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Objetivo: Resolver ecuaciones de segundo grado del tipo $ax^2+c=0$.

", "licence": "None specified"}, "statement": "

1)   ¿Cuáles son las soluciones de la ecuación de segundo grado $8x^2-72=0$?

\n

      A)  $0$  y  $3$

\n

      B)  $0$  y  $-3$

\n

      C)  $-3$  y  $3$

\n

      D)  $-3i$  y  $3i$

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Para resolver la ecuación de segundo grado $8x^2-72=0$ debemos seguir los siguientes pasos:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

$8x^2-72=0$

\n

[[0]]$x^2=$[[1]]

\n		Paso 1: Despejar el término independiente al otro lado de la ecuación con operación contraria (de suma a resta o viceversa según corresponda)
\n

                [[2]]    

\n

$x^2=$ ________

\n

            [[3]]

\n		Paso 2: El coeficiente cuadrático se despeja pasando al otro lado de la ecuación dividiendo.
$x^2=$[[4]]Paso 3: Se desarrolla la división a la derecha de la ecuación.
\n

      

\n

$x=\\pm$[[5]]

\n		Paso 4: Para eliminar el cuadrado, sacamos la raíz cuadrada del valor obtenido en el paso 3, obteniendo las 2 soluciones buscadas.
\n

Por tanto la alternativa correcta es: [[6]]

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Objetivo: Resolver ecuaciones de segundo grado del tipo $ax^2+bx=0$.

", "licence": "None specified"}, "statement": "

2)  Determine las raíces de la ecuación de segundo grado $5x^2+20x=0$

\n

     A)  $0$  y  $4$

\n

     B)  $0$  y  $-4$

\n

     C)  $-2$  y  $2$ 

\n

     D)  $-2i$  y  $2i$

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Para resolver la ecuación de segundo grado $5x^2+20x=0$ debemos seguir los siguientes pasos:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

$5x^2+20x=0$

\n

[[0]]$x \\cdot(x+$[[1]]$)=0$

\n		Paso 1: Debemos factorizar la expresión $5x^2+20x=0$, observando que su factor común es [[0]]$\\cdot x$, además es posible obtener que el otro factor es igual a $(x+$[[1]]$)=0$.
[[0]]$x=0$  y  $(x+$[[1]]$)=0$\n

Paso 2: Como [[0]]$x$ y $(x+$[[1]]$)$ multiplicadas son igual a , por propiedad dos números multiplicados dan  si uno de los dos es , obteniendo así las ecuaciones

\n
\n

$x=$[[2]]   y  $x=$[[3]]

\n		\n

Paso 3: Se resuelven las ecuaciones del paso 2 obteniendo las dos soluciones buscadas.  

\n
\n

La respuesta correcta es la alternativa: [[4]]

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B)  $0$  y  $-4$

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Objetivo: Resolver ecuaciones de segundo grado de la forma $x^2+bx+c=0$ que se pueden factorizar.

", "licence": "None specified"}, "statement": "

3)   Al resolver la ecuación de segundo grado $x^2+4x+3=0$ se obtiene que sus raíces son:

\n

      A)  $-3$  y  $-1$

\n

      B)  $3$  y  $1$

\n

      C)  $-2$  y  $2$

\n

      D)  $2$  y  $-1$

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Para resolver la ecuación de segundo grado $x^2+4x+3=0$ debemos seguir los siguientes pasos:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

$x^2+4x+3=0$

\n		\n

Paso 1: Para factorizar debemos pensar en dos números $a$ y $b$ que multiplicados sean igual a $3$ y sumados sean igual a $4$. Estos números son  $a=$[[0]] y $b=$[[1]]. (Aquí diremos que $a$ es mayor que $b$)

\n
$(x+$[[2]]$)\\cdot(x+$[[3]]$)=0$\n

Paso 2: De los números obtenidos en el paso 1 obtenemos los factores $(x+$[[2]]$)$   y   $(x+$[[3]]$)$ que usaremos para la factorización.

\n
$x+$[[4]]$=0$ y $x+$[[5]]$=0$Paso 3: Como $(x+$[[2]]$)$  y  $(x+$[[3]]$)$  multiplicadas son igual a $0$, por propiedad dos números multiplicados dan $0$ si uno de los dos es $0$, obteniendo así las ecuaciones  $x+$[[4]]$=0$ y $x+$[[5]]$=0$.
$x=$[[6]]  y $x=$[[7]]Paso 4: Se resuelven las ecuaciones del paso 3 obteniendo que                     $x=$[[6]] y $x=$[[7]]
\n

La respuesta correcta es la alternativa: [[8]]

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Objetivo: Determinar la naturaleza de las soluciones de una ecuación de segundo grado, por medio de su discriminante.

", "licence": "None specified"}, "statement": "

4)  ¿Cuál(es) de las siguientes ecuaciones de segundo grado tienen soluciones reales y distintas?

\n

     i.  $2x^2-3x+1=0$

\n

     ii. $4x^2+6x-5=0$  

\n

     A) Solo I

\n

     B) Solo II

\n

     C) I y II

\n

     D) Ninguna  

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Para determinar la naturaleza de las soluciones de una ecuación de segundo grado de la forma $ax^2+bx+c=0$, debemos determinar su discriminante $\\triangle$, cuya fórmula esta dada por la expresión $\\triangle=b^2-4 \\cdot a \\cdot c$.

\n

Recordemos que según los siguientes valores de $\\triangle$, se obtienen la naturaleza de las soluciones:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
CondiciónValorNaturaleza
$\\triangle>0$PositivoReales y Distintas
$\\triangle=0$0Reales e Iguales
$\\triangle<0$NegativoComplejas y Conjugadas
\n
\n

Análisis de i) $2x^2-3x+1=0$.

