Question

Determine o conjunto solução de cada equação em C (números complexos) a) Dois x ao quadrado menos 8x+ 10= 0 b) 3 x ao quadrado menos 18x+30=0 c) 3 x ao quadrado menos 4x+2=0

Answer 1

Explicação passo-a-passo:

a) $\sf 2x^2-8x+10=0$

$\sf \Delta=(-8)^2-4\cdot2\cdot10$

$\sf \Delta=64-80$

$\sf \Delta=-16$

$\sf \Delta=16\cdot(-1)$

$\sf \Delta=16i^2$

$\sf x=\dfrac{-(-8)\pm\sqrt{16i^2}}{2\cdot2}=\dfrac{8\pm4i}{4}$

$\sf x'=\dfrac{8+4i}{4}~\rightarrow~\red{x'=2+i}$

$\sf x"=\dfrac{8-4i}{4}~\rightarrow~\red{x"=2-i}$

$\sf S=\{2-i,2+i\}$

b) $\sf 3x^2-18x+30=0$

$\sf \Delta=(-18)^2-4\cdot3\cdot30$

$\sf \Delta=324-360$

$\sf \Delta=-36$

$\sf \Delta=36\cdot(-1)$

$\sf \Delta=36i^2$

$\sf x=\dfrac{-(-18)\pm\sqrt{36i^2}}{2\cdot3}=\dfrac{18\pm6i}{6}$

$\sf x'=\dfrac{18+6i}{6}~\rightarrow~\red{x'=3+i}$

$\sf x"=\dfrac{18-6i}{6}~\rightarrow~\red{x"=3-i}$

$\sf S=\{3-i,3+i\}$

c) $\sf 3x^2-4x+2=0$

$\sf \Delta=(-4)^2-4\cdot3\cdot2$

$\sf \Delta=16-24$

$\sf \Delta=-8$

$\sf \Delta=8\cdot(-1)$

$\sf \Delta=8i^2$

$\sf x=\dfrac{-(-4)\pm\sqrt{8i^2}}{2\cdot3}=\dfrac{4\pm2\sqrt{2}\cdot i}{6}$

$\sf x'=\dfrac{4+2\sqrt{2}\cdot i}{6}~\rightarrow~\red{x'=\dfrac{2+\sqrt{2}\cdot i}{3}}$

$\sf x"=\dfrac{4-2\sqrt{2}\cdot i}{6}~\rightarrow~\red{x"=\dfrac{2-\sqrt{2}\cdot i}{3}}$

$\sf S=\left\{\dfrac{2-\sqrt{2}\cdot i}{3},\dfrac{2+\sqrt{2}\cdot i}{3}\right\}$