Question

Resolva as equações do segundo grau:

A. X²-8x+12=0
B. X²+2x-8=0
C. x²-5x+8=0
D. -x²+x+12=0
E. 3x²-12=0
F. x²+x-2=0​

Answer 1

$\boxed{\begin{array}{l}\tt a)~\sf x^2-8x+12=0\\\sf s=-\dfrac{b}{a}=-\dfrac{-8}{1}=8\\\sf p=\dfrac{c}{a}=\dfrac{12}{1}=12\\\sf x_1=6~~e~~x_2=2\\\sf pois~6+2=8~~e~6\cdot2=12\\\tt b)~\sf x^2+2x-8=0\\\sf s=-\dfrac{2}{1}=-2\\\sf p=\dfrac{-8}{1}=-8\\\sf x_1=2~~e~~x_2=-4\\\sf pois~2+(-4)=-2~~e~~2\cdot(-4)=-8\\\tt c)~\sf x^2-5x+8=0\\\sf \Delta=b^2-4ac\\\sf\Delta=(-5)^2-4\cdot1\cdot8\\\sf\Delta=25-32\\\sf\Delta=-7<0\implies n\tilde ao~possui~ra\acute izes~reais.\\\sf S=\bigg\{~~\bigg\}\end{array}}$

$\large\boxed{\begin{array}{l}\tt d)~\sf -x^2+x+12=0\\\sf s=-\dfrac{1}{-1}=1\\\sf p=\dfrac{12}{-1}=-12\\\sf x_1=4~~e~~x_2=-3\\\sf pois~4+(-3)=1~~e~~4\cdot(-3)=-12\\\tt e)~\sf3x^2-12=0\\\sf s=\dfrac{0}{1}=0\\\sf p=\dfrac{-12}{3}=-4\\\sf x_1=2~~e~~x_2=-2\\\sf pois~2+(-2)=0~~e~~2\cdot(-2)=-4\\\tt f)~\sf x^2+x-2=0\\\sf s=-\dfrac{1}{1}=-1~~p=\dfrac{-2}{1}=-2\\\sf x_1=1~~e~~x_2=-2\\\sf pois~~1+(-2)=-1~~e~~1\cdot(-2)=-2\end{array}}$

Answer 2

Resposta:

Explicação passo a passo:

A. X²-8x+12=0

A = 1; b = - 8; c= 12

∆= b² - 4ac

∆= (+8)² - 4.1.12

∆= 64 - 48

∆= 16

x = ( - b +/- √∆)/2a

x = [ -(-8) +/- √16]/2.1

x = [ 8 +/- 4]/2

x ' = (8+4)/2= 12/2= 6

X " = (8-4)/2 = 4/2= 2

R.: {6; 2}

____________

B. X²+2x-8=0

a = 1; b = 2; c = - 8

∆= b² - 4ac

∆= 2² - 4.1.(-8)

∆= 4+32

∆= 36

x = ( - b +/- √∆)/2a

x = [ - 2 +/- √36]/2.1

x = [ - 2 +/- 6]/2

x ' = [ - 2 - 6]/2 = - 8/2= - 4

x " = [ - 2 + 6]/2= 4/2= 2

R.: {2; -4}

________________

C. x²-5x+8=0

a = 1; b = - 5; c = 8

∆= b² - 4ac

∆= (-5)² - 4.1.8

∆= 25-32

∆= - 7

(Não há solução para os Números Reais)

∆ < 0

______________

D.

- x²+x+12=0

a = - 1; b = 1; c = 12

∆= b² - 4ac

∆= 1² - 4.(-1).12

∆= 1 + 4.12

∆= 1+48

∆= 49

x = ( - b +/- √∆)/2a

x = (- 1 +/- √49)/2.(-1)

x = (-1 +/- 7)/(-2)

x' = (-1 -7)/(-2)= -8/(-2) = 4

X " = (-1+7)/(-2)= 6/(-2)= - 3

R.: {4; -3}

===========

E. 3x²-12=0

3x² = 12

x² = 12/3

x² = 4

x = √4

x = +/- 2

S = {2; -2}

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

F. x²+x-2=0​

a= 1; b = 1; c = - 2

∆= b² - 4ac

∆= 1² - 4.1.(-2)

∆= 1 + 8

∆= 9

x = (- 1 +/- √9)/2.1

x = (-1 +/- 3)/2

X' = (-1 - 3)/2= -4/2= - 2

X" = (-1+3)/2= 2/2= 1

R.: {-2; 1}