Asignatura: Matemáticas
Autor: mariacasanovaaranda
hace 3 años

x²-6x+5 fórmula general

x²+4x-5 ayuda porfavor

Respuestas

Respuesta dada por: roycroos

╔═════════════════════════════════════════════╗ $\center \mathrm{Una ecuaci\'on cuadr\'atica de la forma:}$

$\mathrm{ax^2 + bx + c=0\:\:donde\:\: a \neq 0}$

$\center \mathrm{Poseer\'a 2 soluciones x_1} \:\mathrm{y\:x_2,\:las\:cuales\:determinaremos\:por:}$

$\boldsymbol{{\mathrm{x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}}} \Rightarrow \boxed{\mathrm{F\'ormula\:general}}$ ╚═════════════════════════════════════════════╝

Problema 1

$\mathsf{\underbrace{\boldsymbol{1}}_{a}x^2\:\underbrace{\boldsymbol{-\:\:\:6}}_{b}x\:+\:\underbrace{\boldsymbol{5}}_{c}=0}$

Entonces a = 1, b = -6, c = 5

Reemplazamos los coeficientes en la fórmula general:

$\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(-6) \pm \sqrt{(-6)^2 - [4(1)(5)]}}{2(1)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{6 \pm \sqrt{36 - (20)}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{6 \pm \sqrt{16}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{6 \pm 4}{2}}$

$\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{6 + 4}{2}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{10}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= 5}}}}$ $\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{6 - 4}{2}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{2}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= 1}}}}$

Problema 2

$\mathsf{\underbrace{\boldsymbol{1}}_{a}x^2\:+\:\underbrace{\boldsymbol{4}}_{b}x\:\underbrace{\boldsymbol{-\:\:\:5}}_{c}=0}$

Entonces a = 1, b = 4, c = -5

Reemplazamos los coeficientes en la fórmula general:

$\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(4) \pm \sqrt{(4)^2 - [4(1)(-5)]}}{2(1)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-4 \pm \sqrt{16 - (-20)}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-4 \pm \sqrt{36}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-4 \pm 6}{2}}$

$\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{-4 + 6}{2}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{2}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= 1}}}}$ $\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{-4 - 6}{2}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{-10}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= -5}}}}$