{"id":62660,"date":"2023-09-08T20:39:00","date_gmt":"2023-09-08T15:09:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=62660"},"modified":"2023-11-29T10:44:58","modified_gmt":"2023-11-29T05:14:58","slug":"rd-sharma-solutions-class-11-maths-chapter-14","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/","title":{"rendered":"RD Sharma Solutions Class 11 Maths Chapter 14 &#8211; Quadratic Equations (Updated For 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-119449\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-14.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 14 \" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-14.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-14-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 14<\/strong> &#8211; Quadratic Equations: The topic quadratic equation might seem familiar right? Here, <span style=\"font-weight: 400;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\">RD Sharma Solutions<\/a><\/span> Class 11 Maths Chapter 14 Quadratic Equations will solve your queries related to real coefficients and real roots. So without any second question grab your <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> Chapter 14 and begin your test preparation.<\/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-69da3d1c6b914\" 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\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69da3d1c6b914\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#download-rd-sharma-solutions-class-11-maths-chapter-14-pdf\" title=\"Download RD Sharma Solutions Class 11 Maths Chapter 14 PDF\">Download RD Sharma Solutions Class 11 Maths Chapter 14 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#exercise-wise-rd-sharma-solutions-class-11-maths-chapter-14-quadratic-equations\" title=\"Exercise-wise: RD Sharma Solutions Class 11 Maths Chapter 14 Quadratic Equations\">Exercise-wise: RD Sharma Solutions Class 11 Maths Chapter 14 Quadratic Equations<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#access-rd-sharma-solutions-class-11-maths-chapter-14\" title=\"Access RD Sharma Solutions Class 11 Maths Chapter 14\">Access RD Sharma Solutions Class 11 Maths Chapter 14<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b\" title=\"[By using the formula,\u00a0a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]\">[By using the formula,\u00a0a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-2\" title=\"[By using the formula,\u00a0a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]\">[By using the formula,\u00a0a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-3\" title=\"[By using the formula,\u00a0a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]\">[By using the formula,\u00a0a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-4\" title=\"[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]\">[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-5\" title=\"[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]\">[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-6\" title=\"[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]\">[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-7\" title=\"[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]\">[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-8\" title=\"[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]\">[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-9\" title=\"[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]\">[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-10\" title=\"[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]\">[By using the formula, a2\u00a0\u2013 b2\u00a0= (a + b) (a \u2013 b)]<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#access-important-chapters-of-rd-sharma-solutions-class-11-maths\" title=\"Access Important Chapters of RD Sharma Solutions Class 11 Maths\u00a0\">Access Important Chapters of RD Sharma Solutions Class 11 Maths\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#faqs-on-rd-sharma-solutions-class-11-maths-chapter-14\" title=\"FAQs on RD Sharma Solutions Class 11 Maths Chapter 14\">FAQs on RD Sharma Solutions Class 11 Maths Chapter 14<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-for-class-11-maths-chapter-14\" title=\"From where can I download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 14?\">From where can I download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 14?