Resolver las siguientes inecuaciones cuadráticas y presentar el conjunto solución en forma de intervalo y gráficamente. 1) X2 < 10 – 3x 2) 2x2 + 3x ≥ 2

Answers

Answer 1

Respuesta:

1) (-5,2)

2) [tex](-\infty,-2]U(\frac{1}{2},\infty)[/tex]

Explicación paso a paso:

1)

[tex]x^{2}<10-3x[/tex]

Para comenzar este problema, debemos moverlo todo al lado izquierdo de la inecuación, por lo que obtenemos:

[tex]x^{2}+3x-10<0[/tex]

Ahora podemos factorizar el lado izquierdo para obtener:

[tex](x+5)(x-2)<0[/tex]

Ahora podemos cambiar el símbolo < por un = para encontrar los valores de x en los cuales la inecuación es igual a cero.

(x+5)(x-2)=0

Y luego despejamos x.

x+5=0

x=-5

y

x-2=0

x=2

Ahora construimos nuestros intervalos posibles.

[tex](-\infty,-5)[/tex]

(-5,2)

y

[tex](2,\infty)[/tex]

Y escogemos algunos valores de prueba. Estos nos ayudarán a determinar si cada intervalo hace que la inecuación sea verdadera o falsa.

[tex](-\infty,-5)[/tex]

Para este intervalo escojamos -6 y evaluemoslo en la inecuación.

(x+5)(x-2)<0

(-6+5)(-6-2)<0

(-1)(-8)<0

8<0

falso, así que este intervalo no es parte de nuestra respuesta.

(-5,2)

para este escojamos x=0 y probémoslo en la inecuación.

(x+5)(x-2)<0

(0+5)(0-2)<0

(5)(-2)<0

-10<0

verdadero, así que este intervalo es parte de nuestra respuesta.

[tex](2,\infty)[/tex]

Para este escojamos 3 y probémoslo en unestra inecuación.

(x+5)(x-2)<0

(3+5)(3-2)<0

(8)(1)<0

8<0

falso, así que este intervalo no es parte de nuestra respuesta.

así que nuestra respuesta es:  (-5,2)

Vea imagen adjunta para representación gráfica.

2)

[tex]2x^{2}+3x\geq2[/tex]

Para resolver este problema, comenzamos moviéndolo todo al lado izquierdo de la inecuación.

[tex]2x^{2}+3x-2\geq0[/tex]

Ahora podemos factorizar el lado izquierdo de la inecuación para obtener:

[tex](2x-1)(x+2)\geq0[/tex]

Ahora podemos cambién el símbolo ≥ por un símbolop de = para obtener los valores de x que hacen que la inecuación sea igual a 0.

(2x-1)(x+2)=0

y ahora despejamos x.

2x-1=0

[tex]x=\frac{1}{2}[/tex]

y

x+2=0

x=-2

Ahora construimos nuestros intervalos posible.

[tex](-\infty,-2][/tex]

[tex][-2,\frac{1}{2}][/tex]

y

[tex][\frac{1}{2},\infty)[/tex]

Ahora escogemos los valores de prueba correspondientes.

[tex](-\infty,-2][/tex]

para este, escojamos -3 y probémoslo en la inecuación.

[tex](2x-1)(x+2)\geq0[/tex]

[tex](2(-3)-1)(-3+2)\geq0[/tex]

[tex](-7)(-1)\geq0[/tex]

[tex]7\geq0[/tex]

verdadero, así que este intervalo es parte de nuestra respuesta.

[tex][-2,\frac{1}{2}][/tex]

para este, utilicemos 0 como valor de prueba.

[tex](2x-1)(x+2)\geq0[/tex]

[tex](2(0)-1)(0+2)\geq0[/tex]

[tex](-1)(2)\geq0[/tex]

[tex]-2\geq0[/tex]

falso, así que este intervalo no es parte de nuestra respuesta.

[tex][\frac{1}{2},\infty)[/tex]

para este, utilicemos 1 como valor de prueba.

[tex](2x-1)(x+2)\geq0[/tex]

[tex](2(1)-1)(1+2)\geq0[/tex]

[tex](1)(3)\geq0[/tex]

[tex]3\geq0[/tex]

verdadero, así que este intervalo es parte de nuestra respuesta.

Así que nuestra respuesta es la unión entre los dos intervalos que resultaron verdadero, por lo que nuestra respuesta es:

[tex](-\infty,-2]U[\frac{1}{2},\infty)[/tex]

Vea la representación gráfica en la imagen adjunta.

Resolver Las Siguientes Inecuaciones Cuadrticas Y Presentar El Conjunto Solucin En Forma De Intervalo

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