
{"id":131625,"date":"2024-06-13T14:48:24","date_gmt":"2024-06-13T11:48:24","guid":{"rendered":"https:\/\/mathema.me\/?p=131625"},"modified":"2024-06-13T14:51:31","modified_gmt":"2024-06-13T11:51:31","slug":"vietas-theorem-and-its-inverse","status":"publish","type":"post","link":"https:\/\/mathema.me\/en\/blog\/vietas-theorem-and-its-inverse\/","title":{"rendered":"Vieta&#8217;s Theorem and Its Inverse: A Guide for 8th Grade Students and Parents"},"content":{"rendered":"\n<p>In this article, Mathema, an educational platform, explores Vieta&#8217;s Theorem and its applications in factored quadratic equations. Designed for 8th-grade students and their parents, this article aims to simplify complex mathematical concepts.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Vieta&#8217;s theorem for factored quadratic equations (a=1)<\/h2>\n\n\n\n<p>These are formulas that express the coefficients of a polynomial in terms of its roots. The theorem is named after Fran\u00e7ois Vi\u00e8te, a French mathematician. Vieta&#8217;s theorem is usually used to solve factored quadratic equations, i.e., when the coefficient a=1. Here&#8217;s how Vieta&#8217;s theorem is formulated for a factored quadratic equation:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p>The sum of the roots of a factored quadratic equation is equal to the second coefficient with the opposite sign, and the product of the roots is equal to the constant term.<\/p>\n<\/blockquote>\n\n\n\n<h2 class=\"wp-block-heading\">Vieta&#8217;s theorem formula for factored quadratic equations<\/h2>\n\n\n\n<p>Here&#8217;s what the Vieta&#8217;s theorem formula looks like for a factored quadratic equation:<\/p>\n\n\n\n\\[x^2+px+q=0,\\;x_1\\cdot x_2=q;\\;x_1+x_2=-p\\]\n\n\n\n<p>This formula of Vieta&#8217;s theorem allows one to compute their sum and product without knowing the roots of the quadratic trinomial.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Application of Vieta&#8217;s theorem <\/h2>\n\n\n\n<p>Let&#8217;s analyze Vieta&#8217;s theorem and the formula of Vieta&#8217;s theorem with a simple example. This will help understand how to correctly solve factored quadratic equations using them.<\/p>\n\n\n\n<p>How to find the roots of an equation using Vieta&#8217;s theorem?<\/p>\n\n\n\n\\[x^2-5x+6=0\\]\n\n\n\n<p><strong>Solution:<\/strong> According to Vieta&#8217;s theorem, we write:<\/p>\n\n\n\n\\[x_1+x_2=5\\]\n\n\n\n\\[x_1\\cdot x_2=6\\]\n\n\n\n<p>We find the values of \ud835\udc65 that satisfy the equation:<\/p>\n\n\n\n\\[x_1=\\;2;\\;x_2=3\\]\n\n\n\n<p><strong>Answer:<\/strong> the roots of the equation are 2 and 3.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Inverse Vieta&#8217;s theorem<\/h2>\n\n\n\n<p>The concept of the inverse Vieta&#8217;s theorem speaks for itself, as it works exactly the opposite way. Using the inverse theorem to Vieta&#8217;s theorem, you can find the roots of factored quadratic equations.<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p>If two numbers are such that their sum equals the second coefficient of the factored quadratic equation taken with the opposite sign, and their product equals its constant term, then these numbers are the roots of this factored quadratic equation.<\/p>\n<\/blockquote>\n\n\n\n<h2 class=\"wp-block-heading\">How to apply the inverse Vieta&#8217;s theorem <\/h2>\n\n\n\n<p><strong>Task: <\/strong>Formulate a quadratic equation, knowing its roots.<\/p>\n\n\n\n\\[x_1=\\;3;\\;x_2=-1\\]\n\n\n\n<p><strong>Solution: <\/strong>Let the sought quadratic equation be:<\/p>\n\n\n\n\\[x^2+px+q=0\\]\n\n\n\n<p>Using the inverse Vieta&#8217;s theorem, we get the following relationships:<\/p>\n\n\n\n\\[x_1\\cdot x_2=q;\\;x_1+x_2=-p\\]\n\n\n\n<p>Now let&#8217;s find the values of \ud835\udc5d and \ud835\udc5e:<\/p>\n\n\n\n\\[p=-(x_1+x_2)=-(3+(-1))=-2\\]\n\n\n\n\\[q=x_1\\cdot x_2\\;=\\;3\\cdot(-1)=-3\\]\n\n\n\n<p><strong>Answer:<\/strong> the equation we were looking for looks like this:<\/p>\n\n\n\n\\[x^2-2x-3=0\\]\n","protected":false},"excerpt":{"rendered":"<p>In this article, Mathema, an educational platform, explores Vieta&#8217;s Theorem and its applications in factored quadratic equations. Designed for 8th-grade [&hellip;]<\/p>\n","protected":false},"author":15,"featured_media":61407,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":true,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[3248,491],"tags":[],"class_list":["post-131625","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-middle-school","category-blog"],"acf":[],"jetpack_featured_media_url":"https:\/\/i0.wp.com\/mathema.me\/wp-content\/uploads\/2024\/02\/post-cover-89.jpg?fit=1080%2C675&ssl=1","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mathema.me\/en\/wp-json\/wp\/v2\/posts\/131625","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathema.me\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathema.me\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathema.me\/en\/wp-json\/wp\/v2\/users\/15"}],"replies":[{"embeddable":true,"href":"https:\/\/mathema.me\/en\/wp-json\/wp\/v2\/comments?post=131625"}],"version-history":[{"count":0,"href":"https:\/\/mathema.me\/en\/wp-json\/wp\/v2\/posts\/131625\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/mathema.me\/en\/wp-json\/wp\/v2\/media\/61407"}],"wp:attachment":[{"href":"https:\/\/mathema.me\/en\/wp-json\/wp\/v2\/media?parent=131625"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathema.me\/en\/wp-json\/wp\/v2\/categories?post=131625"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathema.me\/en\/wp-json\/wp\/v2\/tags?post=131625"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}