\n
Problema (Adaptado de Maclaurin Olympiad<\/strong>, 2003-2009)<\/span> Resolva o sistema formado pelas equa\u00e7\u00f5es [tex]\\boxed{p+pr+pr^2=28}\\;[\/tex] e [tex]\\;\\boxed{p^2r+p^2r^2+p^2r^3=224}\\,.[\/tex]<\/p>\n \n<\/div>\n \n Solu\u00e7\u00e3o <\/p>\n Vamos chamar de [tex]\\textcolor{#800000}{(i)}[\/tex] e [tex]\\textcolor{#800000}{(ii)}[\/tex] as equa\u00e7\u00f5es do sistema:<\/p>\n \n[tex]\\qquad \\qquad S:\\begin{cases} Das equa\u00e7\u00f5es [tex]\\textcolor{#800000}{(i)}\\;[\/tex] e [tex]\\;\\textcolor{#800000}{(ii)}[\/tex] obtemos, respectivamente: Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n \n","protected":false},"excerpt":{"rendered":" Problema (Indicado a partir do 9\u00ba ano do E. F.) (Adaptado de Maclaurin Olympiad, 2003-2009) Resolva o sistema formado pelas equa\u00e7\u00f5es [tex]\\boxed{p+pr+pr^2=28}\\;[\/tex] e [tex]\\;\\boxed{p^2r+p^2r^2+p^2r^3=224}\\,.[\/tex] Solu\u00e7\u00e3o Vamos chamar de [tex]\\textcolor{#800000}{(i)}[\/tex] e [tex]\\textcolor{#800000}{(ii)}[\/tex] as equa\u00e7\u00f5es do sistema: [tex]\\qquad \\qquad S:\\begin{cases} p+pr+pr^2=28\\qquad \\qquad \\quad \\quad \\textcolor{#800000}{(i)}\\\\ p^2r+p^2r^2+p^2r^3=224\\qquad \\quad \\textcolor{#800000}{(ii)} \\end{cases}[\/tex]. Das equa\u00e7\u00f5es [tex]\\textcolor{#800000}{(i)}\\;[\/tex] e [tex]\\;\\textcolor{#800000}{(ii)}[\/tex] obtemos, respectivamente: [tex]\\qquad … <\/p>\n
\n (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n
\n
\n
\np+pr+pr^2=28\\qquad \\qquad \\quad \\quad \\textcolor{#800000}{(i)}\\\\
\np^2r+p^2r^2+p^2r^3=224\\qquad \\quad \\textcolor{#800000}{(ii)}
\n\\end{cases}[\/tex].<\/p>\n
\n[tex]\\qquad p(1+r+r^2)=28\t\\qquad\\qquad \\textcolor{#800000}{(iii)}\\\\
\n\\qquad p^2r(1+r+r^2)=224 \\qquad\\quad \\textcolor{#800000}{(iv)}[\/tex]
\nAgora, como a equa\u00e7\u00e3o [tex]\\textcolor{#800000}{(iii)}\\;[\/tex] n\u00e3o \u00e9 nula, de [tex]\\textcolor{#800000}{(iii)}\\;[\/tex] e [tex]\\textcolor{#800000}{(iv)}\\;[\/tex] segue que:
\n[tex]\\qquad\\dfrac{p^2r(1+r+r^2)}{p(1+r+r^2)}=\\dfrac{224}{28}\\\\
\n\\qquad rp=8 \\\\
\n\\qquad p=\\dfrac{8}{r}, \\qquad \\textcolor{#800000}{(v)}[\/tex]
\nj\u00e1 que [tex]r \\neq 0[\/tex].
\nSubstituindo, agora, [tex]\\textcolor{#800000}{(v)}[\/tex] em [tex]\\textcolor{#800000}{(iii)}[\/tex], temos:
\n[tex]\\qquad p(1+r+r^2)=28\\\\
\n\\qquad \\dfrac{8}{r} \\cdot (1+r+r^2)=28\\\\
\n\\qquad \\dfrac{8}{r}+8+8r=28.[\/tex]
\nMultiplicando essa \u00faltima equa\u00e7\u00e3o por [tex]r[\/tex], segue que:
\n[tex]\\qquad 8+8r+8r^2=28r\\\\
\n\\qquad 8r^2-20r+8=0\\\\
\n\\qquad 2r^2-5r+2=0[\/tex].
\nResolvendo esta \u00faltima equa\u00e7\u00e3o, encontramos:
\n[tex]\\qquad r=\\dfrac{-(-5) \\pm \\sqrt{(-5)^2-4 \\cdot 2 \\cdot 2}}{2 \\cdot 2}=\\dfrac{5 \\pm 3}{4}[\/tex],
\ndonde conclu\u00edmos que [tex]r=2[\/tex] ou [tex]r=\\dfrac{1}{2}[\/tex].
\nPara [tex]r=2[\/tex], temos [tex]p=\\dfrac{8}{2}=4[\/tex] e, para [tex]r=\\dfrac{1}{2}[\/tex], temos [tex]p=\\dfrac{8}{\\dfrac{1}{2}}=16[\/tex].
\nDesta forma, temos duas solu\u00e7\u00f5es para o sistema: [tex]\\,\\fcolorbox{black}{#eee0e5}{$(p_1, r_1)=(4, 2)$}\\;[\/tex] e [tex]\\;\\fcolorbox{black}{#eee0e5}{$(p_2, r_2)=\\left(16, \\dfrac{1}{2}\\right)$}\\,[\/tex].<\/p>\n
\n