{"id":126860,"date":"2023-09-13T17:12:00","date_gmt":"2023-09-13T11:42:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126860"},"modified":"2023-12-06T10:20:35","modified_gmt":"2023-12-06T04:50:35","slug":"rd-sharma-class-10-solutions-chapter-8-exercise-8-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-3\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.3 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126881\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.3.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.3.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.3-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3:\u00a0<\/strong>The fundamental concept covered in this exercise is the factorization approach for solving quadratic equations. Students who are having trouble with any chapter can use the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Solutions Class 10<\/strong><\/a> for free to improve their understanding of the subject. Students can use the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-quadratic-equations\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 8<\/strong><\/a> Exercise 8.3 PDF to help them solve the question.<\/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-69e753449298d\" 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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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-3\/#is-the-rd-sharma-class-10-solutions-chapter-8-exercise-83-available-on-the-kopykitab-website\" title=\"Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3 available on the Kopykitab website?\">Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3 available on the Kopykitab website?<\/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-class-10-solutions-chapter-8-exercise-8-3\/#where-can-i-get-rd-sharma-class-10-maths-solutions-chapter-8-exercise-83-free-pdf\" title=\"Where can I get RD Sharma Class 10 Maths Solutions Chapter 8 Exercise 8.3 Free PDF?\">Where can I get RD Sharma Class 10 Maths Solutions Chapter 8 Exercise 8.3 Free PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-8-exercise-83-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<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\/09\/RD-Sharma-Solution-Class-10-Chapter-8-Exercise-8.3.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-8-Exercise-8.3.pdf\">RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-8-exercise-83-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<article id=\"post-53223\" class=\"post-53223 page type-page status-publish hentry\">\n<p><strong>Solve the following quadratic equation by factorization:<\/strong><\/p>\n<p><strong>1. (x \u2013 4)(x + 2) = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>(x \u2013 4) (x + 2) = 0<\/p>\n<p>So, either x \u2013 4 = 0 \u21d2 x = 4<\/p>\n<p>Or, x + 2 = 0, \u21d2 x = \u2013 2<\/p>\n<p>Thus, the roots of the given quadratic equation are 4 and -2, respectively.<\/p>\n<p><strong>2. (2x + 3) (3x \u2013 7) = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>(2x + 3) (3x \u2013 7) = 0.<\/p>\n<p>So, either 2x + 3 = 0, \u21d2\u00a0x = \u2013 3\/2<\/p>\n<p>Or, 3x -7 = 0, \u21d2 x = 7\/3<\/p>\n<p>Thus, the roots of the given quadratic equation are x = -3\/2 and x = 7\/3, respectively.<\/p>\n<p><strong>3.<\/strong>\u00a0<strong>3x<sup>2\u00a0<\/sup>\u2013 14x \u2013 5 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given.<\/p>\n<p>3x<sup>2\u00a0<\/sup>\u2013 14x \u2013 5 = 0<\/p>\n<p>\u21d2 3x<sup>2\u00a0<\/sup>\u2013 14x \u2013 5 = 0<\/p>\n<p>\u21d2 3x<sup>2\u00a0<\/sup>\u2013 15x + x \u2013 5 = 0<\/p>\n<p>\u21d2 3x(x \u2013 5) + 1(x \u2013 5) = 0<\/p>\n<p>\u21d2 (3x + 1)(x \u2013 5) = 0<\/p>\n<p>Now, either 3x + 1 = 0 \u21d2\u00a0x = -1\/3<\/p>\n<p>Or, x \u2013 5 = 0 \u21d2 x = 5<\/p>\n<p>Thus, the roots of the given quadratic equation are 5 and x = \u2013 1\/3, respectively.