}{2 a}\\)
\n\u21d2 x = \\(\\frac{-(-1) \\pm \\sqrt{(-1)^2-4 \\times 1 \\times(-4)}}{2}\\)
\n\u21d2 x = \\(\\frac{1 \\pm \\sqrt{17}}{2}\\)<\/p>\n

\"OP<\/p>\n

Case-II: When t = 2 \u21d2 x\u00b2 – x – 2 = 0
\n\u21d2 x\u00b2 – 2x + x – 2 = 0
\n\u21d2 x (x – 2) + 1 (x – 2) = 0
\n\u21d2 (x + 1) (x – 2) = 0
\n\u21d2 x = – 1, 2
\nHence x = – 1, 2, \\(\\frac{1 \\pm \\sqrt{17}}{2}\\)<\/p>\n

Question 11.
\n(x – 7) (x – 3) (x + 1) (x + 5) = 1680.
\nSolution:
\nGiven eqn. be
\n(x – 7) (x – 3) (x + 1)(x + 5) = 1680
\n\u21d2 {(x – 7) (x + 5)} {(x-3) (x+ 1)} = 1680
\n\u21d2 {x\u00b2 – 2x – 35} {x\u00b2 – 2x – 3} = 1680 … (1)
\nputting x\u00b2 – 2x = t in eqn. (1); we get
\n(t – 35) (t – 3) = 1680
\n\u21d2 t\u00b2 – 38t + 105 – 1680 = 0
\n\u21d2 t\u00b2 – 38t – 1575 = 0
\n\u2234 t = \\(\\frac{38 \\pm \\sqrt{1444+6300}}{2}=\\frac{38 \\pm 88}{2}\\)
\n\u21d2 t = 63, – 25<\/p>\n

Case-I: When t = 63 \u21d2 x\u00b2 – 2x – 63 = 0
\n\"OP<\/p>\n

Question 12.
\n(2x – 7) (x\u00b2 – 9) (2x + 5) = 91.
\nSolution:
\nGiven equation be,
\n(2x – 7) (x\u00b2 – 9) (2x + 5) = 91
\n\u21d2 {(2x – 7) (x + 3)} {(x – 3)(2x + 5)} = 91
\n\u21d2 (2x\u00b2 – x – 21)(2x\u00b2 – x – 15) = 91 …(1)
\nputting 2x\u00b2 – x = t in eqn. (1) ; we have
\n(t – 21) (t – 15) = 91
\n\u21d2 t\u00b2 – 36t + 224 = 0
\n\u21d2 t\u00b2 – 8t – 28t + 224 = 0
\n\u21d2 t (t – 8) – 28 (t – 8) = 0
\n\u21d2 (t – 8) (t – 28) = 0
\neither t – 8 = 0 or t – 28 = 0
\n\u21d2 t = 8 or t = 28
\nCase-I: When t = 8 \u21d2 2x\u00b2 – x – 8 = 0
\n\u21d2 x = \\(\\frac{-(-1) \\pm \\sqrt{(-1)^2-4 \\times 2 \\times(-8)}}{2 \\times 2}\\)
\n\u21d2 x = \\(\\frac{1 \\pm \\sqrt{65}}{4}\\)<\/p>\n

Case-II: When t = 28
\n\u21d2 2x\u00b2 – x – 28 = 0
\n\u21d2 2x\u00b2 – 8x + 7x – 28 = 0
\n\u21d2 2x (x – 4) + 7 (x – 4) = 0
\n\u21d2 (x – 4) (2x + 7) = 0
\neither x – 4 = 0 or 2x + 7 = 0
\n\u21d2 x = 4 or x = – \\(\\frac { 7 }{ 2 }\\)
\nHence, x = + 4, – \\(\\frac { 7 }{ 2 }\\), \\(\\frac{1 \\pm \\sqrt{65}}{4}\\)<\/p>\n

Question 13.
\nBy substituting y = 2x<\/sup>, or otherwise, solve the equation
\n22x<\/sup> + 2x+2<\/sup> – 4 x 2\u00b3 = 0.
\nSolution:
\nGiven eqn. be,
\n22x<\/sup> + 2x+2<\/sup> – 4 x 2\u00b3 = 0 …(1)
\nputting 22x<\/sup> = y in eqn. (1) ; we have
\ny\u00b2 + y . 2\u00b2 – 32 = 0
\n\u21d2 y\u00b2 + 4y – 32 = 0
\n\u21d2 y\u00b2 + 8y – 4y – 32 = 0
\n\u21d2 y(y + 8) – 4(y + 8) = 0
\n\u21d2 (y + 8)(y – 4) = 0
\neither y + 8 = 0 or y – 4 = 0
\n\u21d2 y = – 8 or y = 4
\n\u21d2 2x<\/sup> = – 8
\nit has no real solution (\u2235 2x<\/sup> > 0)
\nor 2x<\/sup> = 4 = 22<\/sup> \u2234 x = 2
\nThus x = 2<\/p>\n

Question 14.
\n2x\u00b2<\/sup> : 2x<\/sup> = 8 : 1.
\nSolution:
\n\"OP<\/p>\n

Question 15.
\n22x+3<\/sup> + 2x+3<\/sup> = 1 + 2x<\/sup>
\nSolution:
\nGiven eqn. be 22x+3<\/sup> + 2x+3<\/sup> = 1 + 2x<\/sup>
\n\u21d2 (2x<\/sup>)\u00b2 x 2\u00b3 + 2x<\/sup> x 2\u00b3 = 1 + 2x<\/sup> …(1)
\nputting 2x<\/sup> = t in eqn. (1) ;
\nwe have 8t\u00b2 + 8t = 1 + t
\n\u21d2 8t\u00b2 + 7t – 1 = 0
\n\u21d2 8t\u00b2 + 8t – t – 1 = 0
\n\u21d2 8t (t + 1) – 1 (t + 1) = 0
\n\u21d2 (t + 1) (8t – 1) = 0
\neither t + 1 = 0 or 8t – 1 = 0
\n\u21d2 t = – 1 or t = \\(\\frac { 1 }{ 8 }\\)
\n\u21d2 2x<\/sup> = – 1, which is not possible as 2x<\/sup> > 0
\nwhen t = \\(\\frac { 1 }{ 8 }\\) \u21d2 2x<\/sup> = 2-3<\/sup> \u21d2 x = – 3<\/p>\n

Question 16.
\n\\(4^x-3^{x-\\frac{1}{2}}=3^{x+\\frac{1}{2}}-2^{2 x-1}\\)
\nSolution:
\nGiven equation be
\n\"OP<\/p>\n","protected":false},"excerpt":{"rendered":"

The availability of step-by-step OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Ex 10(b) can make challenging problems more manageable. S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(b) Solve the following equations : Question 1. x4 – 5x\u00b2 + 6 = 0. Solution: Given eqn. be, x4 – …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[115871],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/168032"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=168032"}],"version-history":[{"count":3,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/168032\/revisions"}],"predecessor-version":[{"id":168044,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/168032\/revisions\/168044"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=168032"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=168032"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=168032"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}