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Problema Consideremos as fun\u00e7\u00f5es [tex]f[\/tex] e [tex]g[\/tex] definidas por [tex]f(x)=a_1x^2+b_1x+c_1 \\, [\/tex] e [tex] \\, g(x)=a_2x^2+b_2x+c_2[\/tex], com [tex]a_1,b_1,c_1,a_2,b_2,c_2[\/tex] constantes reais e [tex]a_1,\\;a_2\\neq 0[\/tex]. Suponha que existem n\u00fameros reais [tex]\\alpha[\/tex] e [tex]\\beta[\/tex], com [tex]\\alpha\\neq \\beta[\/tex], tais que <\/p>\n [tex]f(\\alpha)=f(\\beta)=g(\\alpha)=g(\\beta)=0.[\/tex]<\/p>\n Mostre que, para [tex]b_1,b_2,c_1,c_2[\/tex] n\u00e3o nulos: <\/p>\n [tex]\\dfrac{a_2}{a_1}=\\dfrac{b_2}{b_1}=\\dfrac{c_2}{c_1}. (Adaptado do livro "Temas e Problemas<\/strong>")<\/p>\n<\/div>\n \n Solu\u00e7\u00e3o<\/p>\n De acordo com o enunciado, temos que [tex]\\alpha[\/tex] e [tex]\\beta[\/tex] s\u00e3o solu\u00e7\u00f5es das equa\u00e7\u00f5es do [tex]2^{\\circ}[\/tex] grau [tex]a_1x^2+b_1x+c_1=0[\/tex] e [tex]a_2x^2+b_2x+c_2=0[\/tex], sendo [tex]a_1\\neq 0[\/tex] e [tex]a_2\\neq 0[\/tex]. Pelas Rela\u00e7\u00f5es de Girard<\/strong> (Se precisar, visite esta Sala<\/strong><\/a>)<\/span>:<\/p>\n \\begin{equation} \\begin{equation} (Pela solu\u00e7\u00e3o apresentada, podemos concluir que: Duas equa\u00e7\u00f5es do 2 graus possuem as mesmas ra\u00edzes se, e somente se, uma \u00e9 m\u00faltipla da outra<\/strong>.)<\/span><\/li>\n<\/ul>\n Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n \n","protected":false},"excerpt":{"rendered":" Problema (Indicado a partir do 2\u00ba ano do E. M.) Consideremos as fun\u00e7\u00f5es [tex]f[\/tex] e [tex]g[\/tex] definidas por [tex]f(x)=a_1x^2+b_1x+c_1 \\, [\/tex] e [tex] \\, g(x)=a_2x^2+b_2x+c_2[\/tex], com [tex]a_1,b_1,c_1,a_2,b_2,c_2[\/tex] constantes reais e [tex]a_1,\\;a_2\\neq 0[\/tex]. Suponha que existem n\u00fameros reais [tex]\\alpha[\/tex] e [tex]\\beta[\/tex], com [tex]\\alpha\\neq \\beta[\/tex], tais que [tex]f(\\alpha)=f(\\beta)=g(\\alpha)=g(\\beta)=0.[\/tex] Mostre que, para [tex]b_1,b_2,c_1,c_2[\/tex] n\u00e3o nulos: [tex]\\dfrac{a_2}{a_1}=\\dfrac{b_2}{b_1}=\\dfrac{c_2}{c_1}. [\/tex] (Adaptado … <\/p>\n
\n (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n
\n
\n[\/tex]<\/p>\n
\n
\n\\boxed{\\dfrac{b_1}{a_1}=- (\\alpha +\\beta)\\;\\;\\;\\; \\mbox{e}\\;\\;\\;\\; \\dfrac{b_2}{a_2}=- (\\alpha +\\beta)
\n \\, \\, }\\;\\;\\;\\; \\mbox{e}\\;\\;\\;\\; \\boxed{\\dfrac{c_1}{a_1}=\\alpha \\cdot \\beta \\;\\;\\;\\; \\mbox{e}\\;\\;\\;\\; \\dfrac{c_2}{a_2}=\\alpha \\cdot \\beta \\, \\, };
\n\\end{equation}
\nassim, segue que:
\n\\begin{equation}
\n \\dfrac{b_1}{a_1}= \\dfrac{b_2}{a_2},\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\textcolor{#800000}{(i)}\\\\
\n \\, \\, \\\\
\n\\dfrac{c_1}{a_1}= \\dfrac{c_2}{a_2}. \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\textcolor{#800000}{(ii)}
\n\\end{equation}
\nComo [tex] b_1,c_1,b_2,c_2[\/tex] s\u00e3o n\u00e3o nulos, podemos reescrever [tex]\\textcolor{#800000}{(i)}[\/tex] e [tex]\\textcolor{#800000}{(ii)}[\/tex]
\n\\begin{equation}
\n\\dfrac{a_2}{a_1}=\\dfrac{b_2}{b_1}\\;\\;\\;\\mbox{e}\\;\\;\\; \\dfrac{a_2}{a_1}= \\dfrac{c_2}{c_1}
\n\\end{equation}
\ne, com isso, chegamos nas igualdades desejadas:
\n\\begin{equation}
\n \\dfrac{a_2}{a_1}= \\dfrac{b_2}{b_1}=\\dfrac{c_2}{c_1}.
\n\\end{equation}
\nNotemos que nos casos em que [tex] b_1=0\\;\\mbox{ou}\\; c_1=0[\/tex], temos:
\n\\begin{equation}
\nb_1=0\\Leftrightarrow \\dfrac{b_1}{a_1}=0 \\Leftrightarrow \\alpha+\\beta=0 \\Leftrightarrow \\dfrac{b_2}{a_2}=0 \\Leftrightarrow b_2=0;
\n\\end{equation}
\nou<\/p>\n
\nc_1=0\\Leftrightarrow \\dfrac{c_1}{a_1}=0 \\Leftrightarrow \\alpha\\cdot\\beta=0 \\Leftrightarrow \\dfrac{c_2}{a_2}=0 \\Leftrightarrow c_2=0.
\n\\end{equation}<\/p>\n
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