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Problema Sendo [tex]\\alpha[\/tex] uma raiz da equa\u00e7\u00e3o [tex]x^2 – x – 1 = 0[\/tex], determine [tex] \\, \\alpha^5 – 5\\alpha \\, [\/tex] sem calcular [tex]\\alpha[\/tex].<\/p>\n \n<\/div>\n \n Solu\u00e7\u00e3o <\/p>\n Se [tex]\\alpha[\/tex] \u00e9 raiz da equa\u00e7\u00e3o [tex]x^2 – x – 1 = 0[\/tex], ent\u00e3o [tex]\\alpha^2 – \\alpha – 1 = 0[\/tex], ou seja, [tex]\\boxed{\\alpha^2 = \\alpha + 1}[\/tex]. \n [tex]\\begin{align*}\\alpha^4 &= \\alpha \\cdot \\alpha^3 \\\\ \n [tex]\\begin{align*}\\alpha^5 &= \\alpha \\cdot \\alpha^4 \\\\ \n Ora, se [tex]\\alpha^5 = 5\\alpha + 3[\/tex], ent\u00e3o [tex]\\fcolorbox{black}{#eee0e5}{$\\alpha^5 – 5\\alpha = 3$}[\/tex].<\/p>\n Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":" Problema (Indicado a partir do 9\u00ba ano do E. F.) Sendo [tex]\\alpha[\/tex] uma raiz da equa\u00e7\u00e3o [tex]x^2 – x – 1 = 0[\/tex], determine [tex] \\, \\alpha^5 – 5\\alpha \\, [\/tex] sem calcular [tex]\\alpha[\/tex]. Solu\u00e7\u00e3o Se [tex]\\alpha[\/tex] \u00e9 raiz da equa\u00e7\u00e3o [tex]x^2 – x – 1 = 0[\/tex], ent\u00e3o [tex]\\alpha^2 – \\alpha – 1 = … <\/p>\n
\n (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n
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\nVamos utilizar esse resultado diversas vezes nos c\u00e1lculos das pot\u00eancias [tex]\\alpha^n[\/tex] para [tex]n = 3, \\,4,\\, 5[\/tex].<\/p>\n\n[tex]\\begin{align*}\\alpha^3 &= \\alpha \\cdot \\alpha^2 \\\\
\n&= \\alpha \\cdot (\\alpha + 1) \\\\
\n&= \\alpha^2 + \\alpha \\\\
\n&= \\alpha + 1 + \\alpha \\\\
\n&= 2\\alpha + 1\\end{align*}[\/tex]<\/li>\n
\n&= \\alpha \\cdot (2\\alpha + 1) \\\\
\n&= 2\\alpha^2 + \\alpha \\\\
\n&= 2\\cdot(\\alpha + 1) + \\alpha \\\\
\n&= 2\\alpha + 2 + \\alpha\\\\
\n&= 3\\alpha + 2\\end{align*}[\/tex]<\/p>\n
\n&= \\alpha \\cdot (3\\alpha + 2) \\\\
\n&= 3\\alpha^2 + 2\\alpha \\\\
\n&= 3\\cdot(\\alpha + 1) + 2\\alpha \\\\
\n&= 3\\alpha + 3 + 2\\alpha\\\\
\n&= 5\\alpha + 3\\end{align*}[\/tex]<\/p>\n
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