Question

Considerando o conjunto dos números complexos, a solução da equação x²-14x+50=0​

Answer 1

Resposta:

$x'=7- i\\\\x''=7+i$

Explicação passo a passo:

$x^{2} -14x+50=0\\\\Para\ ax^{2} +bx+c=0:\\\\a=1\ ,\ b=(-14)\ ,\ c=50\\\\Fo\´rmula\ de\ Bhaskara:$

$x=\frac{-b\pm\sqrt{b^{2} -4ac} }{2a}\\\\x=\frac{-(-14)\pm\sqrt{(-14)^{2} -4(1)(50)} }{2(1)}\\\\x=\frac{14\pm\sqrt{(-14)^{2} -4(50)} }{2}\\\\x=\frac{14\pm\sqrt{196 -200} }{2}\\\\x=\frac{14\pm\sqrt{-4} }{2}$

$x=\frac{14\pm\sqrt{4}\sqrt{-1} }{2}\\\\x=\frac{14\pm\sqrt{2^{2} }i }{2}\\\\x=\frac{14\pm2i }{2}\\\\x=7\pm i$

$x'=7- i\\\\x''=7+i$