【Rpta.】La solución de la ecuación cuadrática es de 2 y 0.5
[tex]\green{{\hspace{50 pt}\above 1.2pt}\boldsymbol{\mathsf{Procedimiento}}{\hspace{50pt}\above 1.2pt}}[/tex]
Recordemos que una ecuación cuadrática es aquella cuyo mayor exponente de la variable es 2 y tiene la siguiente forma.
[tex]\mathsf{ax^2 + bx + c=0\:\:donde\:\: a \neq 0}[/tex]
Por definición sabemos que poseerá 2 soluciones(reales o imaginarias) y la determinaremos por la fórmula general.
[tex]{\boldsymbol{{\mathsf{x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}}}}\atop{\displaystyle \downarrow \atop \boxed{\boldsymbol{\mathsf{F\acute{o}rmula\:general}}}}[/tex]
Entonces de nuestro problema extraemos los coeficientes:
[tex]\mathsf{\underbrace{\boldsymbol{2}}_{a}x^2\:\underbrace{\boldsymbol{-\:\:5}}_{b}x\:+\:\underbrace{\boldsymbol{2}}_{c}=0}[/tex]
Reemplazamos estos valores en la fórmula general:
[tex]\begin{array}{c}\mathsf{x_{1,2}} \mathsf{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\mathsf{x_{1,2}} \mathsf{= \dfrac{-(-5) \pm \sqrt{(-5)^2 - [4(2)(2)]}}{2(2)}}\\\\\\\mathsf{x_{1,2}} \mathsf{= \dfrac{5 \pm \sqrt{25 - (16)}}{4}}\\\\\\\mathsf{x_{1,2}} \mathsf{= \dfrac{5 \pm \sqrt{9}}{4}}\\\\\\\mathsf{x_{1,2}} \mathsf{= \dfrac{5 \pm 3}{4}}\end{array}[/tex]
[tex]\begin{array}{c}\boldsymbol{\Rightarrow}\:\:\:\mathsf{x_{1}} \mathsf{= \dfrac{5 + 3}{4}}\\\\\\\mathsf{\hspace{0 pt}x_{1}} \mathsf{= \dfrac{8}{4}}\\\\\\\mathsf{\hspace{0 pt}\boxed{\boxed{\boldsymbol{\mathsf{x_{1}} \mathsf{= 2}}}}}\end{array}[/tex] [tex]\begin{array}{c}\boldsymbol{\Rightarrow}\:\:\:\mathsf{x_{2}} \mathsf{= \dfrac{5 - 3}{4}}\\\\\\\mathsf{\hspace{0 pt}x_{2}} \mathsf{= \dfrac{2}{4}}\\\\\\\mathsf{\hspace{0 pt}\boxed{\boxed{\boldsymbol{\mathsf{x_{2}} \mathsf{= 0.5}}}}}\end{array}[/tex]
⚠ La gráfica en la imagen solo es para comprobar nuestros resultados.
✠ Tareas similares
➫ https://brainly.lat/tarea/19356545
➫ https://brainly.lat/tarea/20302726
➫ https://brainly.lat/tarea/19393037
[tex]\boxed{\sf{{R}}\quad\raisebox{10pt}{$\sf{\red{O}}$}\!\!\!\!\raisebox{-10pt}{$\sf{\red{O}}$}\quad\raisebox{15pt}{$\sf{{G}}$}\!\!\!\!\raisebox{-15pt}{$\sf{{G}}$}\quad\raisebox{15pt}{$\sf{\red{H}}$}\!\!\!\!\raisebox{-15pt}{$\sf{\red{H}}$}\quad\raisebox{10pt}{$\sf{{E}}$}\!\!\!\!\raisebox{-10pt}{$\sf{{E}}$}\quad\sf{\red{R}}}\hspace{-64.5pt}\rule{10pt}{.2ex}\:\rule{3pt}{1ex}\rule{3pt}{1.5ex}\rule{3pt}{2ex}\rule{3pt}{1.5ex}\rule{3pt}{1ex}\:\rule{10pt}{.2ex}[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Verified answer
【Rpta.】La solución de la ecuación cuadrática es de 2 y 0.5
[tex]\green{{\hspace{50 pt}\above 1.2pt}\boldsymbol{\mathsf{Procedimiento}}{\hspace{50pt}\above 1.2pt}}[/tex]
Recordemos que una ecuación cuadrática es aquella cuyo mayor exponente de la variable es 2 y tiene la siguiente forma.
