Question

resolva as equações abaixo
a.
$6 {x}^{2} - x - 1 = 0$
b.
$25x {}^{2} - 10x + 1 = 0$
c.
$5x {}^{2} + 6x + 1 = 0$

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Answer 1

Resposta:

a)x' = 6/12 = 1/2

x" = -4/12 = -3

b)x'=x" = 1/5

c)x' = -2/10 = -1/5

x" = -10/10 = -1

Explicação passo-a-passo:

a.) 6x² -x - 1 = 0

x = (1 +_V1+24)/12

x = (1+_5)/12

x' = 6/12 = 1/2

x" = -4/12 = -3

b. ) 25x²- 10x +1 = 0

x = (10 +_V100 -100)/50

x = (10 +_0)50

x'=x" = 1/5

c. ) 5x² + 6x + 1 = 0

x = (-6+_V36 -20)/10

x = (-6 +_ 4)/10

x' = -2/10 = -1/5

x" = -10/10 = -1

Answer 2

Resposta:

1) f(x) = 6x² - x - 1 = 0

A = 6

B = -1

C = -1

$x = b \frac{ + }{} \sqrt{ {b}^{2} - 4 \times \frac{a.c}{2.a} } \\ x = - ( - 1) \frac{ + }{} \sqrt{ {1}^{2} - 4 \times \frac{6.( - 1)}{2.6} } \\ x = 1 \frac{ + }{} \sqrt{ \frac{1 + 24}{12} } \\ {x}^{1} = \frac{ 1 + 5}{12 } = \frac{6}{12}= \frac{1}{2} \\ {x}^{2} = \frac{ 1- 5}{12} = \frac{-4}{12} = \frac{1}{3}$

2) f(x) = 25x²-10+1 = 0

R:

$\frac{ - 1}{5} \: \: \: \: \\ \frac{1}{5}$

3) f(x) = 5x² + 6x + 1 = 0

R: 1/5 e 1