{"id":67957,"date":"2023-09-13T18:18:00","date_gmt":"2023-09-13T12:48:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67957"},"modified":"2023-11-30T10:52:50","modified_gmt":"2023-11-30T05:22:50","slug":"rd-sharma-solutions-class-11-maths-chapter-14-exercise-14-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14-exercise-14-2\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 14 Exercise 14.2 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-122946\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-14-Exercise-14.2.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 14 Exercise 14.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-14-Exercise-14.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-14-Exercise-14.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 14 Exercise 14.2:\u00a0<\/strong>You can clear your basics and doubts with the <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths.<\/a> All the solutions designed by the subject matter experts are as per the current CBSE Syllabus. Go and download the Free PDF of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14-quadratic-equations\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 14 <\/a>Exercise 14.2.\u00a0<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69da5397dcd31\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" 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class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-14-exercise-142-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 14 Exercise 14.2 PDF:<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-14-Ex-14.2-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 14 Exercise 14.2<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-14-Ex-14.2-1.pdf\",\"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-for-class-11-maths-chapter-14-exercise-142\"><\/span>Access answers of RD Sharma Solutions for Class 11 Maths Chapter 14 Exercise 14.2<b><\/b><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>1. Solving the following quadratic equations by factorization method:<\/strong><\/p>\n<p><strong>(i) x<sup>2<\/sup>\u00a0+ 10ix \u2013 21 = 0<\/strong><\/p>\n<p><strong>(ii) x<sup>2<\/sup>\u00a0+ (1 \u2013 2i)x \u2013 2i = 0<\/strong><\/p>\n<p><strong>(iii) x<sup>2<\/sup>\u00a0\u2013 (2\u221a3 + 3i) x + 6\u221a3i = 0<\/strong><\/p>\n<p><strong>(iv) 6x<sup>2<\/sup>\u00a0\u2013 17ix \u2013 12 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>x<sup>2<\/sup>\u00a0+ 10ix \u2013 21 = 0<\/p>\n<p>Given: x<sup>2<\/sup>\u00a0+ 10ix \u2013 21 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 10ix \u2013 21 \u00d7 1 = 0<\/p>\n<p>We know, i<sup>2<\/sup>\u00a0= \u20131\u00a0\u21d2\u00a01 = \u2013i<sup>2<\/sup><\/p>\n<p>By substituting 1 = \u2013i<sup>2<\/sup>\u00a0in the above equation, we get<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 10ix \u2013 21(\u2013i<sup>2<\/sup>) = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 10ix + 21i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 3ix + 7ix + 21i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>x(x + 3i) + 7i(x + 3i) = 0<\/p>\n<p>(x + 3i) (x + 7i) = 0<\/p>\n<p>x + 3i = 0 or x + 7i = 0<\/p>\n<p>x = \u20133i or \u20137i<\/p>\n<p>\u2234 The roots of the given equation are \u20133i, \u20137i<\/p>\n<p><strong>(ii)\u00a0<\/strong>x<sup>2<\/sup>\u00a0+ (1 \u2013 2i)x \u2013 2i = 0<\/p>\n<p>Given: x<sup>2<\/sup>\u00a0+ (1 \u2013 2i)x \u2013 2i = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ x \u2013 2ix \u2013 2i = 0<\/p>\n<p>x(x + 1) \u2013 2i(x + 1) = 0<\/p>\n<p>(x + 1) (x \u2013 2i) = 0<\/p>\n<p>x + 1 = 0 or x \u2013 2i = 0<\/p>\n<p>x = \u20131 or 2i<\/p>\n<p>\u2234 The roots of the given equation are \u20131, 2i<\/p>\n<p><strong>(iii)\u00a0<\/strong>x<sup>2<\/sup>\u00a0\u2013 (2\u221a3 + 3i) x + 6\u221a3i = 0<\/p>\n<p>Given: x<sup>2<\/sup>\u00a0\u2013 (2\u221a3 + 3i) x + 6\u221a3i = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 (2\u221a3x + 3ix) + 6\u221a3i = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2\u221a3x \u2013 3ix + 6\u221a3i = 0<\/p>\n<p>x(x \u2013 2\u221a3) \u2013 3i(x \u2013 2\u221a3) = 0<\/p>\n<p>(x \u2013 2\u221a3) (x \u2013 3i) = 0<\/p>\n<p>(x \u2013 2\u221a3) = 0 or (x \u2013 3i) = 0<\/p>\n<p>x = 2\u221a3 or x = 3i<\/p>\n<p>\u2234 The roots of the given equation are 2\u221a3, 3i<\/p>\n<p><strong>(iv)\u00a0<\/strong>6x<sup>2<\/sup>\u00a0\u2013 17ix \u2013 12 = 0<\/p>\n<p>Given: 6x<sup>2<\/sup>\u00a0\u2013 17ix \u2013 12 = 0<\/p>\n<p>6x<sup>2<\/sup>\u00a0\u2013 17ix \u2013 12 \u00d7 1 = 0<\/p>\n<p>We know, i<sup>2<\/sup>\u00a0= \u20131\u00a0\u21d2\u00a01 = \u2013i<sup>2<\/sup><\/p>\n<p>By substituting 1 = \u2013i<sup>2<\/sup>\u00a0in the above equation, we get<\/p>\n<p>6x<sup>2<\/sup>\u00a0\u2013 17ix \u2013 12(\u2013i<sup>2<\/sup>) = 0<\/p>\n<p>6x<sup>2<\/sup>\u00a0\u2013 17ix + 12i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>6x<sup>2<\/sup>\u00a0\u2013 9ix \u2013 8ix + 12i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>3x(2x \u2013 3i) \u2013 4i(2x \u2013 3i) = 0<\/p>\n<p>(2x \u2013 3i) (3x \u2013 4i) = 0<\/p>\n<p>2x \u2013 3i = 0 or 3x \u2013 4i = 0<\/p>\n<p>2x = 3i or 3x = 4i<\/p>\n<p>x = 3i\/2 or x = 4i\/3<\/p>\n<p>\u2234 The roots of the given equation are 3i\/2, 4i\/3<\/p>\n<p><strong>2. Solve the following quadratic equations:<\/strong><\/p>\n<p><strong>(i) x<sup>2<\/sup>\u00a0\u2013 (3\u221a2 + 2i) x + 6\u221a2i = 0<\/strong><\/p>\n<p><strong>(ii) x<sup>2<\/sup>\u00a0\u2013 (5 \u2013 i) x + (18 + i) = 0<\/strong><\/p>\n<p><strong>(iii) (2 + i)x<sup>2<\/sup>\u00a0\u2013 (5- i)x + 2 (1 \u2013 i) = 0<\/strong><\/p>\n<p><strong>(iv) x<sup>2<\/sup>\u00a0\u2013 (2 + i)x \u2013 (1 \u2013 7i) = 0<\/strong><\/p>\n<p><strong>(v) ix<sup>2<\/sup>\u00a0\u2013 4x \u2013 4i = 0<\/strong><\/p>\n<p><strong>(vi) x<sup>2<\/sup>\u00a0+ 4ix \u2013 4 = 0<\/strong><\/p>\n<p><strong>(vii) 2x<sup>2<\/sup>\u00a0+ \u221a15ix \u2013 i = 0\u00a0<\/strong><\/p>\n<p><strong>(viii) x<sup>2<\/sup>\u00a0\u2013 x + (1 + i) = 0<\/strong><\/p>\n<p><strong>(ix) ix<sup>2<\/sup>\u00a0\u2013 x + 12i = 0<\/strong><\/p>\n<p><strong>(x) x<sup>2<\/sup>\u00a0\u2013 (3\u221a2 \u2013 2i)x \u2013 \u221a2i = 0<\/strong><\/p>\n<p><strong>(xi) x<sup>2<\/sup>\u00a0\u2013 (\u221a2 + i)x + \u221a2i = 0<\/strong><\/p>\n<p><strong>(xii) 2x<sup>2<\/sup>\u00a0\u2013 (3 + 7i)x + (9i \u2013 3) = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>x<sup>2<\/sup>\u00a0\u2013 (3\u221a2 + 2i) x + 6\u221a2i = 0<\/p>\n<p>Given: x<sup>2<\/sup>\u00a0\u2013 (3\u221a2 + 2i) x + 6\u221a2i = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 (3\u221a2x + 2ix) + 6\u221a2i = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 3\u221a2x \u2013 2ix + 6\u221a2i = 0<\/p>\n<p>x(x \u2013 3\u221a2) \u2013 2i(x \u2013 3\u221a2) = 0<\/p>\n<p>(x \u2013 3\u221a2) (x \u2013 2i) = 0<\/p>\n<p>(x \u2013 3\u221a2) = 0 or (x \u2013 2i) = 0<\/p>\n<p>x = 3\u221a2 or x = 2i<\/p>\n<p>\u2234 The roots of the given equation are 3\u221a2, 2i<\/p>\n<p><strong>(ii)\u00a0<\/strong>x<sup>2<\/sup>\u00a0\u2013 (5 \u2013 i) x + (18 + i) = 0<\/p>\n<p>Given: x<sup>2<\/sup>\u00a0\u2013 (5 \u2013 i) x + (18 + i) = 0<\/p>\n<p>We shall apply the discriminant rule,<\/p>\n<p>Where, x = (-b\u00a0<strong>\u00b1\u221a<\/strong>(b<sup>2<\/sup>\u00a0\u2013 4ac))\/2a<\/p>\n<p>Here, a = 1, b = -(5-i), c = (18+i)<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-1.