{"id":6115,"date":"2013-07-02T22:55:54","date_gmt":"2013-07-03T01:55:54","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=6115"},"modified":"2023-01-21T23:06:50","modified_gmt":"2023-01-22T02:06:50","slug":"problemao-um-texalphatex-complicado","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-um-texalphatex-complicado\/","title":{"rendered":".Problem\u00e3o: Um [tex]\\alpha[\/tex] complicado"},"content":{"rendered":"
\n

Problema
\n (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/strong><\/p>\n

\n

Seja [tex]\\alpha[\/tex] um n\u00famero real tal que [tex]\\alpha^3-\\alpha-1=0[\/tex].
\nDetermine o valor num\u00e9rico de [tex]\\boxed{\\sqrt[3]{3\\alpha^2-4\\alpha\\,}+\\alpha\\sqrt[4]{2\\alpha^2+3\\alpha+2\\,}\\,}.[\/tex]<\/strong><\/span><\/div>\n

\n

Solu\u00e7\u00e3o 1<\/strong><\/span><\/p>\n

\n

Veja que [tex](\\alpha-1)^3=\\alpha^3-3\\alpha^2+3\\alpha-1[\/tex]. Como [tex]\\alpha^3-\\alpha-1=0[\/tex], temos que [tex]\\alpha=\\alpha^3-1[\/tex].
\nDa\u00ed segue que
\n[tex]\\qquad (\\alpha-1)^3=(\\alpha^3-1)-3\\alpha^2+3\\alpha=-3\\alpha^2+4\\alpha[\/tex],
\nou seja,
\n[tex]\\qquad -(\\alpha-1)^3=3\\alpha^2-4\\alpha[\/tex].
\nPodemos, tamb\u00e9m, escrever que:
\n[tex]\\qquad \\alpha\\sqrt[4]{2\\alpha^2+3\\alpha+2}=\\sqrt[4]{\\alpha^4(2\\alpha^2+3\\alpha+2)}=\\sqrt[4]{2\\alpha^6+3\\alpha^5+2\\alpha^4}[\/tex].
\nComo [tex]\\alpha^3-\\alpha-1=0[\/tex], temos que [tex]\\boxed{\\alpha^3=\\alpha+1}[\/tex].
\nEnt\u00e3o, segue que:
\n[tex]\\qquad \\begin{align*}2\\alpha^6+3\\alpha^5+2\\alpha^4&=2(\\alpha^3)^2+3\\alpha^3\\cdot\\alpha^2+2\\alpha^3\\cdot\\alpha\\\\
\n&= 2(\\alpha+1)^2+3\\alpha^2(\\alpha+1)+2\\alpha(\\alpha+1)\\\\
\n&=(\\alpha+1)\\left[2(\\alpha+1)+3\\alpha^2+2\\alpha\\right]\\\\
\n& =(\\alpha+1)(3\\alpha^2+4\\alpha+2)\\,.\\qquad\\qquad \\textcolor{#800000}{(i)}\\end{align*}[\/tex]
\nVeja que [tex](\\alpha+1)^3=\\alpha^3+3\\alpha^2+3\\alpha+1[\/tex]. Como [tex]\\alpha^3=\\alpha+1[\/tex], ent\u00e3o [tex]\\boxed{(\\alpha+1)^3=3\\alpha^2+4\\alpha+2}[\/tex].
\nAssim, segue de [tex]\\textcolor{#800000}{(i)}[\/tex], que:
\n [tex]\\qquad \\begin{align*}2\\alpha^6+3\\alpha^5+2\\alpha^4 &=(a+1)(a+1)^3\\\\
\n&=(a+1)^4. \\end{align*}[\/tex]
\nDa\u00ed,
\n[tex]\\quad \\begin{align*}\\sqrt[3]{3a^2-4a}+\\alpha\\sqrt[4]{2\\alpha^2+3\\alpha+2}&=\\sqrt[3]{-(\\alpha-1)^3}+\\sqrt[4]{(\\alpha+1)^4}\\\\
\n& =-\\alpha+1+\\alpha+1=2\\,. \\end{align*}[\/tex]
\nLogo o valor da express\u00e3o \u00e9 [tex]2[\/tex].<\/span><\/p>\n

\n

Solu\u00e7\u00e3o enviada pelo Clube Cromossomos Matem\u00e1ticos<\/strong>.<\/p>\n<\/div>\n

