Question

3) Determine as raízes (x' e x'') da função f(t)=-4t+2t² usando a formula resolutiva

Answer 1

Resposta:

Explicação passo a passo:

Fórmula resolutiva ou Bhaskara:

f(t)=-4t+2t²

f(t)=2t²-4t

Para acharmos as raízes faça f(t)=0

2t²-4t=0

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~2t^{2}-4t=0~~e~comparando~com~(a)t^{2}+(b)t+(c)=0,~determinamos~os~coeficientes:~\\a=2{;}~b=-4~e~c=0\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-4)^{2}-4(2)(0)=16-(0)=16\\\\C\acute{a}lculo~das~raizes:&\\t^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-4)-\sqrt{16}}{2(2)}=\frac{4-4}{4}=\frac{0}{4}=0\\\\t^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-4)+\sqrt{16}}{2(2)}=\frac{4+4}{4}=\frac{8}{4}=2\\\\S=\{0,~2\}$