<\/p>\n<p><strong>4. Find the roots of the equation 9x<sup>2\u00a0<\/sup>\u2013 3x \u2013 2 = 0.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>9x<sup>2\u00a0<\/sup>\u2013 3x \u2013 2 = 0.<\/p>\n<p>\u21d2 9x<sup>2\u00a0<\/sup>\u2013 3x \u2013 2 = 0.<\/p>\n<p>\u21d2 9x<sup>2<\/sup>\u00a0\u2013 6x + 3x \u2013 2 = 0<\/p>\n<p>\u21d2 3x (3x \u2013 2) + 1(3x \u2013 2) = 0<\/p>\n<p>\u21d2 (3x \u2013 2)(3x + 1) = 0<\/p>\n<p>Now, either 3x \u2013 2 = 0 \u21d2 x = 2\/3<\/p>\n<p>Or, 3x + 1= 0 \u21d2 x = -1\/3<\/p>\n<p>Thus, the roots of the given quadratic equation are x = 2\/3 and x = -1\/3, respectively.<\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-2.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 1\" \/><\/strong><\/p>\n<p><strong>5.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u00a0<\/p>\n<p>Given,<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-3.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 2\" \/><\/strong><\/p>\n<p>Dividing by 6 on both sides and cross-multiplying, we get<\/p>\n<p>x<sup>2<\/sup>+ 4x \u2013 12 = 0<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ 6x \u2013 2x \u2013 12 = 0<\/p>\n<p>\u21d2 x(x + 6) \u2013 2(x \u2013 6) = 0<\/p>\n<p>\u21d2 (x + 6)(x \u2013 2) = 0<\/p>\n<p>Now, either x + 6 = 0 \u21d2x = -6<\/p>\n<p>Or, x \u2013 2 = 0 \u21d2 x = 2<\/p>\n<p>Thus, the roots of the given quadratic equation are 2 and \u2013 6, respectively.<\/p>\n<p><strong>6. 6x<sup>2\u00a0<\/sup>+ 11x + 3 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is 6x<sup>2\u00a0<\/sup>+ 11x + 3 = 0.<\/p>\n<p>\u21d2 6x<sup>2\u00a0<\/sup>+ 9x + 2x + 3 = 0<\/p>\n<p>\u21d2 3x (2x + 3) + 1(2x + 3) = 0<\/p>\n<p>\u21d2 (2x +3) (3x + 1) = 0<\/p>\n<p>Now, either 2x + 3 = 0 \u21d2\u00a0x = -3\/2<\/p>\n<p>Or, 3x + 1= 0 \u21d2\u00a0x = -1\/3<\/p>\n<p>Thus, the roots of the given quadratic equation are x = -3\/2 and x = -1\/3, respectively.<\/p>\n<p><strong>7. 5x<sup>2\u00a0<\/sup>\u2013 3x \u2013 2 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is 5x<sup>2\u00a0<\/sup>\u2013 3x \u2013 2 = 0.<\/p>\n<p>\u21d2 5x<sup>2\u00a0<\/sup>\u2013 3x \u2013 2 = 0.<\/p>\n<p>\u21d2 5x<sup>2<\/sup>\u00a0\u2013 5x + 2x \u2013 2 = 0<\/p>\n<p>\u21d2 5x(x \u2013 1) + 2(x \u2013 1) = 0<\/p>\n<p>\u21d2 (5x + 2)(x \u2013 1) = 0<\/p>\n<p>Now, either 5x + 2 = 0 \u21d2x = -2\/5<\/p>\n<p>Or, x -1= 0 \u21d2x = 1<\/p>\n<p>Thus, the roots of the given quadratic equation are 1 and x = -2\/5, respectively.<\/p>\n<p><strong>8. 48x<sup>2\u00a0<\/sup>\u2013 13x \u2013 1 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is 48x<sup>2\u00a0<\/sup>\u2013 13x \u2013 1 = 0.<\/p>\n<p>\u21d2 48x<sup>2\u00a0<\/sup>\u2013 13x \u2013 1 = 0.<\/p>\n<p>\u21d2 48x<sup>2\u00a0<\/sup>\u2013 16x + 3x \u2013 1 = 0.<\/p>\n<p>\u21d2 16x(3x \u2013 1) + 1(3x \u2013 1) = 0<\/p>\n<p>\u21d2 (16x + 1)(3x \u2013 1) = 0<\/p>\n<p>Either 16x + 1 = 0 \u21d2\u00a0x = -1\/16<\/p>\n<p>Or, 3x \u2013 1=0 \u21d2\u00a0x = 1\/3<\/p>\n<p>Thus, the roots of the given quadratic equation are x = -1\/16 and x = 1\/3, respectively.<\/p>\n<p><strong>9. 3x<sup>2\u00a0<\/sup>= -11x \u2013 10\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is 3x<sup>2\u00a0<\/sup>= -11x \u2013 10<\/p>\n<p>\u21d2 3x<sup>2\u00a0<\/sup>+ 11x + 10 = 0<\/p>\n<p>\u21d2 3x<sup>2\u00a0<\/sup>+ 6x + 5x + 10 = 0<\/p>\n<p>\u21d2 3x(x + 2) + 5(x + 2) = 0<\/p>\n<p>\u21d2 (3x + 5)(x + 2) = 0<\/p>\n<p>Now, either 3x + 5 = 0 \u21d2\u00a0x = -5\/3<\/p>\n<p>Or, x + 2 = 0 \u21d2 x = -2<\/p>\n<p>Thus, the roots of the given quadratic equation are x = -5\/3 and -2, respectively.