Question

alguém poderia me ajudar pfv é urgente
1)resolva,no universo do numeros complexos,as equações:
a)x²+36=0
b)x²-6x+13=0
c)4x²-4x+5=0
d)x²+49=0​

Answer 1

Explicação passo-a-passo:

a) $x^2+36=0$

$x^2=-36$

$x^2=36\cdot(-1)$

$x^2=36i^2$

$x\pm\sqrt{36}$

$x'=6i$

$x"=-6i$

b) $x^2-6x+13=0$

$\Delta=(-6)^2-4\cdot1\cdot13$

$\Delta=36-52$

$\Delta=-16$

$\Delta=16i^2$

$x=\dfrac{-(-6)\pm\sqrt{16i^2}}{2\cdot1}=\dfrac{6\pm4i}{2}$

$x'=\dfrac{6+4i}{2}~\longrightarrow~x'=3+2i$

$x"=\dfrac{6-4i}{2}~\longrightarrow~x"=3-2i$

c) $4x^2-4x+5=0$

$\Delta=(-4)^2-4\cdot4\cdot5$

$\Delta=16-80$

$\Delta=-64$

$\Delta=64i^2$

$x=\dfrac{-(-4)\pm\sqrt{64i^2}}{2\cdot4}=\dfrac{4\pm8i}{8}$

$x'=\dfrac{4+8i}{8}~\longrightarrow~x'=\dfrac{1+2i}{2}$

$x"=\dfrac{4-8i}{8}~\longrightarrow~x"=\dfrac{1-2i}{2}$

d) $x^2+49=0$

$x^2=-49$

$x^2=49\cdot(-1)$

$x^2=49i^2$

$x\pm\sqrt{49}$

$x'=7i$

$x"=-7i$