Question

calcula⁚(x1+1)(x2+1) si x1 y x2 son raices de la ecuacion ⁚x2-6x-8=0ç

Answer 1

Sin necesidad de resolver la ecuación cuadrática, podemos determinar la suma y producto de raíces.

    $\begin{array}{ccccc}&\begin{array}{c}\sf{ax^2+bx+c=0}\\\downarrow\\\sf{Siendo\ x_1\ y\ x_2\ las\ ra\acute{i}ces}\end{array}&\\\\\boxed{\begin{array}{c}\underline{\boldsymbol{\sf{Suma\ de\ ra\acute{i}ces}}}\\\\\sf{x_1 + x_2=-\dfrac{b}{a}}\end{array}}&&\boxed{\begin{array}{c}\underline{\boldsymbol{\sf{Producto\ de\ ra\acute{i}ces}}}\\\\\sf{x_1 \cdot x_2=\dfrac{c}{a}}\end{array}}\end{array}$

En el problema

   $\begin{array}{c}\sf{\underset{\displaystyle\underset{\,\,a}{\,\,\downarrow}}{\ 1}x^2\underset{\displaystyle\underset{\,\,b}{\,\,\downarrow}}{-\ 6}x\underset{\displaystyle\underset{\,\,c}{\,\,\downarrow}}{-\ 8}=0}\\\\\begin{array}{ccccccccccccccccccc}\boldsymbol{\bigcirc \kern-8pt \triangleright}\quad\sf{a=1}&&&&&&&&&\boldsymbol{\bigcirc \kern-8pt \triangleright}\quad\sf{b=-6}&&&&&&&&&\boldsymbol{\bigcirc \kern-8pt \triangleright}\quad\sf{c=-8}\end{array}\end{array}$

Nos piden

                               $\begin{array}{c}\sf{(x_1+1)(x_2+1)=x_1\cdot x_2+x_1 + x_2+1}\\\\\sf{(x_1+1)(x_2+1)=\underbrace{\sf{x_1\cdot x_2}}_{{Producto}\above 0pt{de\ ra\acute{i}ces}}+\underbrace{\sf{x_1 + x_2}}_{{Suma}\above 0pt{de\ ra\acute{i}ces}}+1}\\\\\sf{(x_1+1)(x_2+1)=\dfrac{c}{a}+\left(-\dfrac{b}{a}\right)+1}\\\\\sf{(x_1+1)(x_2+1)=\dfrac{-8}{1}+\left(-\dfrac{(-6)}{1}\right)+1}\\\\\sf{(x_1+1)(x_2+1)=-8+6+1}\\\\\boxed{\boxed{\boldsymbol{\sf{(x_1+1)(x_2+1)=-1}}}}\end{array}$                               

Rpta. -1.

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