Question

Usando a fórmula de Bhaskara, resolva, em R, a equação dada:
$\frac{x^2+3}{10}$ - $\frac{5x+3}{15}$ = $\frac{3-2x}{6}$

Answer 1

Explicação passo-a-passo:

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$\Large \bf{ \frac{ {x}^{2} + 3}{10} - \frac{5x + 3}{15} = \frac{3 - 2x}{6} } \\ \Large \bf{30 \times ( \frac{ {x}^{2} + 3}{10} - \frac{5x + 3}{15} ) = \cancel{\red {30}} \times \frac{3 - 2x}{\cancel{\red {6}}} } \\ \Large \bf{30 \times \frac{ {x}^{2} + 3}{10} - \cancel{\red {30}} \times \frac{5x + 3}{\cancel{\red {15}}} = 5(3 - 2x) } \\ \Large \bf{3 ( {x}^{2} + 3) - 2(5x + 3) = 5(3 - 2x)} \\ \Large \bf{ {3x}^{2} + 9 \cancel{\red { - 10x}} - 6 = 15 \cancel{\red { - 10x}}} \\ \Large \bf{ {3x}^{2} + 9 - 6 = 15 } \\ \Large \bf{ {3x}^{2} + 3 = 15 } \\ \Large \bf{ {3x}^{2} + 3 - 15 = 0 } \\ \Large \bf{ {3x}^{2} - 12 = 0} \\ \Large \bf{( {3x}^{2} - 12) \div 3 = 0 \div 3} \\ \Large \bf{ {3x}^{2} \div 3 - 12 \div 3 = 0} \\ \Large \bf{ {x}^{2} - 12 \div 3 = 0 } \\ \Large \bf{ {x}^{2} - 4 = 0 } \\ \Large \bf{x = \frac{ - 0 \pm \sqrt{ {0}^{2} - 4 \times 1 \times ( - 4)} }{2 \times 1} } \\ \Large \bf{x = \frac{ \pm \sqrt{ {0}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} } \\ \Large \bf{x = \frac{ \pm \sqrt{ {0}^{2} - 4 \times ( - 4) } }{2} } \\ \Large \bf{x = \frac{ \pm \sqrt{0 - 4 \times ( - 4)} }{2} } \\ \Large \bf{x = \frac{ \pm \sqrt{0 + 16} }{2} } \\ \Large \bf{x = \frac{ \pm \sqrt{16} }{2} } \\ \Large \bf{x = \frac{ \pm4}{2} } \\ \Large \bf{x = \frac{4}{2} } \\ \Large \bf{x = \frac{ - 4}{2} } \\ \Large \bf{x = 2} \\ \Large \bf{x = - 2} \\ \red{ \boxed{ \green{ \boxed{ \pink{ \boxed{ x_{1} = 2 } }} } } } \\ \red{ \boxed{ \green{ \boxed{ \pink{ \boxed{ x_{2} = - 2 } }} } } }$

$\mathcal{ATT : ARMANDO}$