[tex]\mathsf{ax^2 + bx + c=0\:\:donde\:\: a \neq 0}[/tex]
Por definición sabemos que poseerá 2 soluciones(reales o imaginarias) y la determinaremos por la fórmula general.
[tex]{\boldsymbol{{\mathsf{x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}}}}\atop{\displaystyle \downarrow \atop \boxed{\boldsymbol{\mathsf{F\acute{o}rmula\:general}}}}[/tex]
Entonces de nuestro problema extraemos los coeficientes:
[tex]\mathsf{\underbrace{\boldsymbol{2}}_{a}x^2\:\underbrace{\boldsymbol{-\:\:5}}_{b}x\:+\:\underbrace{\boldsymbol{2}}_{c}=0}[/tex]
Reemplazamos estos valores en la fórmula general:
[tex]\begin{array}{c}\mathsf{x_{1,2}} \mathsf{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\mathsf{x_{1,2}} \mathsf{= \dfrac{-(-5) \pm \sqrt{(-5)^2 - [4(2)(2)]}}{2(2)}}\\\\\\\mathsf{x_{1,2}} \mathsf{= \dfrac{5 \pm \sqrt{25 - (16)}}{4}}\\\\\\\mathsf{x_{1,2}} \mathsf{= \dfrac{5 \pm \sqrt{9}}{4}}\\\\\\\mathsf{x_{1,2}} \mathsf{= \dfrac{5 \pm 3}{4}}\end{array}[/tex]
[tex]\begin{array}{c}\boldsymbol{\Rightarrow}\:\:\:\mathsf{x_{1}} \mathsf{= \dfrac{5 + 3}{4}}\\\\\\\mathsf{\hspace{0 pt}x_{1}} \mathsf{= \dfrac{8}{4}}\\\\\\\mathsf{\hspace{0 pt}\boxed{\boxed{\boldsymbol{\mathsf{x_{1}} \mathsf{= 2}}}}}\end{array}[/tex] [tex]\begin{array}{c}\boldsymbol{\Rightarrow}\:\:\:\mathsf{x_{2}} \mathsf{= \dfrac{5 - 3}{4}}\\\\\\\mathsf{\hspace{0 pt}x_{2}} \mathsf{= \dfrac{2}{4}}\\\\\\\mathsf{\hspace{0 pt}\boxed{\boxed{\boldsymbol{\mathsf{x_{2}} \mathsf{= 0.5}}}}}\end{array}[/tex]
⚠ La gráfica en la imagen solo es para comprobar nuestros resultados.
✠ Tareas similares
➫ https://brainly.lat/tarea/19356545
➫ https://brainly.lat/tarea/20302726
➫ https://brainly.lat/tarea/19393037
[tex]\boxed{\sf{{R}}\quad\raisebox{10pt}{$\sf{\red{O}}$}\!\!\!\!\raisebox{-10pt}{$\sf{\red{O}}$}\quad\raisebox{15pt}{$\sf{{G}}$}\!\!\!\!\raisebox{-15pt}{$\sf{{G}}$}\quad\raisebox{15pt}{$\sf{\red{H}}$}\!\!\!\!\raisebox{-15pt}{$\sf{\red{H}}$}\quad\raisebox{10pt}{$\sf{{E}}$}\!\!\!\!\raisebox{-10pt}{$\sf{{E}}$}\quad\sf{\red{R}}}\hspace{-64.5pt}\rule{10pt}{.2ex}\:\rule{3pt}{1ex}\rule{3pt}{1.5ex}\rule{3pt}{2ex}\rule{3pt}{1.5ex}\rule{3pt}{1ex}\:\rule{10pt}{.2ex}[/tex]