Question

responda por favor
3x²+55=0
x²-6x=0
x²-10x+25=0

Answer 1

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

Resolvendo por Bháskara:

$3x^2 + 55 = 0$

a=1, b=0, c=55

Δ=b²−4ac
Δ=(0)²−4*(1)*(55)
Δ=0−220
Δ= −220

Sem solução para o conjunto dos números reais (R)

Solução para o conjunto dos números complexos (C)

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-0 \pm \sqrt{-220}}{2*1} \\ \\ \\ x = \dfrac{ \pm 2\sqrt{55}i}{2} \\ \\ \\ x = \dfrac{ + 2\sqrt{55}i}{2} \\ \\ \\ x = \sqrt{55}i \\ \\ \\ x = \dfrac{ - 2\sqrt{55}i}{2} \\ \\ \\ x = -\sqrt{55}i$


S = {$\sqrt{55}i , \ \ -\sqrt{55}i$}

===================


$x^{2} - 6x = 0$

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$


a=1, b=−6, c=0

Δ=b²−4ac
Δ=(−6)²−4*(1)*(0)
Δ=36+0
Δ=36

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-(-6) \pm \sqrt{36}}{2*1} \\ \\ \\ x = \dfrac{6 \pm 6 }{2} \\ \\ \\ x = \dfrac{6 + 6 }{2} \\ \\ \\ x = \dfrac{12}{2} \\ \\ \\ x' = 6 \\ \\ \\ x'' = \dfrac{6 - 6 }{2} x'' = \dfrac{0}{2} \\ \\ \\ x'' = 0$


S = {6, 0}

====================

$x^2 - 10x + 25 = 0$

a=1, b=−10, c=25

Δ=b²−4ac
Δ=(−10)²−4*(1)*(25)
Δ=100−100
Δ=0

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-(-10) \pm \sqrt{0}}{2*1} \\ \\ \\ x = \dfrac{10 \pm 0 }{2} \\ \\ \\ x = \dfrac{10 + 0 }{2} \\ \\ \\ x = \dfrac{10}{2} \\ \\ \\ x' = 5 \\ \\ \\ x'' = \dfrac{10 - 0 }{2} \\ \\ \\ x'' = \dfrac{10}{2} \\ \\ \\ x'' = 0$


S = {5, 0}