{"id":372,"date":"2020-09-26T17:03:04","date_gmt":"2020-09-26T22:03:04","guid":{"rendered":"http:\/\/blogs.acu.edu\/mu_sigma\/?p=372"},"modified":"2020-10-02T14:31:40","modified_gmt":"2020-10-02T19:31:40","slug":"372","status":"publish","type":"post","link":"https:\/\/blogs.acu.edu\/mu_sigma\/2020\/09\/26\/372\/","title":{"rendered":"Problem 2 &#8211; Fall 2020 &#8211; Symmetric Functions II"},"content":{"rendered":"<p><span style=\"font-size: xx-large\"><b>Symmetric Functions II<\/b><\/span><\/p>\n<p><span style=\"font-size: small\"><i>This problem is from Alexander Karabegov.<\/i><\/span><\/p>\n<p>Solve the following system of equations (you may want to look back at the previous Problem of the Week):<\/p>\n<ul>\n<li><i>x<\/i><sup>3<\/sup> + <i>y<\/i><sup>3<\/sup> = 9<\/li>\n<li>5<i>x<\/i><i>y<\/i> = 2<i>x<\/i><sup>2<\/sup> + 2<i>y<\/i><sup>2<\/sup>.<\/li>\n<\/ul>\n<p><em>Please submit all solutions to mathpotw@acu.edu by 5:00 PM on Thursday.\u00a0<\/em><\/p>\n<p><span style=\"font-size: small\"><i>Correct solutions were submitted by:\u00a0 Ricky Kagoro Lumala<\/i><\/span><\/p>\n<p>There are several approaches that will work. Following are two of them.<\/p>\n<p><b>Approach I .<\/b> We use the results from the previous problem of the week. So we assume that there is a polynomial <i>z<\/i><sup>2<\/sup> + <i>p z<\/i> + <i>q<\/i> with two real roots <i>x<\/i> and <i>y<\/i>. Then we have found that<br \/>\n<i>x<\/i><sup>3<\/sup> + <i>y<\/i><sup>3<\/sup> = &#8211;<i>p<\/i><sup>3<\/sup> + 3<i>pq<\/i>,<br \/>\n<i>x<\/i><sup>2<\/sup> + <i>y<\/i><sup>2<\/sup> = <i>p<\/i><sup>2<\/sup> &#8211; 2<i>q<\/i>, and<br \/>\n<i>x<\/i><i>y<\/i> = <i>q<\/i>.<\/p>\n<p>Substituting these identities into the given system of equations yields (after some simplification) an equivalent system of equations:<br \/>\n&#8211;<i>p<\/i><sup>3<\/sup> + 3<i>pq<\/i> = 9 and<br \/>\n9<i>q<\/i> = 2<i>p<\/i><sup>2<\/sup>.<\/p>\n<p>This system is easy to solve: <i>p<\/i> = -3 and <i>q<\/i> = 2.<\/p>\n<p>Therefore the corresponding quadratic equation is <i>z<\/i><sup>2<\/sup> &#8211; 3<i>z<\/i> + 2 = 0. The roots of this equation are <i>x<\/i> = 1 or <i>x<\/i> = 2. Therefore the solutions of the given equations are:<\/p>\n<p>(<i>x<\/i>, <i>y<\/i>) = (1, 2) and (<i>x<\/i>, <i>y<\/i>) = (2,1).<\/p>\n<p><b>Approach II .<\/b> Divide both sides of the equation 5<i>x<\/i><i>y<\/i> = 2<i>x<\/i><sup>2<\/sup> + 2<i>y<\/i><sup>2<\/sup> by <i>xy<\/i> to get<br \/>\n5 = 2(<i>x\/y<\/i>) + 2(<i>y\/x<\/i>). Now let <i>a<\/i> = <i>x\/y<\/i> which, after rearranging, gives the quadratic equation 2<i>a<\/i><sup>2<\/sup> &#8211; 5<i>a<\/i> + 2 = (2<i>a<\/i> &#8211; 1)(<i>a<\/i> &#8211; 2) = 0. So <i>a<\/i> = 1\/2 or <i>a<\/i> = 2.<\/p>\n<p>This in turn gives 2<i>x<\/i> = <i>y<\/i> or <i>x<\/i> = 2<i>y<\/i>, respectively. Substituting these into the equation<br \/>\n<i>x<\/i><sup>3<\/sup> + <i>y<\/i><sup>3<\/sup> = 9 easily give the same solutions as in the previous approach.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Symmetric Functions II This problem is from Alexander Karabegov. Solve the following system of equations (you may want to look back at the previous Problem of the Week): x3 + y3 = 9 5xy = 2&#215;2 + 2y2. Please submit all solutions to mathpotw@acu.edu by 5:00 PM on Thursday.\u00a0 Correct solutions were submitted by:\u00a0 Ricky [&hellip;]<\/p>\n","protected":false},"author":130,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_et_pb_use_builder":"","_et_pb_old_content":"","_et_gb_content_width":"","footnotes":""},"categories":[181671],"tags":[],"class_list":["post-372","post","type-post","status-publish","format-standard","hentry","category-mu-sigma-events"],"_links":{"self":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/372","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/users\/130"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/comments?post=372"}],"version-history":[{"count":4,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/372\/revisions"}],"predecessor-version":[{"id":376,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/372\/revisions\/376"}],"wp:attachment":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/media?parent=372"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/categories?post=372"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/tags?post=372"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}