{"id":79005,"date":"2017-07-16T02:25:48","date_gmt":"2017-07-16T05:25:48","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=79005"},"modified":"2023-04-07T05:45:17","modified_gmt":"2023-04-07T08:45:17","slug":"problema-solucoes-da-equacao","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-solucoes-da-equacao\/","title":{"rendered":".Problema: Solu\u00e7\u00f5es da equa\u00e7\u00e3o"},"content":{"rendered":"

\n

\n

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

\nDetermine todos os pares [tex](x , y)[\/tex] de n\u00fameros reais tais que [tex](x^2 + y^2 – 1)^2 + (xy)^2 = 0[\/tex].<\/p>\n

\n<\/div>\n

\n

\n

Solu\u00e7\u00e3o <\/p>\n

\nSendo [tex]a[\/tex] e [tex]b[\/tex] n\u00fameros reais, temos que
\n[tex](i)[\/tex]<\/span>[tex]\\,\\,a^2 + b^2 = 0 \\,\\, \\Leftrightarrow \\,\\, a = 0 \\,\\,\\textrm{e}\\,\\, b = 0[\/tex]
\n[tex](ii)[\/tex]<\/span>[tex]\\,a \\cdot b = 0 \\,\\, \\Leftrightarrow \\,\\, a = 0 \\,\\,\\textrm{ou}\\,\\, b = 0[\/tex].
\nDeste modo, temos as seguintes situa\u00e7\u00f5es equivalentes:<\/p>\n

\n

[tex]\\qquad (x^2 + y^2 – 1)^2 + (xy)^2 = 0 \\stackrel{{\\color{#800000}(i)}}{\\Leftrightarrow } \\left\\{\\begin{matrix}
\n x^2 + y^2 – 1 =0\\\\
\nxy = 0
\n\\end{matrix}\\right.
\n\\Leftrightarrow \\left\\{\\begin{matrix}
\n x^2 + y^2 = 1 \\\\
\nxy = 0
\n\\end{matrix}\\right.[\/tex]<\/p>\n

\n

Da segunda equa\u00e7\u00e3o, utilizando [tex](ii)[\/tex]<\/span>, temos [tex]x= 0\\,\\,\\textrm{ou}\\,\\,y=0[\/tex].
\nAgora, substituindo esses valores na primeira equa\u00e7\u00e3o, temos:<\/p>\n