Question

pode se afirmar que as raízes da equação x²-16x+64=0 são
​

Answer 1

Resposta:

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-16x+64=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-16~e~c=64\\\\C\'alculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-16)^{2}-4(1)(64)=256-(256)=0\\\\C\'alculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-16)-\sqrt{0}}{2(1)}=\frac{16-0}{2}=\frac{16}{2}=8\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-16)+\sqrt{0}}{2(1)}=\frac{16+0}{2}=\frac{16}{2}=8\\\\S=\{8,~8\}$

Answer 2

A=(b)²-4 (a)(c)=(-16)²-4(1)(64)=256-(256)=0