<\/p>\n<p><strong>10. 25x(x + 1) = \u2013 4\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is 25x(x + 1) = -4<\/p>\n<p>25x(x + 1) = -4<\/p>\n<p>\u21d2 25x<sup>2\u00a0<\/sup>+ 25x + 4 = 0<\/p>\n<p>\u21d2 25x<sup>2\u00a0<\/sup>+ 20x + 5x + 4 = 0<\/p>\n<p>\u21d2 5x (5x + 4) + 1(5x + 4) = 0<\/p>\n<p>\u21d2 (5x + 4)(5x + 1) = 0<\/p>\n<p>Now, either 5x + 4 = 0 therefore\u00a0x = \u2013 4\/5<\/p>\n<p>Or, 5x + 1 = 0 therefore\u00a0x = -1 \/5<\/p>\n<p>Thus, the roots of the given quadratic equation are x = \u2013 4\/5 and x = -1\/5, respectively.<\/p>\n<p><strong>11. 16x \u2013 10\/x = 27<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>16x \u2013 10\/x = 27<\/p>\n<p>On multiplying x on both sides we have,<\/p>\n<p>\u21d2 16x<sup>2<\/sup>\u00a0\u2013 10 = 27x<\/p>\n<p>\u21d2 16x<sup>2<\/sup>\u00a0\u2013 27x \u2013 10 = 0<\/p>\n<p>\u21d2 16x<sup>2<\/sup>\u00a0\u2013 32x + 5x \u2013 10 = 0<\/p>\n<p>\u21d2 16x(x \u2013 2) +5(x \u2013 2) = 0<\/p>\n<p>\u21d2 (16x + 5) (x \u2013 2) = 0<\/p>\n<p>Now, either 16x + 5 = 0 \u21d2 x = -5\/16<\/p>\n<p>Or, x \u2013 2 = 0 \u21d2 x = 2<\/p>\n<p>Thus, the roots of the given quadratic equation are x = \u2013 5\/16 and x = 2, respectively.<\/p>\n<p><strong>12.<\/strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-4.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 3\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u00a0<\/p>\n<p>Given equation is,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-5.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 4\" \/><\/p>\n<p>On cross multiplying on both the sides we get,<\/p>\n<p>2 = 3x(x \u2013 2)<\/p>\n<p>2 = 3x<sup>2\u00a0<\/sup>\u2013 6x<\/p>\n<p>3x<sup>2<\/sup>\u2013 6x \u2013 2 = 0<\/p>\n<p>\u21d2 3x<sup>2<\/sup>\u2013 3x \u2013 3x \u2013 2 = 0<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-6.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 5\" \/><\/p>\n<p>Now, either<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-7.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 6\" \/><\/strong><\/p>\n<p>Thus,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-8.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 7\" \/>are the solutions of the given quadratic equations.<\/p>\n<p><strong>13. x \u2013 1\/x = 3, x \u2260 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>x \u2013 1\/x = 3<\/p>\n<p>On multiplying x on both sides, we have,<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0\u2013 1 = 3x<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0\u2013 3x \u2013 1 = 0<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-9.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 8\" \/><br \/><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-10.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 9\" \/><\/p>\n<p><strong>14.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u00a0<\/p>\n<p>Given,<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-11.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 10\" \/><\/strong><\/p>\n<p>Dividing by 11 both the sides and cross-multiplying, we get,<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 3x \u2013 28 = \u2013 30<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 3x \u2013 2 = 0<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 2x \u2013 x \u2013 2 = 0<\/p>\n<p>\u21d2 x(x \u2013 2) \u2013 1(x \u2013 2) = 0<\/p>\n<p>\u21d2 (x \u2013 2)(x \u2013 1) = 0<\/p>\n<p>Now, either x \u2013 2 = 0 \u21d2 x = 2<\/p>\n<p>Or, x \u2013 1 = 0 \u21d2 x = 1<\/p>\n<p>Thus, the roots of the given quadratic equation are 1 and 2, respectively.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-12.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 11\" \/><\/p>\n<p><strong>15.