Question

Determine o conjunto solução da equação a seguir: (obs: números complexos)

x² - x + 4 = 0​

Answer 1

Explicação passo-a-passo:

$\sf x^2-x+4=0$

$\sf \Delta=(-1)^2-4\cdot1\cdot4$

$\sf \Delta=1-16$

$\sf \Delta=-15$

$\sf \Delta=15\cdot(-1)$

$\sf \Delta=15i^2$

$\sf x=\dfrac{-(-1)\pm\sqrt{15i^2}}{2\cdot1}=\dfrac{1\pm i\sqrt{15}}{2}$

$\sf x'=\dfrac{1+i\sqrt{15}}{2}$

$\sf x"=\dfrac{1-i\sqrt{15}}{2}$

O conjunto solução é:

$\sf \red{S=\Big\{\dfrac{1-i\sqrt{15}}{2},\dfrac{1+i\sqrt{15}}{2}\Big\}}$

Answer 2

Oie, Td Bom?!

$x {}^{2} - x + 4 = 0$

• Coeficientes:

$a = 1 \: , \: b = - 1 \: ,\: c = 4$

• Fórmula resolutiva:

$x = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a}$

$x = \frac{ - ( - 1) ± \sqrt{( - 1) {}^{2} - 4 \: . \: 1 \: . \: 4} }{2 \: . \: 1}$

$x = \frac{1± \sqrt{1 - 16} }{2}$

$x = \frac{1± \sqrt{ - 15} }{2}$

$x = \frac{1± \sqrt{15} i}{2}$

• $x = \frac{1 + \sqrt{15} i}{2}$

$x = \frac{1}{2} + \frac{ \sqrt{15} i}{2}$

$x = \frac{1}{2} + \frac{ \sqrt{15} }{2} i$

• $x = \frac{1 - \sqrt{15}i }{2}$

$x = \frac{1}{2} - \frac{ \sqrt{15}i }{2}$

$x = \frac{1}{2} - \frac{ \sqrt{15} }{2} i$

$S = \left \{ \frac{1}{2} + \frac{ \sqrt{15} }{2} i \:, \: \frac{1}{2} - \frac{ \sqrt{15} }{2} i\right \}$

Att. Makaveli1996