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 1\" \/><\/p>\n<p>We can write 48 + 14i = 49 \u2013 1 + 14i<\/p>\n<p>So,<\/p>\n<p>48 + 14i = 49 + i<sup>2<\/sup>\u00a0+ 14i [\u2235\u00a0i<sup>2<\/sup>\u00a0= \u20131]<\/p>\n<p>= 7<sup>2<\/sup>\u00a0+ i<sup>2<\/sup>\u00a0+ 2(7)(i)<\/p>\n<p>= (7 + i)<sup>2<\/sup>\u00a0[Since,\u00a0(a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ 2ab]<\/p>\n<p>By using the result 48 + 14i = (7 + i)<sup>\u00a02<\/sup>, we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-2.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 2\" \/><\/p>\n<p>x = 2 + 3i or 3 \u2013 4i<\/p>\n<p>\u2234 The roots of the given equation are 3 \u2013 4i, 2 + 3i<\/p>\n<p><strong>(iii)<\/strong>\u00a0(2 + i)x<sup>2<\/sup>\u00a0\u2013 (5- i)x + 2 (1 \u2013 i) = 0<\/p>\n<p>Given: (2 + i)x<sup>2<\/sup>\u00a0\u2013 (5- i)x + 2 (1 \u2013 i) = 0<\/p>\n<p>We shall apply the discriminant rule,<\/p>\n<p>Where, x = (-b\u00a0<strong>\u00b1\u221a<\/strong>(b<sup>2<\/sup>\u00a0\u2013 4ac))\/2a<\/p>\n<p>Here, a = (2+i), b = -(5-i), c = 2(1-i)<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-3.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 3\" \/><\/p>\n<p>We have i<sup>2<\/sup>\u00a0= \u20131<\/p>\n<p>By substituting \u20131 = i<sup>2<\/sup>\u00a0in the above equation, we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-4.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 4\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-5.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 5\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-6.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 6\" \/><\/p>\n<p>x = (1 \u2013 i) or 4\/5 \u2013 2i\/5<\/p>\n<p>\u2234 The roots of the given equation are (1 \u2013 i), 4\/5 \u2013 2i\/5<\/p>\n<p><strong>(iv)\u00a0<\/strong>x<sup>2<\/sup>\u00a0\u2013 (2 + i)x \u2013 (1 \u2013 7i) = 0<\/p>\n<p>Given: x<sup>2<\/sup>\u00a0\u2013 (2 + i)x \u2013 (1 \u2013 7i) = 0<\/p>\n<p>We shall apply the discriminant rule,<\/p>\n<p>Where, x = (-b\u00a0<strong>\u00b1\u221a<\/strong>(b<sup>2<\/sup>\u00a0\u2013 4ac))\/2a<\/p>\n<p>Here, a = 1, b = -(2+i), c = -(1-7i)<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-7.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 7\" \/><\/p>\n<p>We can write 7 \u2013 24i = 16 \u2013 9 \u2013 24i<\/p>\n<p>7 \u2013 24i = 16 + 9(\u20131) \u2013 24i<\/p>\n<p>= 16 + 9i<sup>2<\/sup>\u00a0\u2013 24i [\u2235\u00a0i<sup>2<\/sup>\u00a0= \u20131]<\/p>\n<p>= 4<sup>2<\/sup>\u00a0+ (3i)<sup>2<\/sup>\u00a0\u2013 2(4) (3i)<\/p>\n<p>= (4 \u2013 3i)<sup>2<\/sup>\u00a0[\u2235\u00a0(a \u2013 b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0+ 2ab]<\/p>\n<p>By using the result 7 \u2013 24i = (4 \u2013 3i)<sup>2<\/sup>, we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-8.