Question

resuelve este ejercicio ​

Answer 1

Respuesta:

$\large{ \bold{ \underline{ \color{aqua} \clubs{} \color{lime}ecuaciones \color{aqua} \clubs}}} \\ \large{ \bold{ \underline{\color{orange} \bigstar{} \color{blue}cuadr \acute{a}ticas. \color{orange} \bigstar}}}$

$\sf{ \underline {\orange{ \bigstar}\blue{sabemos \: \: que \: \: una \: \: ecuaci \acute{o}n}} }\\ \sf{ \underline{ \blue{cuadr \acute{a}tica \: \: ( \color{springgreen}de \: \: segundo \: \: grado \blue{)}}}} \\ \sf{ \underline{ \blue{tiene \: \: dos \: \: soluciones \: \: las \: \: cuales}}} \\ \sf{ \underline{ \blue{en \: \: esta \: \: ecuacion \: \: son:}}}$

$\boxed{\boxed{\bold{ \color{red} \real{espuesta \rightarrow{} \boxed{ \bold{ \color{blue}x_{1} \color{black}= \color{blue}1}}} \: \: \: \: \: \: \boxed{ \bold{ \color{blue}x_{2} \color{black}= \color{blue}0.665}}}}}$

⭐Explicación paso a paso:

Resolvemos esta ecuación cuadratica por "fórmula general", idenitificando los valores de "a", "b" y "c"; y los sustituimos en la fórmula:

$\bold{\color{deepskyblue} {2x}^{2} \color{lawngreen} -\frac{10}{3}x \color{orange} + \frac{4}{3} \color{black} =\color{red}0} \\ \\ \\ \bold{\color{deepskyblue}a = 2\: \: \: \: \: \: \: \: \color{lawngreen}b = - \frac{10}{3} \: \: \: \: \: \: \: \: \color{orange}c= \frac{4}{3} } \\ \\ \\ \bold{\color{blue}x \color{black} = \frac{ - \color{lawngreen}b \color{black} \pm{} \sqrt{ \color{lawngreen}b \color{black}^{2}-4 \color{deepskyblue}a \color{orange}c} }{2 \color{deepskyblue}a} } \\ \\ \\ \bold{ \color{blue}x \color{black}= \frac{ - ( \color{lawngreen} - \frac{10}{3} \color{black}) \pm{} \sqrt{( \color{lawngreen} - \frac{10}{3} \color{black})^{2} -4( \color{deepskyblue}2 \color{black})( \color{orange} \frac{4}{3} \color{black})} }{2 (\color{deepskyblue}2 \color{black})} } \\ \\ \\ \bold{ \color{blue}x \color{black} = \frac{ \color{lawngreen} \frac{10}{3} \color{black} \pm{} \sqrt{ \color{lawngreen}11.11 \color{black} - \color{springgreen}10.66} }{ \color{lime}4} } \\ \\ \\ \bold{ \color{blue}x \color{black} = \frac{ \color{lawngreen} \frac{10}{3} \color{black} \pm{} \sqrt{ \color{springgreen}0.45} }{ \color{lime}4} } \\ \\ \\ \bold{ \color{blue}x_{1} \color{black} = \frac{ \color{lawngreen} \frac{10}{3} \color{black} + \color{springgreen}0.67}{ \color{lime}4} } \\ \\ \\ \bold{ \color{blue}x_{1} \color{black} = \frac{ \color{lawngreen}3.33 \color{black} + \color{springgreen}0.67}{ \color{lime}4} } \\ \\ \\ \bold{ \color{blue}x_{1} \color{black} = \frac{ \color{turquoise}4}{ \color{lime}4}} \\ \\ \\ \color{red}\boxed{\boxed{\bold{ \color{blue}x_{1} \color{black} = \color{blue}1}}}$

$\bold{\color{blue}x_{2} \color{black} = \frac{ \color{lawngreen}3.33 \color{black} - \color{springgreen}0.67}{ \color{lime}4} } \\ \\ \\ \bold{ \color{blue}x_{2} \color{black}= \frac{ \color{turquoise}2.66}{ \color{lime}4} } \\ \\ \\ \color{red}\boxed{\boxed{\bold{ \color{blue}x_{2} \color{black} = \color{blue}0.665}}} \\ \\ \\ \color{red}\boxed{\boxed{\bold{\color{blue}x_{1} \color{black} = \color{blue}1}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\boxed{\bold{ \color{blue}x_{2} \color{black} =\color{blue}0.66}}}$

$\bold{ \color{lime} \spades\colorbox{aqua}{ \color{blue}jnjok60 \: de \: brainly.lat} \spades}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \color{palegreen} \ddag{} \color{dodgerblue} \texttt{suerte} \color{palegreen} \ddag$

$\color{blue}\boxed {\color{aqua}\boxed{\color{aquamarine}\boxed{\color{turquoise}\boxed{\bold{ \color{yellow}\bigstar{} \color{cyan}\dag{} \color{lime}\clubs{} \color{deepskyblue}e \color{lawngreen}s \color{orange}p \color{red}e \color{blue}r \color{springgreen}o \: \: \: \purple{l} \color{deepskyblue}e \: \: \: \color{lawngreen}s \color{orange}i \color{red}r \color{blue}v \color{springgreen}a \: \: \: \purple{m} \color{deepskyblue}i \: \: \: \color{lawngreen}\real{} \color{orange}e \color{red}s \color{blue}p \color{springgreen}u \purple{e} \color{deepskyblue}s \color{lawngreen}t \color{orange}a \color{red}. \color{lime}\clubs{} \color{cyan}\dag{} \color{yellow}\bigstar}}}}}$