Matemáticas, pregunta formulada por anaissaavve, hace 9 meses

1.
Determina las soluciones de las siguientes ecuaciones de segundo grado.
a) –6x2 + 4x = 0
b) x2 - 4x + 4 = 0

Respuestas a la pregunta

Contestado por roycroos

Recordemos que una ecuación cuadrática tiene la siguiente forma:

$\boxed{\boldsymbol{ax^2+bx+c=0}}$

Por la fórmula general tenemos que:

$\boldsymbol{\boxed{x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}}$

a)

En el problema tenemos que: a = -6, b = 4, c = 0

Entonces reemplazamos

$\center x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\\center x_{1,2}=\dfrac{-(4)\pm \sqrt{(4)^2-[4(-6)(0)]}}{2(-6)}\\\\\\\center x_{1,2}=\dfrac{-4\pm \sqrt{16-(0)}}{-12}\\\\\\\center x_{1,2}=\dfrac{-4\pm \sqrt{16}}{-12}\\\\\\\center x_{1,2}=\dfrac{-4\pm4}{-12}\\\\\\\\$

$\center \Rightarrow\:x_{1}=\dfrac{-4+4}{-12}\\\\\\\center x_{1}=\dfrac{0}{-12}\\\\\\\center \boxed{\boxed{\boldsymbol{x_{1}= 0}}}\\\\\\\\\center \Rightarrow\:x_{2}=\dfrac{-4-4}{-12}\\\\\\\center x_{2}=\dfrac{-8}{-12}\\\\\\\center \boxed{\boxed{\boldsymbol{x_{2}=0.666667}}}$

b)

En el problema tenemos que: a = 1, b = -4, c = 4

Entonces reemplazamos

$\center x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\\center x_{1,2}=\dfrac{-(-4)\pm \sqrt{(-4)^2-[4(1)(4)]}}{2(1)}\\\\\\\center x_{1,2}=\dfrac{4\pm \sqrt{16-(16)}}{2}\\\\\\\center x_{1,2}=\dfrac{4\pm \sqrt{0}}{2}\\\\\\\center x_{1,2}=\dfrac{4\pm0}{2}\\\\\\\\$

$\center \Rightarrow\:x_{1}=\dfrac{4+0}{2}\\\\\\\center x_{1}=\dfrac{4}{2}\\\\\\\center \boxed{\boxed{\boldsymbol{x_{1}=2}}}\\\\\\\\\Rightarrow\:x_{2}=\dfrac{4-0}{2}\\\\\\\center x_{2}=\dfrac{4}{2}\\\\\\\center \boxed{\boxed{\boldsymbol{x_{2}=2}}}$