Question

Quantos 1ue e 5×^(2)+4×-8=0

Answer 1

Resposta:

Explicação passo a passo:

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~5x^{2}+4x-8=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=5{;}~b=4~e~c=-8\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(4)^{2}-4(5)(-8)=16-(-160)=176$

Fatorar 176

176 | 2

88 | 2

44 | 2

22 | 2

11 | 11

1 | 1

176 = 2 . 2 . 2 . 2 . 11 = 2⁴.11

√176 = √2⁴.11 = 2²√11=4√11

$\displaystyle C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(4)-\sqrt{176}}{2(5)}=\frac{-(4)-4\sqrt{11}}{2(5)}=\frac{-2-2\sqrt{11}}{5}\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(4)+\sqrt{176}}{2(5)}=\frac{-(4)+4\sqrt{11}}{2(5)}=\frac{-2+2\sqrt{11}}{5}\\\\S=\{\frac{-2-2\sqrt{11}}{5},~\frac{-2+2\sqrt{11}}{5}\}$