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-17\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-14\" title=\"How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 14?\">How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 14?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-18\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#can-i-access-the-rd-sharma-solutions-for-class-11-maths-chapter-14-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 11 Maths Chapter 14\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 11 Maths Chapter 14\u00a0PDF offline?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-19\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/#is-it-even-beneficial-to-study-rd-sharma-solutions-class-11-maths-chapter-14\" title=\"Is it even beneficial to study RD Sharma Solutions Class 11 Maths Chapter 14?\">Is it even beneficial to study RD Sharma Solutions Class 11 Maths Chapter 14?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-14-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 14 PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-14-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 14<\/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\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-14-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-11-maths-chapter-14-quadratic-equations\"><\/span>Exercise-wise: RD Sharma Solutions Class 11 Maths Chapter 14 Quadratic Equations<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14-exercise-14-1\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Exercise 14.1<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14-exercise-14-2\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Exercise 14.2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-14\"><\/span><span style=\"font-size: 30px; font-weight: bold; background-color: initial;\">Access RD Sharma Solutions Class 11 Maths Chapter 14<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>EXERCISE 14.1 PAGE NO: 14.5<\/p>\n<p><strong>Solve the following quadratic equations by factorization method only:<\/strong><\/p>\n<p><strong>1. x<sup>2<\/sup>\u00a0+ 1 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: x<sup>2<\/sup>\u00a0+ 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\u2013 i<sup>2<\/sup>\u00a0= 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b\"><\/span>[By using the formula,\u00a0a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0= (a + b) (a \u2013 b)]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>(x + i) (x \u2013 i) = 0<\/p>\n<p>x + i = 0 or x \u2013 i = 0<\/p>\n<p>x = \u2013i or x = i<\/p>\n<p><em>\u2234 The roots of the given equation are i, -i<\/em><\/p>\n<p><strong>2. 9x<sup>2<\/sup>\u00a0+ 4 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: 9x<sup>2<\/sup>\u00a0+ 4 = 0<\/p>\n<p>9x<sup>2<\/sup>\u00a0+ 4 \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>So,<\/p>\n<p>9x<sup>2<\/sup>\u00a0+ 4(\u2013i<sup>2<\/sup>) = 0<\/p>\n<p>9x<sup>2<\/sup>\u00a0\u2013 4i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(3x)<sup>2<\/sup>\u00a0\u2013 (2i)<sup>2<\/sup>\u00a0= 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-2\"><\/span>[By using the formula,\u00a0a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0= (a + b) (a \u2013 b)]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>(3x + 2i) (3x \u2013 2i) = 0<\/p>\n<p>3x + 2i = 0 or 3x \u2013 2i = 0<\/p>\n<p>3x = \u20132i or 3x = 2i<\/p>\n<p>x = -2i\/3 or x = 2i\/3<\/p>\n<p><em>\u2234 The roots of the given equation are 2i\/3, -2i\/3<\/em><\/p>\n<p><strong>3. x<sup>2<\/sup>\u00a0+ 2x + 5 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: x<sup>2<\/sup>\u00a0+ 2x + 5 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 2x + 1 + 4 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 2(x) (1) + 1<sup>2<\/sup>\u00a0+ 4 = 0<\/p>\n<p>(x + 1)<sup>2<\/sup>\u00a0+ 4 = 0 [since,\u00a0(a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ 2ab + b<sup>2<\/sup>]<\/p>\n<p>(x + 1)<sup>2<\/sup>\u00a0+ 4 \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 + 1)<sup>2<\/sup>\u00a0+ 4(\u2013i<sup>2<\/sup>) = 0<\/p>\n<p>(x + 1)<sup>2<\/sup>\u00a0\u2013 4i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(x + 1)<sup>2<\/sup>\u00a0\u2013 (2i)<sup>2<\/sup>\u00a0= 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-3\"><\/span>[By using the formula,\u00a0a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0= (a + b) (a \u2013 b)]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>(x + 1 + 2i)(x + 1 \u2013 2i) = 0<\/p>\n<p>x + 1 + 2i = 0 or x + 1 \u2013 2i = 0<\/p>\n<p>x = \u20131 \u2013 2i or x = \u20131 + 2i<\/p>\n<p><em>\u2234 The roots of the given equation are -1+2i, -1-2i<\/em><\/p>\n<p><strong>4. 