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 8\" \/><\/p>\n<p>x = 3 \u2013 i or -1 + 2i<\/p>\n<p>\u2234 The roots of the given equation are (-1 + 2i), (3 \u2013 i)<\/p>\n<p><strong>(v)<\/strong>\u00a0ix<sup>2<\/sup>\u00a0\u2013 4x \u2013 4i = 0<\/p>\n<p>Given: ix<sup>2<\/sup>\u00a0\u2013 4x \u2013 4i = 0<\/p>\n<p>ix<sup>2<\/sup>\u00a0+ 4x(\u20131) \u2013 4i = 0 [We know, i<sup>2<\/sup>\u00a0= \u20131]<\/p>\n<p>So by substituting \u20131 = i<sup>2<\/sup>\u00a0in the above equation, we get<\/p>\n<p>ix<sup>2<\/sup>\u00a0+ 4xi<sup>2<\/sup>\u00a0\u2013 4i = 0<\/p>\n<p>i(x<sup>2<\/sup>\u00a0+ 4ix \u2013 4) = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 4ix \u2013 4 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 4ix + 4(\u20131) = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 4ix + 4i<sup>2<\/sup>\u00a0= 0\u00a0[Since,\u00a0i<sup>2<\/sup>\u00a0= \u20131]<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 2ix + 2ix + 4i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>x(x + 2i) + 2i(x + 2i) = 0<\/p>\n<p>(x + 2i) (x + 2i) = 0<\/p>\n<p>(x + 2i)<sup>2<\/sup>\u00a0= 0<\/p>\n<p>x + 2i = 0<\/p>\n<p>x = \u20132i, -2i<\/p>\n<p>\u2234 The roots of the given equation are \u20132i, \u20132i<\/p>\n<p><strong>(vi)\u00a0<\/strong>x<sup>2<\/sup>\u00a0+ 4ix \u2013 4 = 0<\/p>\n<p>Given: x<sup>2<\/sup>\u00a0+ 4ix \u2013 4 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 4ix + 4(\u20131) = 0 [We know, i<sup>2<\/sup>\u00a0= \u20131]<\/p>\n<p>So by substituting \u20131 = i<sup>2<\/sup>\u00a0in the above equation, we get<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 4ix + 4i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 2ix + 2ix + 4i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>x(x + 2i) + 2i(x + 2i) = 0<\/p>\n<p>(x + 2i) (x + 2i) = 0<\/p>\n<p>(x + 2i)<sup>2<\/sup>\u00a0= 0<\/p>\n<p>x + 2i = 0<\/p>\n<p>x = \u20132i, -2i<\/p>\n<p>\u2234 The roots of the given equation are \u20132i, \u20132i<\/p>\n<p><strong>(vii)\u00a0<\/strong>2x<sup>2<\/sup>\u00a0+ \u221a15ix \u2013 i = 0<strong>\u00a0<\/strong><\/p>\n<p>Given: 2x<sup>2<\/sup>\u00a0+ \u221a15ix \u2013 i = 0<strong>\u00a0<\/strong><\/p>\n<p>We shall apply the discriminant rule,<\/p>\n<p>Where, x = (-b\u00a0<strong>\u00b1\u221a<\/strong>(b<sup>2<\/sup>\u00a0\u2013 4ac))\/2a<\/p>\n<p>Here, a = 2, b = \u221a15i, c = -i<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-9.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 9\" \/><\/p>\n<p>We can write 15 \u2013 8i = 16 \u2013 1 \u2013 8i<\/p>\n<p>15 \u2013 8i = 16 + (\u20131) \u2013 8i<\/p>\n<p>= 16 + i<sup>2<\/sup>\u00a0\u2013 8i [\u2235\u00a0i<sup>2<\/sup>\u00a0= \u20131]<\/p>\n<p>= 4<sup>2<\/sup>\u00a0+ (i)<sup>2<\/sup>\u00a0\u2013 2(4)(i)<\/p>\n<p>= (4 \u2013 i)<sup>2<\/sup>\u00a0[Since,\u00a0(a \u2013 b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0+ 2ab]<\/p>\n<p>By using the result 15 \u2013 8i = (4 \u2013 i)<sup>2<\/sup>, we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-10.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 10\" \/><\/p>\n<p>\u2234 The roots of the given equation are [1+ (4 \u2013 \u221a15)i\/4] , [-1 -(4 + \u221a15)i\/4]<\/p>\n<p><strong>(viii)<\/strong>\u00a0x<sup>2<\/sup>\u00a0\u2013 x + (1 + i) = 0<\/p>\n<p>Given: x<sup>2<\/sup>\u00a0\u2013 x + (1 + i) = 0<\/p>\n<p>We shall apply the discriminant rule,<\/p>\n<p>Where, x = (-b\u00a0<strong>\u00b1\u221a<\/strong>(b<sup>2<\/sup>\u00a0\u2013 4ac))\/2a<\/p>\n<p>Here, a = 1, b = -1, c = (1+i)<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-11.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 11\" \/><br \/><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-12.