Question

Fórmula general con procedimiento por favor.. x2+11x+24=0, 3x-5x+2=0, x2-100=0

Answer 1

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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 = 1, b = 11, c = 24  

   

Entonces reemplazamos

                                    $x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x_{1,2}=\dfrac{-(11)\pm \sqrt{(11)^2-[4(1)(24)]}}{2(1)}\\\\\\x_{1,2}=\dfrac{-11\pm \sqrt{121-(96)}}{2}\\\\\\x_{1,2}=\dfrac{-11\pm \sqrt{25}}{2}\\\\\\x_{1,2}=\dfrac{-11\pm5}{2}\\\\\\\\\Rightarrow\:x_{1}=\dfrac{-11+5}{2}\\\\\\x_{1}=\dfrac{-6}{2}\\\\\\\boxed{\boxed{\boldsymbol{x_{1}=-3}}}\\\\\\\\\Rightarrow\:x_{2}=\dfrac{-11-5}{2}\\\\\\x_{2}=\dfrac{-16}{2}\\\\\\\boxed{\boxed{\boldsymbol{x_{2}=-8}}}$

b) En el problema tenemos que: a = 3, b = -5, c = 2  

   

Entonces reemplazamos

                                    $x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x_{1,2}=\dfrac{-(-5)\pm \sqrt{(-5)^2-[4(3)(2)]}}{2(3)}\\\\\\x_{1,2}=\dfrac{5\pm \sqrt{25-(24)}}{6}\\\\\\x_{1,2}=\dfrac{5\pm \sqrt{1}}{6}\\\\\\x_{1,2}=\dfrac{5\pm1}{6}\\\\\\\\\Rightarrow\:x_{1}=\dfrac{5+1}{6}\\\\\\x_{1}=\dfrac{6}{6}\\\\\\\boxed{\boxed{\boldsymbol{x_{1}=1}}}\\\\\\\\\Rightarrow\:x_{2}=\dfrac{5-1}{6}\\\\\\x_{2}=\dfrac{4}{6}\\\\\\\boxed{\boxed{\boldsymbol{x_{2}=0.666667}}}$

c) En el problema tenemos que: a = 1, b = 0, c = -100  

   

Entonces reemplazamos

                                    $x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x_{1,2}=\dfrac{-(0)\pm \sqrt{(0)^2-[4(1)(-100)]}}{2(1)}\\\\\\x_{1,2}=\dfrac{-0\pm \sqrt{0-(-400)}}{2}\\\\\\x_{1,2}=\dfrac{-0\pm \sqrt{400}}{2}\\\\\\x_{1,2}=\dfrac{-0\pm20}{2}\\\\\\\\\Rightarrow\:x_{1}=\dfrac{-0+20}{2}\\\\\\x_{1}=\dfrac{20}{2}\\\\\\\boxed{\boxed{\boldsymbol{x_{1}=10}}}\\\\\\\\\Rightarrow\:x_{2}=\dfrac{-0-20}{2}\\\\\\x_{2}=\dfrac{-20}{2}\\\\\\\boxed{\boxed{\boldsymbol{x_{2}=-10}}}$