4x<sup>2<\/sup>\u00a0\u2013 12x + 25 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: 4x<sup>2<\/sup>\u00a0\u2013 12x + 25 = 0<\/p>\n<p>4x<sup>2<\/sup>\u00a0\u2013 12x + 9 + 16 = 0<\/p>\n<p>(2x)<sup>2<\/sup>\u00a0\u2013 2(2x)(3) + 3<sup>2<\/sup>\u00a0+ 16 = 0<\/p>\n<p>(2x \u2013 3)<sup>2<\/sup>\u00a0+ 16 = 0 [Since,\u00a0(a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ 2ab + b<sup>2<\/sup>]<\/p>\n<p>(2x \u2013 3)<sup>2<\/sup>\u00a0+ 16 \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>(2x \u2013 3)<sup>2<\/sup>\u00a0+ 16(\u2013i<sup>2<\/sup>) = 0<\/p>\n<p>(2x \u2013 3)<sup>2<\/sup>\u00a0\u2013 16i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(2x \u2013 3)<sup>2<\/sup>\u00a0\u2013 (4i)<sup>2<\/sup>\u00a0= 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-4\"><\/span>[By using the formula, a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0= (a + b) (a \u2013 b)]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>(2x \u2013 3 + 4i) (2x \u2013 3 \u2013 4i) = 0<\/p>\n<p>2x \u2013 3 + 4i = 0 or 2x \u2013 3 \u2013 4i = 0<\/p>\n<p>2x = 3 \u2013 4i or 2x = 3 + 4i<\/p>\n<p>x = 3\/2 \u2013 2i or x = 3\/2 + 2i<\/p>\n<p><em>\u2234 The roots of the given equation are 3\/2 + 2i, 3\/2 \u2013 2i<\/em><\/p>\n<p><strong>5. x<sup>2<\/sup>\u00a0+ x + 1 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: x<sup>2<\/sup>\u00a0+ x + 1 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ x + \u00bc + \u00be = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 2 (x) (1\/2) + (1\/2)<sup>2<\/sup>\u00a0+ \u00be = 0<\/p>\n<p>(x + 1\/2)<sup>2<\/sup>\u00a0+ \u00be = 0 [Since,\u00a0(a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ 2ab + b<sup>2<\/sup>]<\/p>\n<p>(x + 1\/2)<sup>2<\/sup>\u00a0+ \u00be \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 + \u00bd)<sup>2<\/sup>\u00a0+ \u00be (-1)<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(x + \u00bd)<sup>2<\/sup>\u00a0+ \u00be i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(x + \u00bd)<sup>2<\/sup>\u00a0\u2013 (<strong>\u221a<\/strong>3i\/2)<sup>2<\/sup>\u00a0= 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-5\"><\/span>[By using the formula, a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0= (a + b) (a \u2013 b)]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>(x + \u00bd +\u00a0<strong>\u221a<\/strong>3i\/2) (x + \u00bd \u2013\u00a0<strong>\u221a<\/strong>3i\/2) = 0<\/p>\n<p>(x + \u00bd +\u00a0<strong>\u221a<\/strong>3i\/2) = 0 or (x + \u00bd \u2013\u00a0<strong>\u221a<\/strong>3i\/2) = 0<\/p>\n<p>x = -1\/2 \u2013\u00a0<strong>\u221a<\/strong>3i\/2 or x = -1\/2 +\u00a0<strong>\u221a<\/strong>3i\/2<\/p>\n<p><em>\u2234 The roots of the given equation are -1\/2 +\u00a0<strong>\u221a<\/strong>3i\/2, -1\/2 \u2013\u00a0<strong>\u221a<\/strong>3i\/2<\/em><\/p>\n<p><strong>6. 4x<sup>2<\/sup>\u00a0+ 1 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: 4x<sup>2<\/sup>\u00a0+ 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>4x<sup>2<\/sup>\u00a0\u2013 i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(2x)<sup>2<\/sup>\u00a0\u2013 i<sup>2<\/sup>\u00a0= 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-6\"><\/span>[By using the formula, a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0= (a + b) (a \u2013 b)]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>(2x + i) (2x \u2013 i) = 0<\/p>\n<p>2x + i = 0 or 2x \u2013 i = 0<\/p>\n<p>2x = \u2013i or 2x = i<\/p>\n<p>x = -i\/2 or x = i\/2<\/p>\n<p><em>\u2234 The roots of the given equation are i\/2, -i\/2<\/em><\/p>\n<p><strong>7. x<sup>2<\/sup>\u00a0\u2013 4x + 7 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: x<sup>2<\/sup>\u00a0\u2013 4x + 7 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 4x + 4 + 3 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2(x) (2) + 2<sup>2<\/sup>\u00a0+ 3 = 0<\/p>\n<p>(x \u2013 2)<sup>2<\/sup>\u00a0+ 3 = 0 [Since,\u00a0(a \u2013 b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0\u2013 2ab + b<sup>2<\/sup>]<\/p>\n<p>(x \u2013 2)<sup>2<\/sup>\u00a0+ 