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 12\" \/><\/p>\n<p>We can write 3 + 4i = 4 \u2013 1 + 4i<\/p>\n<p>3 + 4i = 4 + i<sup>2<\/sup>\u00a0+ 4i [\u2235\u00a0i<sup>2<\/sup>\u00a0= \u20131]<\/p>\n<p>= 2<sup>2<\/sup>\u00a0+ i<sup>2<\/sup>\u00a0+ 2(2) (i)<\/p>\n<p>= (2 + i)<sup>2<\/sup>\u00a0[Since,\u00a0(a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ 2ab]<\/p>\n<p>By using the result 3 + 4i = (2 + i)<sup>2<\/sup>, we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/09\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-13.jpg\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 13\" \/><\/p>\n<p>x = 2i\/2 or (2 \u2013 2i)\/2<\/p>\n<p>x = i or 2(1-i)\/2<\/p>\n<p>x = i or (1 \u2013 i)<\/p>\n<p>\u2234 The roots of the given equation are (1-i), i<\/p>\n<p><strong>(ix)\u00a0<\/strong>ix<sup>2<\/sup>\u00a0\u2013 x + 12i = 0<\/p>\n<p>Given: ix<sup>2<\/sup>\u00a0\u2013 x + 12i = 0<\/p>\n<p>ix<sup>2<\/sup>\u00a0+ x(\u20131) + 12i = 0 [We know, i<sup>2<\/sup>\u00a0= \u20131]<\/p>\n<p>so by substituting \u20131 = i<sup>2<\/sup>\u00a0in the above equation, we get<\/p>\n<p>ix<sup>2<\/sup>\u00a0+ xi<sup>2<\/sup>\u00a0+ 12i = 0<\/p>\n<p>i(x<sup>2<\/sup>\u00a0+ ix + 12) = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ ix + 12 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ ix \u2013 12(\u20131) = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ ix \u2013 12i<sup>2<\/sup>\u00a0= 0\u00a0[Since,\u00a0i<sup>2<\/sup>\u00a0= \u20131]<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 3ix + 4ix \u2013 12i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>x(x \u2013 3i) + 4i(x \u2013 3i) = 0<\/p>\n<p>(x \u2013 3i) (x + 4i) = 0<\/p>\n<p>x \u2013 3i = 0 or x + 4i = 0<\/p>\n<p>x = 3i or \u20134i<\/p>\n<p>\u2234 The roots of the given equation are -4i, 3i<\/p>\n<p><strong>(x)<\/strong>\u00a0x<sup>2<\/sup>\u00a0\u2013 (3<strong>\u221a<\/strong>2 \u2013 2i)x \u2013\u00a0<strong>\u221a<\/strong>2i = 0<\/p>\n<p>Given: x<sup>2<\/sup>\u00a0\u2013 (3<strong>\u221a<\/strong>2 \u2013 2i)x \u2013\u00a0<strong>\u221a<\/strong>2i = 0<\/p>\n<p>We shall apply the discriminant rule,<\/p>\n<p>Where, x = (-b\u00a0<strong>\u00b1\u221a<\/strong>(b<sup>2<\/sup>\u00a0\u2013 4ac))\/2a<\/p>\n<p>Here, a = 1, b = -(3<strong>\u221a<\/strong>2 \u2013 2i), c = \u2013<strong>\u221a<\/strong>2i<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-14.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 14\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-15.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 15\" \/><\/p>\n<p><strong>(xi)<\/strong>\u00a0x<sup>2<\/sup>\u00a0\u2013 (<strong>\u221a<\/strong>2 + i)x +\u00a0<strong>\u221a<\/strong>2i = 0<\/p>\n<p>Given: x<sup>2<\/sup>\u00a0\u2013 (<strong>\u221a<\/strong>2 + i)x +\u00a0<strong>\u221a<\/strong>2i = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 (<strong>\u221a<\/strong>2x + ix) +\u00a0<strong>\u221a<\/strong>2i = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013\u00a0<strong>\u221a<\/strong>2x \u2013 ix +\u00a0<strong>\u221a<\/strong>2i = 0<\/p>\n<p>x(x \u2013\u00a0<strong>\u221a<\/strong>2) \u2013 i(x \u2013\u00a0<strong>\u221a<\/strong>2) = 0<\/p>\n<p>(x \u2013\u00a0<strong>\u221a<\/strong>2) (x \u2013 i) = 0<\/p>\n<p>(x \u2013\u00a0<strong>\u221a<\/strong>2) = 0 or (x \u2013 i) = 0<\/p>\n<p>x =\u00a0<strong>\u221a<\/strong>2 or x = i<\/p>\n<p>\u2234 The roots of the given equation are i,\u00a0<strong>\u221a<\/strong>2<\/p>\n<p><strong>(xii)\u00a0<\/strong>2x<sup>2<\/sup>\u00a0\u2013 (3 + 7i)x + (9i \u2013 3) = 0<\/p>\n<p>Given: 2x<sup>2<\/sup>\u00a0\u2013 (3 + 7i)x + (9i \u2013 3) = 0<\/p>\n<p>We shall apply the discriminant rule,<\/p>\n<p>Where, x = (-b\u00a0<strong>\u00b1\u221a<\/strong>(b<sup>2<\/sup>\u00a0\u2013 4ac))\/2a<\/p>\n<p>Here, a = 2, b = -(3\u00a0<strong>+ 7<\/strong>i), c = (9i \u2013 3)<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-16.