\n

Solu\u00e7\u00e3o 2<\/strong><\/span><\/p>\n

\n

Como [tex]\\alpha^3-\\alpha-1=0[\/tex], temos que [tex]\\boxed{\\alpha=\\alpha^3-1}[\/tex], assim:
\n[tex]\\quad\\begin{align*} \\sqrt[3]{3\\alpha^2-4\\alpha}&=\\sqrt[3]{3\\alpha^2-3\\alpha-\\alpha }\\\\
\n&=\\sqrt[3]{3\\alpha^2-3\\alpha-(\\alpha^3-1)}\\\\
\n&=\\sqrt[3]{3\\alpha^2-3\\alpha-\\alpha^3+1 }\\\\
\n&=\\sqrt[3]{(1-\\alpha)^3}\\\\
\n&=1-\\alpha \\;.\\qquad \\textcolor{#800000}{(i)}\\end{align*}[\/tex]
\nAgora, observando que [tex]\\boxed{\\alpha^3=\\alpha+1}[\/tex], a segunda parcela da soma a ser calculada pode ser simplificada assim:
\n[tex]\\quad\\begin{align*} \\alpha\\sqrt[4]{2\\alpha^2+3\\alpha+2}&= \\alpha\\sqrt[4]{\\alpha^2+\\alpha^2+\\alpha+2\\alpha+2}\\\\
\n&= \\alpha\\sqrt[4]{\\alpha^2+\\alpha^2+\\alpha+2(\\alpha+1)}\\\\
\n&= \\alpha\\sqrt[4] {\\alpha^2+\\alpha(\\alpha+1)+2(\\alpha+1)}\\\\
\n&= \\alpha\\sqrt[4] {\\alpha^2+\\alpha\\alpha^3+2\\alpha^3}\\\\
\n&=\\alpha\\sqrt[4]{\\alpha^2+\\alpha^4+2\\alpha^3}\\\\
\n&=\\alpha\\sqrt[4]{(\\alpha^3+\\alpha^2)+(\\alpha^4+\\alpha^3)}\\\\
\n&= \\alpha\\sqrt[4]{\\alpha^2(\\alpha+1)+\\alpha^3(\\alpha+1)}\\\\
\n&= \\alpha\\sqrt[4]{\\alpha^2 \\alpha^3+\\alpha^3\\alpha^3~}\\\\
\n&= \\alpha\\sqrt[4]{\\alpha^5+\\alpha^6\\;}\\\\
\n&=\\alpha\\sqrt[4]{\\alpha^5(\\alpha+1)\\;}\\\\
\n&= \\alpha\\sqrt[4]{\\alpha^8}\\\\
\n&=\\alpha^3\\\\
\n&=\\alpha+1\\,.\\qquad \\textcolor{#800000}{(ii)}\\end{align*}[\/tex]
\nDeste modo, por [tex]\\textcolor{#800000}{(i)}[\/tex] e [tex]\\textcolor{#800000}{(ii)}[\/tex], temos que:
\n[tex]\\quad\\begin{align*}\\sqrt[3]{3\\alpha^2-4\\alpha\\,}+\\alpha\\sqrt[4]{2\\alpha^2+3\\alpha+2\\,}&=\\left(1-\\alpha\\right)+\\left(\\alpha+1\\right)\\\\
\n&=1-\\alpha+\\alpha+1\\\\
\n&=2\\;.\\end{align*}[\/tex]
\nPortanto, [tex]\\fcolorbox{black}{#eee0e5}{$ \\sqrt[3]{3\\alpha^2-4\\alpha\\,}+\\alpha\\sqrt[4]{2\\alpha^2+3\\alpha+2\\,}=2$}\\,.[\/tex]<\/p>\n

\n

Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n

Participou da discuss\u00e3o do problema o Clube Cromossomos Matem\u00e1ticos. <\/strong><\/div>\n","protected":false},"excerpt":{"rendered":"

Problema (Indicado a partir do 9\u00ba ano do E. F.) Seja [tex]\\alpha[\/tex] um n\u00famero real tal que [tex]\\alpha^3-\\alpha-1=0[\/tex]. Determine o valor num\u00e9rico de [tex]\\boxed{\\sqrt[3]{3\\alpha^2-4\\alpha\\,}+\\alpha\\sqrt[4]{2\\alpha^2+3\\alpha+2\\,}\\,}.[\/tex] Solu\u00e7\u00e3o 1 Veja que [tex](\\alpha-1)^3=\\alpha^3-3\\alpha^2+3\\alpha-1[\/tex]. Como [tex]\\alpha^3-\\alpha-1=0[\/tex], temos que [tex]\\alpha=\\alpha^3-1[\/tex]. Da\u00ed segue que [tex]\\qquad (\\alpha-1)^3=(\\alpha^3-1)-3\\alpha^2+3\\alpha=-3\\alpha^2+4\\alpha[\/tex], ou seja, [tex]\\qquad -(\\alpha-1)^3=3\\alpha^2-4\\alpha[\/tex]. Podemos, tamb\u00e9m, escrever que: [tex]\\qquad \\alpha\\sqrt[4]{2\\alpha^2+3\\alpha+2}=\\sqrt[4]{\\alpha^4(2\\alpha^2+3\\alpha+2)}=\\sqrt[4]{2\\alpha^6+3\\alpha^5+2\\alpha^4}[\/tex]. Como [tex]\\alpha^3-\\alpha-1=0[\/tex], temos que [tex]\\boxed{\\alpha^3=\\alpha+1}[\/tex]. Ent\u00e3o, … <\/p>\n

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