3 \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 \u2013 2)<sup>2<\/sup>\u00a0+ 3(\u2013i<sup>2<\/sup>) = 0<\/p>\n<p>(x \u2013 2)<sup>2<\/sup>\u00a0\u2013 3i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(x \u2013 2)<sup>2<\/sup>\u00a0\u2013 (<strong>\u221a<\/strong>3i)<sup>2<\/sup>\u00a0= 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-7\"><\/span>[By using the formula, a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0= (a + b) (a \u2013 b)]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>(x \u2013 2 +\u00a0<strong>\u221a<\/strong>3i) (x \u2013 2 \u2013\u00a0<strong>\u221a<\/strong>3i) = 0<\/p>\n<p>(x \u2013 2 +\u00a0<strong>\u221a<\/strong>3i) = 0 or (x \u2013 2 \u2013\u00a0<strong>\u221a<\/strong>3i) = 0<\/p>\n<p>x = 2 \u2013\u00a0<strong>\u221a<\/strong>3i or x = 2 +\u00a0<strong>\u221a<\/strong>3i<\/p>\n<p>x = 2\u00a0<strong>\u00b1<\/strong>\u00a0<strong>\u221a<\/strong>3i<\/p>\n<p><em>\u2234 The roots of the given equation are 2\u00a0<strong>\u00b1<\/strong>\u00a0<strong>\u221a<\/strong>3i<\/em><\/p>\n<p><strong>8. x<sup>2<\/sup>\u00a0+ 2x + 2 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: x<sup>2<\/sup>\u00a0+ 2x + 2 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 2x + 1 + 1 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 2(x)(1) + 1<sup>2<\/sup>\u00a0+ 1 = 0<\/p>\n<p>(x + 1)<sup>2<\/sup>\u00a0+ 1 = 0 [\u2235\u00a0(a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ 2ab + b<sup>2<\/sup>]<\/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 + 1)<sup>2<\/sup>\u00a0+ (\u2013i<sup>2<\/sup>) = 0<\/p>\n<p>(x + 1)<sup>2<\/sup>\u00a0\u2013 i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(x + 1)<sup>2<\/sup>\u00a0\u2013 (i)<sup>2<\/sup>\u00a0= 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-8\"><\/span>[By using the formula, a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0= (a + b) (a \u2013 b)]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>(x + 1 + i) (x + 1 \u2013 i) = 0<\/p>\n<p>x + 1 + i = 0 or x + 1 \u2013 i = 0<\/p>\n<p>x = \u20131 \u2013 i or x = \u20131 + i<\/p>\n<p>x = -1\u00a0<strong>\u00b1\u00a0<\/strong>i<\/p>\n<p><em>\u2234 The roots of the given equation are -1\u00a0<strong>\u00b1\u00a0<\/strong>i<\/em><\/p>\n<p><strong>9. 5x<sup>2<\/sup>\u00a0\u2013 6x + 2 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: 5x<sup>2<\/sup>\u00a0\u2013 6x + 2 = 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 = 5, b = -6, c = 2<\/p>\n<p>So,<\/p>\n<p>x = (-(-6)<strong>\u00a0\u00b1\u221a<\/strong>(-6<sup>2<\/sup>\u00a0\u2013 4 (5)(2)))\/ 2(5)<\/p>\n<p>= (6\u00a0<strong>\u00b1 \u221a<\/strong>(36-40))\/10<\/p>\n<p>= (6\u00a0<strong>\u00b1 \u221a<\/strong>(-4))\/10<\/p>\n<p>= (6\u00a0<strong>\u00b1 \u221a<\/strong>4(-1))\/10<\/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>x = (6\u00a0<strong>\u00b1 \u221a<\/strong>4i<sup>2<\/sup>)\/10<\/p>\n<p>= (6\u00a0<strong>\u00b1\u00a0<\/strong>2i)\/10<\/p>\n<p>= 2(3<strong>\u00b1<\/strong>i)\/10<\/p>\n<p>= (3<strong>\u00b1<\/strong>i)\/5<\/p>\n<p>x = 3\/5\u00a0<strong>\u00b1<\/strong>\u00a0i\/5<\/p>\n<p><em>\u2234 The roots of the given equation are 3\/5\u00a0<strong>\u00b1<\/strong>\u00a0i\/5<\/em><\/p>\n<p><strong>10. 21x<sup>2<\/sup>\u00a0+ 9x + 1 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: 21x<sup>2<\/sup>\u00a0+ 9x + 1 = 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 = 21, b = 9, c = 1<\/p>\n<p>So,<\/p>\n<p>x = (-9<strong>\u00a0\u00b1\u221a<\/strong>(9<sup>2<\/sup>\u00a0\u2013 4 (21)(1)))\/ 2(21)<\/p>\n<p>= (-9\u00a0<strong>\u00b1 \u221a<\/strong>(81-84))\/42<\/p>\n<p>= (-9\u00a0<strong>\u00b1 \u221a<\/strong>(-3))\/42<\/p>\n<p>= (-9\u00a0<strong>\u00b1 \u221a<\/strong>3(-1))\/42<\/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>x = (-9\u00a0<strong>\u00b1 \u221a<\/strong>3i<sup>2<\/sup>)\/42<\/p>\n<p>= (-9\u00a0<strong>\u00b1 \u221a<\/strong>(<strong>\u221a<\/strong>3i)<sup>2<\/sup>\/42<\/p>\n<p>= (-9\u00a0<strong>\u00b1 \u221a<\/strong>3i)\/42<\/p>\n<p>= -9\/42\u00a0<strong>\u00b1 \u221a<\/strong>3i\/42<\/p>\n<p>= -3\/14\u00a0<strong>\u00b1 \u221a<\/strong>3i\/42<\/p>\n<p><em>\u2234 The roots of the given equation are -3\/14\u00a0<strong>\u00b1 \u221a<\/strong>3i\/42<\/em><\/p>\n<p><strong>11. x<sup>2<\/sup>\u00a0\u2013 x + 1 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: x<sup>2<\/sup>\u00a0\u2013 x + 1 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 x + \u00bc + \u00be = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2 (x) (1\/2) + (1\/2)<sup>2<\/sup>\u00a0+ \u00be = 0<\/p>\n<p>(x \u2013 1\/2)<sup>2<\/sup>\u00a0+ \u00be = 0 [Since,\u00a0(a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ 2ab + b<sup>2<\/sup>]<\/p>\n<p>(x \u2013 1\/2)<sup>2<\/sup>\u00a0+ \u00be \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 \u2013 \u00bd)<sup>2<\/sup>\u00a0+ \u00be (-1)<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(x \u2013 \u00bd)<sup>2<\/sup>\u00a0+ \u00be (-i)<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(x \u2013 \u00bd)<sup>2<\/sup>\u00a0\u2013 (<strong>\u221a<\/strong>3i\/2)<sup>2<\/sup>\u00a0= 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-9\"><\/span>[By using the formula, a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0= (a + b) (a \u2013 b)]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>(x \u2013 \u00bd +\u00a0<strong>\u221a<\/strong>3i\/2) (x \u2013 \u00bd \u2013\u00a0<strong>\u221a<\/strong>3i\/2) = 0<\/p>\n<p>(x \u2013 \u00bd +\u00a0<strong>\u221a<\/strong>3i\/2) = 0 or (x \u2013 \u00bd \u2013\u00a0<strong>\u221a<\/strong>3i\/2) = 0<\/p>\n<p>x = 1\/2 \u2013\u00a0<strong>\u221a<\/strong>3i\/2 or x = 1\/2 +\u00a0<strong>\u221a<\/strong>3i\/2<\/p>\n<p><em>\u2234 The roots of the given equation are 1\/2 +\u00a0<strong>\u221a<\/strong>3i\/2, 1\/2 \u2013\u00a0<strong>\u221a<\/strong>3i\/2<\/em><\/p>\n<p><strong>12. x<sup>2<\/sup>\u00a0+ x + 1 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: x<sup>2<\/sup>\u00a0+ x + 1 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ x + \u00bc + \u00be = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 2 (x) (1\/2) + (1\/2)<sup>2<\/sup>\u00a0+ \u00be = 0<\/p>\n<p>(x + 1\/2)<sup>2<\/sup>\u00a0+ \u00be = 0 [Since,\u00a0(a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ 2ab + b<sup>2<\/sup>]<\/p>\n<p>(x + 1\/2)<sup>2<\/sup>\u00a0+ \u00be \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 + \u00bd)<sup>2<\/sup>\u00a0+ \u00be (-1)<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(x + \u00bd)<sup>2<\/sup>\u00a0+ \u00be i<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(x + \u00bd)<sup>2<\/sup>\u00a0\u2013 (<strong>\u221a<\/strong>3i\/2)<sup>2<\/sup>\u00a0= 0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"by-using-the-formula-a2-%e2%80%93-b2-a-b-a-%e2%80%93-b-10\"><\/span>[By using the formula, a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>\u00a0= (a + b) (a \u2013 b)]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>(x + \u00bd +\u00a0<strong>\u221a<\/strong>3i\/2) (x + \u00bd \u2013\u00a0<strong>\u221a<\/strong>3i\/2) = 0<\/p>\n<p>(x + \u00bd +\u00a0<strong>\u221a<\/strong>3i\/2) = 0 or (x + \u00bd \u2013\u00a0<strong>\u221a<\/strong>3i\/2) = 0<\/p>\n<p>x = -1\/2 \u2013\u00a0<strong>\u221a<\/strong>3i\/2 or x = -1\/2 +\u00a0<strong>\u221a<\/strong>3i\/2<\/p>\n<p><em>\u2234 The roots of the given equation are -1\/2 +\u00a0<strong>\u221a<\/strong>3i\/2, -1\/2 \u2013\u00a0<strong>\u221a<\/strong>3i\/2<\/em><\/p>\n<p><strong>13. 17x<sup>2<\/sup>\u00a0\u2013 8x + 1 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: 17x<sup>2<\/sup>\u00a0\u2013 8x + 1 = 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 = 17, b = -8, c = 1<\/p>\n<p>So,<\/p>\n<p>x = (-(-8)<strong>\u00a0\u00b1\u221a<\/strong>(-8<sup>2<\/sup>\u00a0\u2013 4 (17)(1)))\/ 2(17)<\/p>\n<p>= (8\u00a0<strong>\u00b1 \u221a<\/strong>(64-68))\/34<\/p>\n<p>= (8\u00a0<strong>\u00b1 \u221a<\/strong>(-4))\/34<\/p>\n<p>= (8\u00a0<strong>\u00b1 \u221a<\/strong>4(-1))\/34<\/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>x = (8\u00a0<strong>\u00b1 \u221a<\/strong>(2i)<sup>2<\/sup>)\/34<\/p>\n<p>= (8\u00a0<strong>\u00b1\u00a0<\/strong>2i)\/34<\/p>\n<p>= 2(4<strong>\u00b1<\/strong>i)\/34<\/p>\n<p>= (4<strong>\u00b1<\/strong>i)\/17<\/p>\n<p>x = 4\/17\u00a0<strong>\u00b1<\/strong>\u00a0i\/17<\/p>\n<p><em>\u2234 The roots of the given equation are 4\/17\u00a0<strong>\u00b1<\/strong>\u00a0i\/17<\/em><\/p>\n<p><span style=\"font-size: 30px; font-weight: bold; background-color: initial;\">Important Topics This Chapter Includes-<\/span><\/p>\n<p>Going through the important topics before you actually begin your exam preparation is one intelligent move. Thus to simplify the situation we compiled the important topics to learn from the chapter-<\/p>\n<ul>\n<li>Some useful definitions and results.