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-13.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 12\" \/><\/p>\n<p>On cross multiplying we get,<\/p>\n<p>\u21d2 x(3x \u2013 8) = 8(x \u2013 3)(x \u2013 2)<\/p>\n<p>\u21d2 3x<sup>2<\/sup>\u00a0\u2013 8x = 8(x<sup>2<\/sup>\u00a0\u2013 5x + 6)<\/p>\n<p>\u21d2 8x<sup>2<\/sup>\u00a0\u2013 40x + 48 \u2013 (3x<sup>2<\/sup>\u00a0\u2013 8x) = 0<\/p>\n<p>\u21d2 5x<sup>2<\/sup>\u00a0\u2013 32x + 48 = 0<\/p>\n<p>\u21d2 5x<sup>2<\/sup>\u00a0\u2013 20x \u2013 12x + 48 = 0<\/p>\n<p>\u21d2 5x(x \u2013 4) \u2013 12(x \u2013 4) = 0<\/p>\n<p>\u21d2 (x \u2013 4)(5x \u2013 12) = 0<\/p>\n<p>Now, either x \u2013 4 = 0 \u21d2 x = 4<\/p>\n<p>Or, 5x \u2013 12 = 0 \u21d2 x = 12\/5<\/p>\n<p>Thus, the roots of the given quadratic equation are 12\/5 and 4, respectively.<\/p>\n<p><strong>16. a<sup>2<\/sup>x<sup>2\u00a0<\/sup>\u2013 3abx + 2b<sup>2\u00a0<\/sup>= 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is a<sup>2<\/sup>x<sup>2\u00a0<\/sup>\u2013 3abx + 2b<sup>2\u00a0<\/sup>= 0<\/p>\n<p>\u21d2 a<sup>2<\/sup>x<sup>2\u00a0<\/sup>\u2013 abx \u2013 2abx + 2b<sup>2\u00a0<\/sup>= 0<\/p>\n<p>\u21d2 ax(ax \u2013 b) \u2013 2b(ax \u2013 b) = 0<\/p>\n<p>\u21d2 (ax \u2013 b)(ax \u2013 2b) = 0<\/p>\n<p>Now, either ax \u2013 b = 0 \u21d2 x = b\/a<\/p>\n<p>Or, ax \u2013 2b = 0 \u21d2\u00a0x = 2b\/a<\/p>\n<p>Thus, the roots of the quadratic equation are x = 2b\/a and x = b\/a, respectively.<\/p>\n<p><strong>17. 9x<sup>2<\/sup>\u00a0\u2013 6b<sup>2<\/sup>x \u2013 (a<sup>4<\/sup>\u00a0\u2013 b<sup>4<\/sup>) = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-14.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 13\" \/><\/p>\n<p>Thus, the roots of the quadratic equation are\u00a0x = (b<sup>2\u00a0<\/sup>\u2013 a<sup>2<\/sup>)\/3 and\u00a0x = (a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>)\/3, respectively.<\/p>\n<p><strong>18. 4x<sup>2\u00a0<\/sup>+ 4bx \u2013 (a<sup>2\u00a0<\/sup>\u2013 b<sup>2<\/sup>) = 0\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>4x<sup>2\u00a0<\/sup>+ 4bx \u2013 (a<sup>2\u00a0<\/sup>\u2013 b<sup>2<\/sup>) = 0<\/p>\n<p>For factorizing,<\/p>\n<p>4(a<sup>2\u00a0<\/sup>\u2013 b<sup>2<\/sup>) = -4(a \u2013 b) (a + b) = [-2(a-b)] [2(a + b)]<\/p>\n<p>\u21d2 2(b \u2013 a)*2(b + a)<\/p>\n<p>\u21d2 4x<sup>2<\/sup>+ (2(b \u2013 a) + 2(b + a)) \u2013 (a \u2013 b)(a + b) = 0<\/p>\n<p>So, now<\/p>\n<p>4x<sup>2\u00a0\u00a0<\/sup>+ 2(b \u2013 a)x++ 2(b + a)x + (b \u2013 a)(a + b) = 0<\/p>\n<p>\u21d2 2x(2x + (b \u2013 a)) +(a + b)(2x + (b \u2013 a)) = 0<\/p>\n<p>\u21d2 (2x + (b \u2013 a))(2x + b + a) = 0<\/p>\n<p>Now, either (2x + (b \u2013 a)) = 0 \u21d2x = (a \u2013 b)\/2<\/p>\n<p>Or, (2x + b + a) = 0 \u21d2\u00a0x = -(a + b)\/2<\/p>\n<p>Thus, the roots of the given quadratic equation are x = -(a + b)\/2 and x = (a \u2013 b)\/2, respectively.<\/p>\n<p><strong>19. ax<sup>2\u00a0<\/sup>+ (4a<sup>2\u00a0<\/sup>\u2013 3b)x \u2013 12ab = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is ax<sup>2\u00a0<\/sup>+ (4a<sup>2\u00a0<\/sup>\u2013 3b)x \u2013 12ab = 0<\/p>\n<p>\u21d2 ax<sup>2\u00a0<\/sup>+ 4a<sup>2<\/sup>x \u2013 3bx \u2013 12ab = 0<\/p>\n<p>\u21d2 ax(x + 4a) \u2013 3b(x + 4a) = 0<\/p>\n<p>\u21d2 (x + 4a)(ax \u2013 3b) = 0<\/p>\n<p>Now, either x + 4a = 0 \u21d2 x = -4a<\/p>\n<p>Or, ax \u2013 3b = 0 \u21d2 x = 3b\/a<\/p>\n<p>Thus, the roots of the given quadratic equation are x = 3b\/a and -4a, respectively.<\/p>\n<p><strong>20. 2x<sup>2<\/sup>\u00a0+ ax \u2013 a<sup>2<\/sup>\u00a0= 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u00a0<\/p>\n<p>Given,<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-15.