Question

As raízes da equação -8x³+40x²-48x=0 são

A) 0,5 e 6
B) 0,5 e 1
C) 0,2 e 3
D) 0,-2 e -3
E) 0,-5 e -1

Answer 1

Equação do 3º:

Como falta o termo d , podemos colocar x em evidencia e tornar uma equação do 2º:

$\mathtt{-8x^3+40x^2-48x=0} \\ \\ \mathtt{x(-8x^2+ 40x- 48) = 0} \\ \\ \mathtt{x'= 0} \\ \\ \\ \mathtt{- 8x^2+40x-48 = 0} \\ \\ \mathtt{a = -8} \\ \mathtt{b = 40} \\ \mathtt{c = -48} \\ \\ \mathtt{\Delta=b^2-4~. ~a~.~c } \\ \mathtt{\Delta= 40^2-4~.~(-8)~.~(-48)} \\ \mathtt{\Delta= 1.600+32~.~(-48) } \\ \mathtt{\Delta= 1.600-1.536} \\ \mathtt{\Delta=64 }$

$\mathtt{\dfrac{-b+-~\sqrt{\Delta}}{2~.~a}~~=~~\dfrac{-40+-~\sqrt{64}}{2~.~(-8)}~~=~~\dfrac{-40+-~8}{-16}} \\ \\ \\ \mathtt{x' = \dfrac{-40+8}{-16}~~=~~\dfrac{-32}{-16}~~=~~\dfrac{32}{16}~~=~~2} \\ \\ \\ \mathtt{x''= \dfrac{-40-8}{-16}~~=~~\dfrac{-48}{-16}~~=~~\dfrac{48}{16}~~=~~3} \\ \\ \\ \boxed{\boxed{\mathtt{Resposta: Letra~C)~~~x = 0~~~~x'= 2~~~~x''' = 3}}}$