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 16\" \/><\/p>\n<p>We can write 16 + 30i = 25 \u2013 9 + 30i<\/p>\n<p>16 + 30i = 25 + 9(\u20131) + 30i<\/p>\n<p>= 25 + 9i<sup>2<\/sup>\u00a0+ 30i [\u2235\u00a0i<sup>2<\/sup>\u00a0= \u20131]<\/p>\n<p>= 5<sup>2<\/sup>\u00a0+ (3i)<sup>2<\/sup>\u00a0+ 2(5)(3i)<\/p>\n<p>= (5 + 3i)<sup>2<\/sup>\u00a0[\u2235\u00a0(a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ 2ab]<\/p>\n<p>By using the result 16 + 30i = (5 + 3i)<sup>2<\/sup>, we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-17.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 17\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations-image-18.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 14 \u2013 Quadratic Equations image - 18\" \/><\/p>\n<p>This is the complete blog of RD Sharma Solutions for Class 11 Maths Chapter 14 Exercise 14.2. To Know more about the <a href=\"http:\/\/cbse.nic.in\" target=\"_blank\" rel=\"noopener\">CBSE<\/a>\u00a0Class 11 Maths exam, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-of-rd-sharma-solutions-for-class-11-maths-chapter-14-exercise-142\"><\/span>FAQs of\u00a0 RD Sharma Solutions for Class 11 Maths Chapter 14 Exercise 14.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630500381842\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-for-class-11-maths-chapter-14-exercise-142\"><\/span>How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 14 Exercise 14.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 2 questions in\u00a0RD Sharma Solutions Class 11 Maths Chapter 14 Exercise 14.2.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630500440343\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-14-ex-142\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 14 Ex 14.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630500456288\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-11-maths-chapter-14-exercise-142\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 14 Exercise 14.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 14 Exercise 14.2:\u00a0You can clear your basics and doubts with the RD Sharma Solutions Class 11 Maths. All the solutions designed by the subject matter experts are as per the current CBSE Syllabus. Go and download the Free PDF of RD Sharma Solutions Class 11 Maths Chapter 14 &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 14 Exercise 14.2 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14-exercise-14-2\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 14 Exercise 14.2 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":122946,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67957"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=67957"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67957\/revisions"}],"predecessor-version":[{"id":514514,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67957\/revisions\/514514"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/122946"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67957"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67957"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67957"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}