Question

¿cuales son las ecuaciones cuadráticas mediante fórmula gral de a). 2x²+4x+2=0.

b). 3x²+x+1=0


c) x²-x-20=0​

Answer 1

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

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

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

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

a) 2x² + 4x + 2 = 0

Tenemos que: a = 2, b = 4, c = 2  

Reemplazamos los coeficientes en la fórmula general:

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

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

b) 3x² + x + 1 = 0

Tenemos que: a = 3, b = 1, c = 1  

Reemplazamos los coeficientes en la fórmula general:

                                    $\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(1) \pm \sqrt{(1)^2 - [4(3)(1)]}}{2(3)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-1 \pm \sqrt{1 - (12)}}{6}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-1 \pm \sqrt{-11}}{6}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-1 \pm 3.31662i}{6}}$

$\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{-1 + 3.31662i}{6}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{-1}{6} + \dfrac{3.31662i}{6}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= -0.166667 + 0.552771i}}}}$             $\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{-1 - 3.31662i}{6}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{-1}{6} - \dfrac{3.31662i}{6}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= -0.166667 - 0.552771i}}}}$                        

c) x² - x - 20 = 0​

Tenemos que: a = 1, b = -1, c = -20  

Reemplazamos los coeficientes en la fórmula general:

                                   $\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(-1) \pm \sqrt{(-1)^2 - [4(1)(-20)]}}{2(1)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{1 \pm \sqrt{1 - (-80)}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{1 \pm \sqrt{81}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{1 \pm 9}{2}}$

                              $\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{1 + 9}{2}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{10}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= 5}}}}$                      $\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{1 - 9}{2}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{-8}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= -4}}}}$