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Resolver el siguiente ecuación:

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\\[\\begin{eqnarray*}\\simplify[std]{x^4-{a^2+b^2}x^2+{(a*b)^2}}&=&{0}\\end{eqnarray*}\\]

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Observación: Para ingresar su respuesta, hágalo en el siguiente orden:

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$x_1<x_2<x_3<x_4$

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$x_1=$ [[0]] ; $x_2=$ [[1]] ; $x_3=$[[2]] ; $x_4=$[[3]]  

\n

\n

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Input your answer as a fraction and not a decimal.

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Utilicemos la sustitución $\\,x^2=t\\,$ en la ecuación:

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\\[\\begin{eqnarray*}\\simplify[std]{x^4-{a^2+b^2}x^2+{(a*b)^2}}&=&{0}\\end{eqnarray*}\\]

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Nos queda:

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\\[\\begin{eqnarray*}\\simplify[std]{t^2-{a^2+b^2}t+{(a*b)^2}}&=&{0}\\end{eqnarray*}\\]

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Esta ecuación la podemos resolver factorizando.

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\\begin{align}
(\\simplify{t-{a^2}})(\\simplify{t-{b^2}})&=\\,{0}
\\end{align}

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\\[\\simplify{(t-{a^2}={0})} \\,\\,\\,\\,;\\,\\,\\, \\simplify{(t-{b^2}={0})} \\]

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\\[\\simplify{(t={a^2})} \\,\\,\\,\\,;\\,\\,\\, \\simplify{(t={b^2})} \\]

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Ahora nos devolvemos con la sustitución $\\,x^2=t\\,$, de modo de encontrar los valores de la variable $x$:

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\\[\\simplify{(x^2={a^2})} \\,\\,\\,\\,;\\,\\,\\, \\simplify{(x^2={b^2})} \\]

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Estas ecuaciones las podemos resolver nuevamente factorizando:

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\\[\\simplify{(x^2-{a^2}={0})} \\,\\,\\,\\,;\\,\\,\\, \\simplify{(x^2-{b^2}={0})} \\]

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\\begin{align}
(\\simplify{x-{a}})(\\simplify{x+{a}})&=\\,{0}\\,\\,\\,\\,;\\,\\,\\,(\\simplify{x-{b}})(\\simplify{x+{b}})=\\,{0}
\\end{align}

\n

\\begin{align}
\\simplify{(x={a})} \\,\\,\\,\\,;\\,\\,\\, \\simplify{x={-a}}\\,\\,\\,\\,;\\,\\,\\,\\simplify{(x={b})} \\,\\,\\,\\,;\\,\\,\\, \\simplify{x={-b}}
\\end{align}

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Finalmente ordenamos las soluciones de menor a mayor como lo pide el enunciado:

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\\begin{align}
\\simplify{(x_1={-b})} \\,\\,\\,\\,;\\,\\,\\, \\simplify{x_2={-a}}\\,\\,\\,\\,;\\,\\,\\,\\simplify{(x_3={a})} \\,\\,\\,\\,;\\,\\,\\, \\simplify{x_4={b}}
\\end{align}

\n

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