Question

resuelve las siguientes ecuaciones de segundo grado

a) 3x²-5x+1=0

b) x2+3x+2=0

c) x²+6x-7=0

d) 2x²-5x+3=0

ayudenme por favor es urgente...​

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) 3x² - 5x + 1 = 0

Del problema tenemos que: a = 3, b = -5, c = 1      

Reemplazamos los coeficientes en la fórmula general:

                                    $\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(-5) \pm \sqrt{(-5)^2 - [4(3)(1)]}}{2(3)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{5 \pm \sqrt{25 - (12)}}{6}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{5 \pm \sqrt{13}}{6}}$                                                                       $\center \Rightarrow\: \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= \dfrac{5+\sqrt{13}}{6}}}}}$                      $\center \Rightarrow\: \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= \dfrac{5-\sqrt{13}}{6}}}}}$

 

b) x² + 3x + 2 = 0

Del problema tenemos que: a = 1, b = 3, c = 2

Reemplazamos los coeficientes en la fórmula general:

                                     $\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(3) \pm \sqrt{(3)^2 - [4(1)(2)]}}{2(1)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-3 \pm \sqrt{9 - (8)}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-3 \pm \sqrt{1}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-3 \pm 1}{2}}$                                

                        $\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{-3 + 1}{2}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{-2}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= -1}}}}$                     $\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{-3 - 1}{2}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{-4}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= -2}}}}$

c) x² + 6x - 7 = 0

Del problema tenemos que: a = 1, b = 6, c = -7

Reemplazamos los coeficientes en la fórmula general:

                                    $\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(6) \pm \sqrt{(6)^2 - [4(1)(-7)]}}{2(1)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-6 \pm \sqrt{36 - (-28)}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-6 \pm \sqrt{64}}{2}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-6 \pm 8}{2}}$            

                          $\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{-6 + 8}{2}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{2}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= 1}}}}$                     $\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{-6 - 8}{2}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{-14}{2}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= -7}}}}$

d) 2x² - 5x + 3 = 0

Del problema tenemos que: a = 2, b = -5, c = 3

Reemplazamos los coeficientes en la fórmula general:

                                    $\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(-5) \pm \sqrt{(-5)^2 - [4(2)(3)]}}{2(2)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{5 \pm \sqrt{25 - (24)}}{4}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{5 \pm \sqrt{1}}{4}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{5 \pm 1}{4}}$

                               $\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{5 + 1}{4}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{6}{4}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= 1.5}}}}$                    $\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{5 - 1}{4}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{4}{4}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= 1}}}}$

                                                                                                          〆ʀᴏɢʜᴇʀ ✌