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 14\" \/><\/strong><\/p>\n<p>Thus, the roots of the given quadratic equation are x = a\/2 and -a, respectively.<\/p>\n<p><strong>21. 16\/x \u2013 1 = 15\/(x + 1), x \u2260 0, -1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-16.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 15\" \/><\/strong><\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-17.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 16\" \/><\/strong><\/p>\n<p>Thus, the roots of the given quadratic equation are x = 4 and -4, respectively.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-18.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 17\" \/><\/p>\n<p><strong>22. , x \u2260 -2, 3\/2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u00a0<\/p>\n<p>Given,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-19.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 18\" \/><\/p>\n<p>On cross-multiplying we get,<\/p>\n<p>(x + 3)(2x \u2013 3) = (x + 2)(3x \u2013 7)<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>\u2013 3x + 6x \u2013 9 = 3x<sup>2\u00a0<\/sup>\u2013 x \u2013 14<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>+ 3x \u2013 9 = 3x<sup>2\u00a0<\/sup>\u2013 x \u2013 14<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 3x \u2013 x \u2013 14 + 9 = 0<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 5x + x \u2013 5 = 0<\/p>\n<p>\u21d2 x(x \u2013 5) + 1(x \u2013 5) = 0<\/p>\n<p>\u21d2 (x \u2013 5)(x + l) \u2013 0<\/p>\n<p>Now, either x \u2013 5 = 0 or x + 1 = 0<\/p>\n<p>\u21d2 x = 5 and x = -1<\/p>\n<p>Thus, the roots of the given quadratic equation are 5 and -1, respectively.<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-20.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 19\" \/>23.<\/strong><\/p>\n<p><strong>, x \u2260 3, 4<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given equation is<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-21.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 20\" \/><\/p>\n<p>On cross-multiplying, we have<\/p>\n<p>3(4x<sup>2\u00a0<\/sup>\u2013 19x + 20) = 25(x<sup>2\u00a0<\/sup>\u2013 7x + 12)<\/p>\n<p>\u21d2 12x<sup>2\u00a0<\/sup>\u2013 57x + 60 = 25x<sup>2<\/sup>\u00a0\u2013 175x + 300<\/p>\n<p>\u21d213x<sup>2\u00a0<\/sup>\u2013 78x \u2013 40x + 240 = 0<\/p>\n<p>\u21d213x<sup>2\u00a0<\/sup>\u2013 118x + 240 = 0<\/p>\n<p>\u21d213x<sup>2\u00a0<\/sup>\u2013 78x \u2013 40x + 240 = 0<\/p>\n<p>\u21d213x(x \u2013 6) \u2013 40(x \u2013 6) = 0<\/p>\n<p>\u21d2 (x \u2013 6)(13x \u2013 40) = 0<\/p>\n<p>Now, either x \u2013 6 = 0 \u21d2x = 6<\/p>\n<p>Or, 13x \u2013 40 = 0 \u21d2x =\u00a040\/13<\/p>\n<p>Thus, the roots of the given quadratic equation are 6 and 40\/13, respectively.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-22.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 21\" \/><strong>24. x \u2260 0, 2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-23.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 22\" \/><\/p>\n<p>On cross-multiplying, we get<\/p>\n<p>4(2x<sup>2\u00a0<\/sup>+ 2) = 17(x<sup>2\u00a0<\/sup>\u2013 2x)<\/p>\n<p>\u21d2 8x<sup>2\u00a0<\/sup>+ 8 = 17x<sup>2\u00a0<\/sup>\u2013 34x<\/p>\n<p>\u21d2 9x<sup>2\u00a0<\/sup>\u2013 34x \u2013 8 = 0<\/p>\n<p>\u21d2 9x<sup>2\u00a0<\/sup>\u2013 36x + 2x \u2013 8 = 0<\/p>\n<p>\u21d2 9x(x \u2013 4) + 2(x \u2013 4) = 0<\/p>\n<p>\u21d2 9x + 2)(x \u2013 4) = 0<\/p>\n<p>Now, either 9x + 2 = 0 \u21d2x = -2\/9<\/p>\n<p>Or, x \u2013 4 = 0 \u21d2 x = 4<\/p>\n<p>Thus, the roots of the given quadratic equation are x = -2\/9 and 4, respectively.<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-24.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 23\" \/>25.