Question

quais são as raízes das equações: F(x)=x ao quadrado-7x+12 e F(x)=x(x+3)-40 ?

Answer 1

x^2-7x+12=0
delta=b^2-4ac
delta=(-7)^2-4.1.12
delta=49-48
delta=1
x=-b-/+ raiz de delta/2a
x=-(-7)-1/2.1
x=7-1/2
x=6/2
x=3
x=-(-7)+1/2.1
x=7+1/2
x=8/2
x=4
(3,4)
x(x+3)-40
x^2+3x-40=0
delta=b^2-4ac
delta=3^2-4.1.(-40)
delta=9+160
delta=169
x=-b-/+ raiz de delta/2a
x=-3-13/2.1
x=-16/2
x=-8
x=-3+13/2.1
x=10/2
x=5
(-8, 5)

Answer 2

a)

f(x) = x² - 7x + 12
0 = x² - 7x + 12
x² - 7x + 12 = 0
a = 1; b = - 7; b = 12

Δ = b² - 4ac
Δ = (-7)² - 4.1.12
Δ = 49 - 48
Δ = 1

x = - b +/- √Δ = - ( - 7) +/- √1
           2a                2.1

x = 7 + 1 = 8/2 = 4 
        2

x = 7 - 1 = 6/2 = 3
        2

Resp.: x = 3 e x = 4
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b)

f(x) = x.(x + 3) - 40
0 = x² + 3x - 40
x² + 3x - 40 = 0
a = 1; b = 3; c = - 40

Δ = b² - 4ac
Δ = 3² - 4.1.(-40)
Δ = 9 + 160
Δ = 169

x = - b +/- √Δ = - 3 +/- √169
           2a              2.1

x = - 3 + 13 = 10/2 = 5
           2

 x = - 3 - 13 = - 16/2 = - 8
           2 

R.: x = 5 e x = - 8