Question

Encontre as raízes das equações:
a) x2 - 7x +6=0
b) x2 + 3x -18 =0
c) 2y2 - 6y - 20=0
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Answer 1

Resposta:

a) x² - 7x + 6 = 0

Δ = -7² - 4.a.6

Δ = 49 - 24

Δ = 25

x = -(-7) +- √25

            2. 1

x' = 7 + 5  >> x' = 12/2 >> x' = 6

         2

x'' = 7-5  >> x'' = 2/2 >> x'' = 1

        2

b) x² + 3x -18 = 0

Δ = 3² - 4.1.(-18)

Δ = 9 + 72

Δ = 81

x = -(3) +- √81

         2.1

x' = -3 + 9 >> x' = 6/2 >> x' = 3

          2

x'' = -3 -9 >> x'' = -12/2 >> x'' = -6

         2

c) 2y² - 6y - 20=0

Δ = (-6)² - 4.2.(-20)

Δ = 36 + 160

Δ = 196

y = -(-6) +- √ 196

           2.2

y' = 6 + 14 >> y' = 20/4 >> y' = 5

        4

y'' = 6 - 14 >> y'' = -8/4 >> y'' = -2

        4

Answer 2

Item a) x²-7x+6 = 0

coeficiente a= +1

coeficiente b= -7

coeficiente c= +6

∆=b²-4ac

∆=(-7)²-4(+1)(+6)

∆=49-24

∆=25

x= -b±√∆/2a

$x1 = \frac{ - ( - 7) + \sqrt{25} }{2(1)} = \frac{ + 7 + 5}{2} = \frac{12}{2} = 6$

$x2 = \frac{ + 7 - 5}{2} = \frac{2}{2} = 1$

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Item b) x²+3x-18=0

coeficiente a= +1

coeficiente b= +3

coeficiente c= -18

∆=(3)²-4(+1)(-18)

∆=9+72

∆=81

$x1 = \frac{- ( + 3) + \sqrt{81} }{2(1)} = \frac{ - 3 + 9}{2} = \frac{ + 6}{2} = 3$

$x2 = \frac{ - 3 - 9}{2} = \frac{ - 12}{2} = - 6$

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Item c) 2y²-6y-20

coeficiente a= +2

coeficiente b= -6

coeficiente c= -20

∆=(-6)²-4(+2)(-20)

∆=36+160

∆=196

x1= -(-6)+14/2(2) =+6+14/4= 20/4 = 5

x2= +6-14/4 = -8/4 = -2