<\/li>\n<li>Quadratic equations with real coefficients.<\/li>\n<li>Quadratic equations with complex coefficients.<\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"access-important-chapters-of-rd-sharma-solutions-class-11-maths\"><\/span>Access Important Chapters of RD Sharma Solutions Class 11 Maths\u00a0<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 1 &#8211; Sets<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 2 &#8211; Relations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 3 &#8211; Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 4 &#8211; Measurement of Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 5 &#8211; Trigonometric Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 6 &#8211; Graphs of Trigonometric Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 7 &#8211; Trigonometric Ratios of Compound Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 8 &#8211; Transformation Formulae<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 9 &#8211; Trigonometric Ratios of Multiple and Sub-Multiple Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 10 &#8211; Sine and Cosine Formulae and Their Applications<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 11 &#8211; Trigonometric Equations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 12 &#8211; Mathematical Induction<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 13 &#8211; Complex Numbers<\/a><\/li>\n<\/ul>\n<p>Well, this is everything that you need from the RD Sharma Solutions Class 11 Maths Chapter 14 Quadratic Equations. If you find any issue concerning the same make sure to comment down. Best for your <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener noreferrer\">CBSE<\/a> class 11 maths exam.\u00a0<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-11-maths-chapter-14\"><\/span>FAQs on RD Sharma Solutions Class 11 Maths Chapter 14<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-1629818362987\" 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-for-class-11-maths-chapter-14\"><\/span>From where can I download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 14?<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-1629818398922\" 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-class-11-maths-chapter-14\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 14?<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 id=\"faq-question-1629818423963\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-for-class-11-maths-chapter-14-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 11 Maths Chapter 14\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629818449374\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-even-beneficial-to-study-rd-sharma-solutions-class-11-maths-chapter-14\"><\/span>Is it even beneficial to study RD Sharma Solutions Class 11 Maths Chapter 14?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, your preparation will be strengthened with this amazing help book. All your questions will be answered by this book.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 14 &#8211; Quadratic Equations: The topic quadratic equation might seem familiar right? Here, RD Sharma Solutions Class 11 Maths Chapter 14 Quadratic Equations will solve your queries related to real coefficients and real roots. So without any second question grab your RD Sharma Solutions Class 11 Maths Chapter &#8230; <a title=\"RD Sharma Solutions Class 11 Maths Chapter 14 &#8211; Quadratic Equations (Updated For 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-14\/\" aria-label=\"More on RD Sharma Solutions Class 11 Maths Chapter 14 &#8211; Quadratic Equations (Updated For 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":119449,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,73411,73410],"tags":[3428,73334],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62660"}],"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=62660"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62660\/revisions"}],"predecessor-version":[{"id":513797,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62660\/revisions\/513797"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/119449"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=62660"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=62660"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=62660"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}