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is,<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-25.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 24\" \/><\/strong><\/p>\n<p>On cross-multiplying, we get<\/p>\n<p>7(-12x) = 48(x<sup>2<\/sup>\u00a0\u2013 9)<\/p>\n<p>\u21d2 -84x = 48x<sup>2<\/sup>\u00a0\u2013 432<\/p>\n<p>\u21d2 48x<sup>2<\/sup>\u00a0+ 84x \u2013 432 = 0<\/p>\n<p>\u21d2 4x<sup>2<\/sup>\u00a0+ 7x \u2013 36 = 0 dividing by 12]<\/p>\n<p>\u21d2 4x<sup>2<\/sup>\u00a0+ 16x \u2013 9x \u2013 36 = 0<\/p>\n<p>\u21d2 4x(x + 4) \u2013 9(x \u2013 4) = 0<\/p>\n<p>\u21d2 (4x \u2013 9)(x + 4) = 0<\/p>\n<p>Now, either 4x \u2013 9 = 0 \u21d2x = 9\/4<\/p>\n<p>Or, x + 4 = 0 \u21d2 x = -4<\/p>\n<p>Thus, the roots of the given quadratic equation are x = 9\/4 and -4, respectively.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-26.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 25\" \/><strong>26.<\/strong><\/p>\n<p>, x\u00a0<strong>\u2260 0<\/strong><\/p>\n<p><strong>Solution:<br \/><\/strong><\/p>\n<p>Given equation is,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-27.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 26\" \/><br \/><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-28.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 27\" \/><\/p>\n<p>On cross multiplying, we have<\/p>\n<p>x(3x \u2013 5) = 6(x<sup>2\u00a0<\/sup>\u2013 3x + 2)<\/p>\n<p>\u21d2 3x<sup>2\u00a0<\/sup>\u2013 5x = 6x<sup>2\u00a0<\/sup>\u2013 18x + 12<\/p>\n<p>\u21d2 3x<sup>2\u00a0<\/sup>\u2013 13x + 12 = 0<\/p>\n<p>\u21d2 3x<sup>2\u00a0<\/sup>\u2013 9x \u2013 4x + 12 = 0<\/p>\n<p>\u21d2 3x(x \u2013 3) \u2013 4(x \u2013 3) = 0<\/p>\n<p>\u21d2 (x \u2013 3)(3x \u2013 4) = 0<\/p>\n<p>Now, either x \u2013 3 = 0 \u21d2 x = 3<\/p>\n<p>Or, 3x \u2013 4 = 0 \u21d2 4\/3.<\/p>\n<p>Thus, the roots of the given quadratic equation are 3 and 4\/3, respectively.<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-29.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 28\" \/>27.<\/strong><\/p>\n<p><strong>, x \u2260 1, -1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given equation is,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-30.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 29\" \/><\/p>\n<p>On cross \u2013 multiplying we have,<\/p>\n<p>\u21d2 6(4x) = 5(x<sup>2\u00a0<\/sup>\u2013 1) = 24x<\/p>\n<p>\u21d2 5x<sup>2\u00a0<\/sup>\u2013 5 = 5x<sup>2\u00a0<\/sup>\u2013 24x \u2013 5 =0<\/p>\n<p>\u21d2 5x<sup>2\u00a0<\/sup>\u2013 25x + x \u2013 5 = 0<\/p>\n<p>\u21d2 5x(x \u2013 5) + 1(x \u2013 5) = 0<\/p>\n<p>\u21d2 (5x + 1)(x \u2013 5) = 0<\/p>\n<p>Now, either x \u2013 5 = 0 \u21d2 x = 5<\/p>\n<p>Or, 5x + 1 = 0 \u21d2\u00a0x = \u22121\/5<\/p>\n<p>Thus, the roots of the given quadratic equation are x = \u22121\/5 and 5, respectively.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-31.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 30\" \/><strong>28.<\/strong><\/p>\n<p><strong>, x \u2260 1, -1\/2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given equation is,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-32.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 31\" \/><\/p>\n<p>On cross \u2013 multiplying we have,<\/p>\n<p>\u21d2 2(5x<sup>2\u00a0<\/sup>+ 2x + 2) = 5(2x<sup>2\u00a0<\/sup>\u2013 x \u2013 1)<\/p>\n<p>\u21d2 10x<sup>2\u00a0<\/sup>+ 4x + 4 = 10x<sup>2\u00a0<\/sup>\u2013 5x \u2013 5<\/p>\n<p>\u21d2 4x + 5x + 4 + 5 = 0<\/p>\n<p>\u21d2 9x + 9 = 0<\/p>\n<p>\u21d2 9x = -9<\/p>\n<p>Thus, x = -1 is the only root of the given equation.<\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-33.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 32\" \/>29.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-34.