\n

Los valores de $a$, $b$ y $c$ son $a=$[[0]], $b=$[[1]] y $c=$[[2]]. 

\n

El discriminante de esta ecuación es:

\n

$\\triangle=$ $($ [[1]] $)$$^2$ $-4 \\cdot$ [[0]] $\\cdot$[[2]]

\n

$\\triangle=$ [[3]] [[4]] [[5]]

\n

$\\triangle=$[[6]]

\n

Para esta ecuación el $\\triangle=$ es [[7]], por tanto las soluciones son [[8]].

\n
\n

Análisis de i) $4x^2+6x-5=0$.

\n

Los valores de $a$, $b$ y $c$ son $a=$[[9]], $b=$[[10]] y $c=$[[11]]. 

\n

El discriminante de esta ecuación es:

\n

$\\triangle=$ $($ [[10]] $)$$^2$ $-4 \\cdot$ [[9]] $\\cdot$[[11]]

\n

$\\triangle=$ [[12]] [[13]] [[14]]

\n

$\\triangle=$[[15]]

\n

Para esta ecuación el $\\triangle=$ es [[16]], por tanto las soluciones son [[17]].

\n
\n

La respuesta correcta es la alternativa: [[18]] 

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Objetivo: Determina la suma de las soluciones de una ecuación de segundo grado.

", "licence": "None specified"}, "statement": "

5) Determine la suma de las raíces de la ecuación de segundo grado $5x^2-20x+15=0$.

\n

    A) $-4$

\n

    B) $-3$

\n

    C) $3$

\n

    D) $4$

\n

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Para determinar la suma de las soluciones de una ecuación de segundo grado de la forma $ax^2+bx+c=0$, debemos resolver la expresión $x_1+x_2= - \\Large \\frac{b}{a}$.

\n

Primero determinemos para la ecuación $5x^2-20x+15=0$ los valores  de $a$ y $b$, los cuales son $a=$[[0]]  y $b=$[[1]].

\n

Ahora reemplazamos en la expresión $x= - \\Large \\frac{b}{a}$ obteniendo:

\n

$x_1+x_2= - \\Large \\frac{b}{a}$.

\n

                                     [[2]]      

\n

$x_1+x_2= -$ _______ 

\n

                                 [[3]] 

\n

$x_1+x_2=$[[4]]

\n

La respuesta correcta es la alternativa: [[5]]

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Objetivo: Determinar las soluciones de una ecuación de segundo grado usando la fórmula general $x= \\Large\\frac{-b \\pm \\sqrt{b^2-4 \\cdot a \\cdot c}}{2 \\cdot a}$

", "licence": "None specified"}, "statement": "

6)  Determine las raíces de la ecuación de segundo grado $10x^2+5x-5=0$                                  

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Para determinar las soluciones de una ecuación de segundo grado de la forma $ax^2+bx+c=0$, usaremos la fórmula general de las soluciones la cual es $x= \\Large\\frac{-b \\pm \\sqrt{b^2-4 \\cdot a \\cdot c}}{2 \\cdot a}$.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$10x^2+5x-5=0$Paso 1: Primero debemos determinar los valores de $a$, $b$ y $c$ en la ecuación $10x^2+5x-5=0$, obteniendo que estos valores son $a=$[[0]],    $b=$[[1]]  y      $c=$[[2]].
\n

$($[[1]]$)$$^2$$-4 \\cdot$ [[0]] $\\cdot$[[2]]

\n

[[3]][[5]][[4]]

\n

[[6]]

\n		\n

Paso 2: Vamos a determinar el valor de la expresión $b^2-4ac$.

\n

 -  Calculamos $b^2$

\n

 -  Determinamos el valor de $-4ac$ y su signo

\n

 -  Reducimos los valores anteriores

\n
[[7]]Paso 3: Sacamos la raíz cuadrada del valor obtenido al final del paso 2, para así obtener cuanto vale $\\sqrt{b^2-4ac}$.
Paso 4:  Reemplazamos el valor del paso 3 en la formula general obteniendo así las dos soluciones pedidas:Observación: Como ya sabemos que $\\sqrt{b^2-4ac}=$[[7]], entonces ahora reemplazamos en:
\n

         $-$[[1]]$+$[[7]]

\n

$x_1=$ _________________

\n

         $2\\cdot$[[0]] 

\n

Paso 5: Ahora debes desarrollar el numerador y denominador obteniendo:

\n		\n

       $-$[[1]]$-$[[7]]

\n

$x_2=$ _________________

\n

         $2\\cdot$[[0]] 

\n

Paso 5: Ahora debes desarrollar el numerador y denominador obteniendo:

\n
\n

             [[8]]

\n

$x_1=$ _______

\n

              [[10]]

\n		\n

            [[9]]

\n

$x_2=$ _______

\n

             [[10]]

\n
Paso 6: Simplificar las fracciones obtenidas en el paso 5, dejarla como entero en caso de serlo, y si es decimal dejarlo escrito con una sola cifra decimalPaso 6: Simplificar las fracciones obtenidas en el paso 5, dejarla como entero en caso de serlo, y si es decimal dejarlo escrito con una sola cifra decimal
$x_1=$[[11]]\n

$x_2=$[[12]]

\n
\n

    

\n

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Aun no ha contestado toda la pregunta que se ha solicitado responder.

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Ah pasado el tiempo del examen.

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Quedan 5 minutos para finalizar el examen.

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Estimado(a) estudiante por medio del siguiente examen se podrá usted ayudar para resolver la guía 4 de aprendizaje de NM2, sobre la resolución de ecuaciones de segundo grado.

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