Question

resolva as seguintes equações
a) x²-5x+6=0
b) x²+5x-6=0
c)x²-x-12=0
d)x²-14x-32=0
e)x²-12x+36=0
f)x²+4x+4=0

Answer 1

Resolva as seguintes equaçõesa) x²-5x+6=0
a = 1
b = - 5
c = 6
Δ = b² - 4ac
Δ = (-5)² - 4(1)(6)
Δ=  + 25 - 24
Δ= 1 -----------------------------> √Δ = 1 porque √1 = 1
se
Δ > 0(DUAS raízes diferentes)
(baskara)
x = -b + √Δ/2a

x' = - (-5) + √1/2(1)
x' = + 5 + 1/2
x = 6/2
x' = 3
e
x"= -(-5) - √1/2(1)
x" = + 5 - 1/2
x" = 4/2
x" = 2

x' = 3
x" = 2
 
b) x²+5x-6=0
a = 1
b = 5
c = - 6
Δ = b² - 4ac
Δ = 5² - 4(1)(-6)
Δ = 25 + 24
Δ = 49 ----------------------------->√Δ = 7 porque √49 = 7
Δ > 0(DUAS raízes diferentes)
(baskara)
x = -b + √Δ/2a

x' = -5 + √49/2(1)
x' = - 5 + 7/2
x' = 2/2
x' = 1
e
x" = -5 - √49/2(1)
x" = - 5 - 7/2
x" = - 12/2
x" = - 6

x' = 1
x" = -6

c)x²-x-12=0
a = 1
b = - 1
c = - 12
Δ = b² - 4ac
Δ = (-1)² - 4(1)(-12)
Δ = + 1 + 48
Δ = 49 ------------------> √Δ = 7 porque √49 = 7
Δ > 0(DUAS raízes diferentes)
(baskara)
x = -b + √Δ/2a

x' = -(-1) + √49/2(1)
x' = + 1 + 7/2
x' = 8/2
x' = 4
e
x" = -(-1) - √49/2(1)
x" = + 1 -7/2
x" = - 6/2
x" = - 3

x' = 4
x" = - 3

d)x²-14x-32=0
a = 1
b = - 14
c = - 32
Δ = b² - 4ac
Δ = (-14)² - 4(1)(-32)
Δ = + 196 + 128
Δ = 324 -----------------------------> √Δ = 18  porque √324 = 18
Δ > 0(DUAS raízes diferentes)
(baskara)
x = -b + √Δ/2a

x' = -(-14) + √324/2(1)
x' = + 14 + 18/2
x' = 32/2
x' = 16
e
x" = -(-14) - √324/2(1)
x" = + 14 - 18/2
x" = - 4/2
x" = - 2

x' = 16
x" = -2

e)x²-12x+36=0
a = 1
b = - 12
c = 36
Δ = b² - 4ac
Δ = (-12)² - 4(1)(36)
Δ = + 144 - 144
Δ = 0
se
Δ = 0
x = - b/2a
x = -(-12)/2(1)
x' = + 12/2
x = 6

f)x²+4x+4=0
a = 1
b = 4
c = 4
Δ = b² - 4ac
Δ = 4² - 4(1)(4)
Δ = 16 - 16
Δ = 0
se
Δ = 0
então

x = - b2a
x = -4/2(1)
x = -4/2
x = -2

Answer 2

a)
x² - 5x + 6 = 0
a = 1; b = - 5; c = 6

Δ = b² - 4ac
Δ = (-5)² - 4.1.6
Δ = 25 - 24
Δ = 1

x = - b +/- √Δ = - ( - 5) +/- √1
           2a              2.1

x = 5 + 1 = 6/2 = 3
        2

x = 5 - 1 = 4/2 = 2
        2

R.: x = 2 e x = 3
******************************
b)
x² + 5x - 6 = 0
a = 1; b = 5; c = - 6

Δ = b² - 4ac
Δ = 5² - 4.1.(-6)
Δ = 25 + 24
Δ = 49

x = - b +/- √Δ  = - 5 +/- √49
           2a              2.1

x = - 5 + 7 = 2/2 = 1
           2

x = - 5 - 7 = - 12/2 = - 6
         2

R.: x = 1 e x = - 6
******************************
c)
x² - x - 12 = 0
a = 1; b = - 1; c = - 12

Δ = b² - 4ac
Δ = (-1)² - 4.1.(-12)
Δ = 1 - 4.(- 12)
Δ = 1 + 48
Δ = 49

x = - b +/- √Δ  = - ( - 1) +/- √49
          2a                  2.1

x = 1 + 7 = 8/2 = 4
         2

x = 1 - 7 = - 6/2 = - 3
        2

R.: x = 4 e x = - 3
******************************
d)
x² - 14x - 32 = 0
a = 1; b = - 14; c = - 32

Δ = b² - 4ac
Δ = (-14)² - 4.1.(-32)
Δ = 196 + 128
Δ = 324

x = - b +/- √Δ  = - ( -14) +/- √324
           2a                 2.1

x = 14 + 18  = 32/2 = 16
          2

x = 14 - 18 = - 4/2 = - 2
          2

R.: x = 16 e x = - 2
***************************************
e)
x² - 12x + 36 = 0
a = 1; b = - 12; c = 36

Δ = b² - 4ac
Δ = (-12)² - 4.1.36
Δ = 144 - 144
Δ = 0

x = - b +/- √Δ = - ( -12) +/- 0
           2a                  2.1

x = 12  = 6
       2

R.: x = 6

-----------------------------------------------------------
f)
x² + 4x + 4 = 0
a = 1; b = 4; c = 4

Δ = b² - 4ac
Δ = 4² - 4.1.4
Δ = 16 - 16
Δ = 0

x = - b +/- √Δ  = - 4 +/- 0
        2a                  2.1

x = - 4 = - 2
       2

R.: x = - 2