Question

equação de 2° grau x2 + 18x + 17 =0​

Answer 1

Resposta:

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+18x+17=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=18~e~c=17\\\\\Delta=(b)^{2}-4(a)(c)=(18)^{2}-4(1)(17)=324-(68)=256\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(18)-\sqrt{256}}{2(1)}=\frac{-18-16}{2}=\frac{-34}{2}=-17\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(18)+\sqrt{256}}{2(1)}=\frac{-18+16}{2}=\frac{-2}{2}=-1\\\\S=\{-17,~-1\}$

Answer 2

${x}^{2} + 18x + 17 = 0$

Delta = b^2 - 4 . a . c

Delta = 18^2 - 4 . 1 . 17

Delta = 324 - 68

Delta = 256

x = -b +- raiz de delta / 2 . a

x = -18 +- raiz de 256 / 2 . 1

x = -18 +- 16 / 2

x' = -18 + 16 / 2

x' = -2 / 2 = -1

x'' = -18 - 16 / 2

x'' = -34 / 2 = - 17