╔═════════════════════════════════════════════╗ [tex]\center \mathrm{Una ecuaci\'on cuadr\'atica de la forma:}[/tex]
[tex]\mathrm{ax^2 + bx + c=0\:\:donde\:\: a \neq 0}[/tex]
[tex]\center \mathrm{Poseer\'a 2 soluciones x_1} \:\mathrm{y\:x_2,\:las\:cuales\:determinaremos\:por:}[/tex]
[tex]\boldsymbol{{\mathrm{x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}}} \Rightarrow \boxed{\mathrm{F\'ormula\:general}}[/tex]╚═════════════════════════════════════════════╝
En el problema
[tex]\mathsf{\underbrace{\boldsymbol{1}}_{a}x^2\:\underbrace{\boldsymbol{-\:\:\:6}}_{b}x\:+\:\underbrace{\boldsymbol{8}}_{c}=0}[/tex]
Entonces a = 1, b = -6, c = 8
Reemplazamos los coeficientes en la fórmula general:
[tex]\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(-6) \pm \sqrt{(-6)^2 - [4(1)(8)]}}{2(1)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{6 \pm \sqrt{36 - (32)}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{6 \pm \sqrt{4}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{6 \pm 2}{2}}[/tex]
[tex]\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{6 + 2}{2}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{8}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= 4}}}}[/tex] [tex]\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{6 - 2}{2}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{4}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= 2}}}}[/tex]
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╔═════════════════════════════════════════════╗ [tex]\center \mathrm{Una ecuaci\'on cuadr\'atica de la forma:}[/tex]
[tex]\mathrm{ax^2 + bx + c=0\:\:donde\:\: a \neq 0}[/tex]
[tex]\center \mathrm{Poseer\'a 2 soluciones x_1} \:\mathrm{y\:x_2,\:las\:cuales\:determinaremos\:por:}[/tex]
[tex]\boldsymbol{{\mathrm{x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}}} \Rightarrow \boxed{\mathrm{F\'ormula\:general}}[/tex]╚═════════════════════════════════════════════╝
En el problema
[tex]\mathsf{\underbrace{\boldsymbol{1}}_{a}x^2\:\underbrace{\boldsymbol{-\:\:\:6}}_{b}x\:+\:\underbrace{\boldsymbol{8}}_{c}=0}[/tex]
Entonces a = 1, b = -6, c = 8
Reemplazamos los coeficientes en la fórmula general:
[tex]\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(-6) \pm \sqrt{(-6)^2 - [4(1)(8)]}}{2(1)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{6 \pm \sqrt{36 - (32)}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{6 \pm \sqrt{4}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{6 \pm 2}{2}}[/tex]
[tex]\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{6 + 2}{2}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{8}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= 4}}}}[/tex] [tex]\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{6 - 2}{2}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{4}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= 2}}}}[/tex]
〆ʀᴏɢʜᴇʀ ✌
Tareas similares:
✦ https://brainly.lat/tarea/19356545
✦ https://brainly.lat/tarea/20302726
✦ https://brainly.lat/tarea/19393037
✦ https://brainly.lat/tarea/19390997
✦ https://brainly.lat/tarea/19356545