Respostas

2014-07-16T22:00:30-03:00
A)   2x²-3x+1=0
a = 2
b = - 3
c = 1
Δ = b² - 4ac
Δ = (-3)² - 4(2)(1)
Δ =  9 - 8
Δ = 1 -----------------------------√1 = 1
se
Δ > 0
então (baskara)
x = - b + √Δ/2a
x' = -(-3) - √1/2(2)
x' = + 3 - 1/4
x' = 2/4
x' = 1/2
e
x" = -(-3) + √1/2(2)
x" = + 3 + 1/4
x" = 4/4 
x" = 1

b)   x²-2x-3=0
a = 1
b = -2
c = - 3
Δ = b² - 4ac
Δ = (-2)² - 4(1)(-3)
Δ = + 4 + 12
Δ = 16 ---------------------------------√16 = 4
se
Δ > 0
então (baskara)
x = - b + √Δ/2a
x' = - (-2) - √16/2(1)
x' = + 2 - 4/2
x' = -2/2
x' = - 1
e
x" = -(-2) + √16/2(2)
x" = + 2 + 4/4
x" = 6/4
x" = 3/2

c)   -3x²+10x-3=0
a = -3
b = 10
c = - 3
Δ= b² - 4ac
Δ = (10)² - 4(-3)(-3)
Δ= 100 - 36
Δ= 64-----------------------------√64 = 8
se
Δ > 0
então (baskara)
x = -b +√Δ/2a
x' = -10 - √64/2(-3)
x' = -10 - 8/-6
x' = - 18/-6
x' = + 18/6
x' = 3
e
x" = -10 + √64/2(-3)
x" = -10 + 8/-6
x" = -2/-6
x"=+ 2/6
x" = 1/3


d    x²+x+2+0
x² + x + 2 = 0
a = 1
b = 1
c = 2
Δ = b² - 4ac
Δ = (1)² -4(1)(2)
Δ = 1 - 8
Δ = - 7
se
Δ < 0  não existe ZERO REAIS

ENTÃO
Δ = - 7
√ - 7  = -7(1) = √7i² = + √7i

x
 = - √7i
e
x₂ = + √7i
2014-07-16T22:02:08-03:00
A) a= 2 , b = -3 , c= 1
delta = b^2 -4 ac
" = (-3)^2 -4 x 2 x 1
" = 9 - 8
" = 1
x' = -(-3) + raiz quadrada de 1 / 2x2
x' = 3 +1 / 4
x' = 4/4 ; x'= 1

x" = -(-3) - raiz quadrada de 1/ 2 x 2
x" = 3 -1 / 4 ; x" = 2/4 ; x"= 1/2

b) a= 1 ; b= -2 ; c= -3
delta = (-2)^2 -4 x 1 x (-3)
" = 4 + 12
" = 16
x' = -(-2) + raiz quadrada de 16/ 2 x 1
x'= 2 + 4/ 2 ; x'= 6/2 ; x'= 3

x"= -(-2) - raiz quadrada de 16 / 2 x 1
x"= 2 - 4/ 2 ; x" = -2/ 2 ; x" = -1

c) a= -3 ; b = 10 ; c=-3
delta = 10^2 -4 x (-3) x (-3)
" = 100 - 36
" = 64
x'= -10 + raiz quadrada de 64/ 2 x (-3)
x' = -10 + 8 / -6 ; x' = -2/ -6 ; x' = -1/-3

x" = -10 - raiz quadrada de 64 / 2 x (-3)
x" = -10 - 8 / -6 ; x" = -18 / -6 ; x"= 3