{"id":2795,"date":"2022-07-25T23:51:00","date_gmt":"2022-07-26T03:51:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=2795"},"modified":"2024-10-25T11:09:20","modified_gmt":"2024-10-25T15:09:20","slug":"algebra-challenge-2022-07-25","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=2795","title":{"rendered":"Algebra Challenge 2022\/07\/25"},"content":{"rendered":"\n<p>Let $x$ and $a$ are real numbers, and $a$ is a constant with $a&gt;=0$, and $x^2=a(x-\\lfloor x \\rfloor)$. Find the number of solutions for $x$, in terms of $a$. Click <a href=\"javascript:toggle_visibility('algebra-chall-2022-07-25');\">here<\/a> for the solution.<\/p>\n\n\n\n<div id=\"algebra-chall-2022-07-25\" style=\"display:none\">\n\n\n\n<p><strong>Solution<\/strong>: Let $y=x-\\lfloor x \\rfloor$ and $z=\\lfloor x \\rfloor$. Therefore $0\\le y&lt;1$, $z$ is an integer, $$(y+z)^2=ay\\tag{1}$$ and $$y^2+(2z-a)y+z^2=0\\tag{2}$$<\/p>\n\n\n\n<p>Solve $y$ in equation (2), we have: $$y=\\dfrac{a-2z\\pm\\sqrt{a^2-4az}}{2}\\tag{3}$$ Because $y$ is a real number, therefore $$a^2-4az\\ge 0$$ Because $a\\ge 0$, we have $$z\\le \\dfrac{a}{4} \\tag{4}$$ <\/p>\n\n\n\n<p>Additionally, because $$4z^2\\ge0$$ $$ a^2-4az+4z^2\\ge a^2-4az$$ $$(a-2z)^2\\ge a^2-4az$$ Because (4), $a-2z\\ge 0$, we have $$a-2z\\ge\\sqrt{a^2-4az}$$ Therefore, $y&gt;=0$ for all $a\\ge 0$.<\/p>\n\n\n\n<p>Because $y&lt;1$, according to (3), we have $$a-2z \\pm\\sqrt{a^2-4az} &lt;2\\tag{5}$$ Solve inequality (5), we have $$-\\sqrt{a}-1&lt;z&lt;\\sqrt{a}-1 \\tag{6}$$ Because $$\\dfrac{a}{4}-\\sqrt{a}+1=(\\dfrac{a}{2}-1)^2\\ge 0$$ Inequality (6) implies inequality (4). Therefore integer $z$ are in $(-\\sqrt{a}-1, \\sqrt{a}-1)$. Because each integer $z$ value corresponds to a unique real value of $x$, therefore the number of solutions for $x$ is the number of integers in the range of $(-\\sqrt{a}-1, \\sqrt{a}-1) \\boxed{\\phantom{|}}$<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Let $x$ and $a$ are real numbers, and $a$ is a constant with $a&gt;=0$, and $x^2=a(x-\\lfloor x \\rfloor)$. Find the number of solutions for $x$, in terms of $a$. Click here for the solution. Solution: Let $y=x-\\lfloor x \\rfloor$ and &hellip; <a href=\"https:\/\/mathfun4kids.com\/mlog\/?p=2795\">Continue reading <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2795"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=2795"}],"version-history":[{"count":30,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2795\/revisions"}],"predecessor-version":[{"id":4602,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2795\/revisions\/4602"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2795"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2795"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2795"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}