Question

pergunta de matemática
$6 \times {}^{2} - \times - 1 = 0$
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Answer 1

$\large\boxed{\begin{array}{l} \rm \: 6x {}^{2} - x - 1 = 0 \\ \\ \Rightarrow \begin{cases} \rm \: a = 6 \\ \rm \: b = - 1 \\ \rm \: c = - 1\end{cases} \\ \\ \rm\Delta = b {}^{2} - 4ac \\ \Delta = ( - 1) {}^{2} - 4 \: . \: 6 \: . \: ( - 1) \\ \Delta = 1 + 24 \\ \Delta = 25 \\ \\ \rm \: x = \dfrac{ - b \pm \sqrt{\Delta} }{2a} \\ \\ \rm \: x = \dfrac{ - ( - 1) \pm \sqrt{25} }{2.6} \\ \\ \rm \: x = \dfrac{1 \pm5}{12} \begin{cases} \rm \: x_1 = \dfrac{1 + 5}{12} = \dfrac{6}{12} = \boxed{ \dfrac{3}{6} } \\ \\ \rm \: x_2 = \dfrac{1 - 5}{12} = \dfrac{ - 4}{12} = \boxed{ \frac{ - 2}{6} }\end{cases} \end{array}}$

Answer 2

Resposta:

$6 {x}^{2} - x - 1 = 0 \\ x = \frac{ - ( - 1) ± \sqrt{ {( - 1)}^{2} - 4 \times 6 \times ( - 1) } }{2 \times 6} \\ x = \frac{1 ± \sqrt{1 + 24} }{12} \\ x = \frac{1 ± \sqrt{25} }{12} \\ x = \frac{1±5 }{12}$

$x' = \frac{1 - 5}{12} = \frac{ - 4}{12} = - \frac{1}{3}$

$x'' = \frac{1 + 5}{12} = \frac{6}{12} = \frac{1}{2}$