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 33\" \/><\/p>\n<p>Thus, the roots of the given quadratic equation are x = 1 and x = -2, respectively.<\/p>\n<p><strong>30.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given equation is,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-35.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 34\" \/><\/p>\n<p>On cross-multiplying, we have,<\/p>\n<p>3(2x<sup>2<\/sup>\u00a0\u2013 22x + 58) = 10(x<sup>2<\/sup>\u00a0\u2013 12x + 35)<\/p>\n<p>\u21d2 6x<sup>2<\/sup>\u00a0\u2013 66x + 174 = 10x<sup>2\u00a0<\/sup>\u2013 120x + 350<\/p>\n<p>\u21d2 4x<sup>2<\/sup>\u00a0\u2013 54x + 176 = 0<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>\u2013 27x + 88 = 0<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>\u2013 16x \u2013 11x + 88 = 0<\/p>\n<p>\u21d2 2x(x \u2013 8) \u2013 11(x + 8) = 0<\/p>\n<p>\u21d2 (x \u2013 8)(2x \u2013 11) = 0<\/p>\n<p>Now, either x \u2013 8 = 0 \u21d2 x = 8<\/p>\n<p>Or, 2x \u2013 11 = 0 \u21d2\u00a0x = 11\/2<\/p>\n<p>Thus, the roots of the given quadratic equation are x = 11\/2 and 8, respectively.<\/p>\n<\/article>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-8-exercise-83\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3<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-1631524413597\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-is-rd-sharma-class-10-solutions-chapter-8-exercise-83-helpful-for-board-exams\"><\/span>How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3 helpful for board exams?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>For self-evaluation, RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3 provides solutions with thorough descriptions as per term limits specified by the Board. Students will gain valuable experience solving these problems, allowing them to complete the assignment on time.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631526248403\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-the-rd-sharma-class-10-solutions-chapter-8-exercise-83-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3 available on the Kopykitab website?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631526469532\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-maths-solutions-chapter-8-exercise-83-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Maths Solutions Chapter 8 Exercise 8.3 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Maths Solutions Chapter 8 Exercise 8.3 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.3:\u00a0The fundamental concept covered in this exercise is the factorization approach for solving quadratic equations. Students who are having trouble with any chapter can use the RD Sharma Solutions Class 10 for free to improve their understanding of the subject. Students can use the RD Sharma Solutions &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.3 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-3\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.3 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126881,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126860"}],"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\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126860"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126860\/revisions"}],"predecessor-version":[{"id":517660,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126860\/revisions\/517660"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126881"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126860"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126860"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126860"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}