Question

as raízes da função f(x)= x^2+ax+b são 4 e -8. Nessas condições, calcule os valores de a e b

Answer 1

Vamos considerar:
$f(x)=ax^2+bx+c$

Desta forma, temos que calcular os coeficientes 'b' e 'c'.


$x'=4\qquad\qquad\qquad{x''}=-8\\\\\\x'+x''=\dfrac{-b+\sqrt\Delta}{2a}+\dfrac{-b-\sqrt\Delta}{2a}\\\\\\(4)+(-8)=\dfrac{(-b+\sqrt\Delta)+(-b-\sqrt\Delta)}{2a}\\\\\\4-8=\dfrac{-b+\sqrt{\Delta}-b-\sqrt\Delta}{2a}\\\\\\-4=\dfrac{-2b}{2a}\\\\\\-4=-\dfrac{b}{a}\qquad\qquad\qquad\text{Observe que na fun\c{c}}\mathrm{\tilde{a}o\ do\ enunciado:\ }a=1\\\\\\-4=-\dfrac{b}{a}\\\\\\-4=-\dfrac{b}{1}\\\\\\-4=-b\\\\\boxed{b=4}$



$x'\cdot{x}''=\dfrac{-b+\sqrt\Delta}{2a}\cdot\dfrac{-b-\sqrt{\Delta}}{2a}\\\\\\(4)\cdot(-8)=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\cdot\dfrac{-b-\sqrt{b^2-4ac}}{2a}\\\\\\-32=\dfrac{(-b+\sqrt{b^2-4ac})\cdot(-b-\sqrt{b^2-4ac})}{(2a)\cdot(2a)}\\\\\\-32=\dfrac{(-b)^2-(\sqrt{b^2-4ac})^2}{4a^2}\\\\\\-32=\dfrac{b^2-b^2-4ac}{4a^2}\\\\\\-32=\dfrac{-4ac}{4a^2}\\\\\\-32=-\dfrac{c}{a}\\\\\\-32=\dfrac{c}{1}\\\\\\\boxed{-32=c}$