{"id":4804,"date":"2025-01-05T16:17:00","date_gmt":"2025-01-05T20:17:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=4804"},"modified":"2025-02-24T16:19:29","modified_gmt":"2025-02-24T20:19:29","slug":"algebra-challenge-5","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=4804","title":{"rendered":"Algebra Challenge – 5"},"content":{"rendered":"\n

Find the real solutions for the following equations: <\/p>\n\n\n\n$$a^2+b^2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =1\\tag{1}$$\n$$b^2+c^2+\\sqrt{3}bc=1\\tag{2}$$\n$$c^2+a^2+\\ \\ \\ \\ \\ ca=1\\tag{3}$$\n\n\n\n

Click here<\/a> for the solution.<\/p>\n\n\n\n

\n\n\n\n

Solution: <\/strong>If $c=0$, based on equation $(2)$ and $(3)$, we have $a^2=1$ and $b^2=1$. Therefore, $a^2+b^2=2$, which is in conflict with equation $(1)$. Therefore $c\\ne 0$.<\/p>\n\n\n\n

Because $c\\ne 0$, based on equation $(1)$ and $(3)$, we have:<\/p>\n\n\n\n

$$a=\\dfrac{b^2-c^2}{c}\\tag{4}$$<\/p>\n\n\n\n

Based on equation $(1)$ and $(2)$, we have:<\/p>\n\n\n\n

$$a^2=c^2+\\sqrt{3}bc\\tag{5}$$<\/p>\n\n\n\n

Replacing $a$ in equation $(5)$ with equation $(3)$, we have:<\/p>\n\n\n\n

$$\\Big(\\dfrac{b^2-c^2}{c}\\Big)^2=c^2+\\sqrt{3}bc\\tag{6}$$<\/p>\n\n\n\n

Simpifying equation $(6)$, we have: $$b(b^3-2bc^2-\\sqrt{3}c^3)=0\\tag{7}$$<\/p>\n\n\n\n

If $b=0$, we have $a^2=1$, $c^2=1$, $c^2+a^2+ca=1$, which leads solutions of $(a, b, c)$ as $(-1, 0, 1)$ and $(1, 0, -1)$.<\/p>\n\n\n\n

If $(b^3-2bc^2-\\sqrt{3}c^3=0$, as $c\\ne 0$, we have: $$\\Big(\\dfrac{b}{c}\\Big)^3-2\\Big(\\dfrac{b}{c}\\Big)-\\sqrt{3}=0\\tag{8}$$<\/p>\n\n\n\n

The rational solution for $\\dfrac{b}{c}$ in equation $(8)$ is $\\sqrt{3}$. Therefore $$b=\\sqrt{3}c$$<\/p>\n\n\n\n

Applying the above result in equation $(2)$, we have $$c=\\pm\\dfrac{\\sqrt{7}}{7}$$<\/p>\n\n\n\n

which also results in $$b=\\pm\\dfrac{\\sqrt{21}}{7}$$ and $$a=\\pm\\dfrac{2\\sqrt{7}}{7}$$<\/p>\n\n\n\n

Combining the all cases, there are 4 real solutions for $(a, b, c)$:<\/p>\n\n\n\n

$(-1,0,1)$, $(1,0,-1)$, $\\Big(\\dfrac{2\\sqrt{7}}{7},\\dfrac{\\sqrt{21}}{7},\\dfrac{\\sqrt{7}}{7}\\Big)$, and $\\Big(-\\dfrac{2\\sqrt{7}}{7},-\\dfrac{\\sqrt{21}}{7},-\\dfrac{\\sqrt{7}}{7}\\Big)$<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Find the real solutions for the following equations: $$a^2+b^2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =1\\tag{1}$$ $$b^2+c^2+\\sqrt{3}bc=1\\tag{2}$$ $$c^2+a^2+\\ \\ \\ \\ \\ ca=1\\tag{3}$$ Click here for the solution. Solution: If $c=0$, based on equation … Continue reading →<\/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\/4804"}],"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=4804"}],"version-history":[{"count":32,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4804\/revisions"}],"predecessor-version":[{"id":4836,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4804\/revisions\/4836"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4804"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4804"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4804"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}