Question

Usando a fórmula de Bhaskara, determine o conjunto solução das equações abaixo: a) x2 − 5x + 6 = 0 b) 2x2+ 12x − 14 = 0 c) −x² − 4x + 5 = 0

Answer 1

Resposta:

a) S = {2; 3}

b) S = {-7; 1}

c) S = {-5; 1}

Explicação passo-a-passo:

$a) \: {x}^{2} - 5x + 6 = 0 \\ \\ a = 1 \\ b = ( - 5) \\ c = 6 \\ \\ \Delta = {b}^{2} - 4ac \\ \Delta = {( - 5)}^{2} - 4 \cdot1 \cdot6 \\ \Delta = 25 - 24 \\ \Delta = 1 \\ \\ x = \frac{ - b \pm \sqrt{\Delta} }{2a} \\ \\ x = \frac{ - ( - 5) \pm \sqrt{1} }{2 \cdot1 } \\ \\ x = \frac{5 \pm1}{2} \\ \\ x_{1} = \frac{5 + 1}{2} = \frac{6}{2} = \bf{3} \\ \\ x_{2} = \frac{5 - 1}{2} = \frac{4}{2} = \bf{2}$

$b) \: 2 {x}^{2} + 12x - 14 = 0 \: \bf{( \div 2)} \\ {x}^{2} + 6x - 7 = 0 \\ \\ a = 1 \\ b = 6 \\ c = ( - 7) \\ \\ \Delta = {b}^{2} - 4ac \\ \Delta = {6}^{2} - 4 \cdot1 \cdot( - 7)\\ \Delta = 36 + 28\\ \Delta = 64 \\ \\ x = \frac{ - b \pm \sqrt{\Delta} }{2a} \\ \\ x = \frac{ - 6 \pm \sqrt{64} }{2 \cdot1 } \\ \\ x = \frac{ - 6 \pm8}{2} \\ \\ x_{1} = \frac{ - 6 + 8}{2} = \frac{2}{2} = \bf{1} \\ \\ x_{2} = \frac{ - 6- 8}{2} = - \frac{14}{2} = \bf{ - 7}$

$c) \: - {x}^{2} - 4x + 5 = 0 \: \cdot( - 1) \\ {x}^{2} + 4x - 5 = 0 \\ \\ a = 1 \\ b = 4\\ c = ( - 5) \\ \\ \Delta = {b}^{2} - 4ac \\ \Delta = {4}^{2} - 4 \cdot1 \cdot( - 5) \\ \Delta = 16 + 20 \\ \Delta = 36 \\ \\ x = \frac{ - b \pm \sqrt{\Delta} }{2a} \\ \\ x = \frac{ - 4 \pm \sqrt{36} }{2 \cdot1 } \\ \\ x = \frac{ - 4 \pm6}{2} \\ \\ x_{1} = \frac{ - 4 + 6}{2} = \frac{2}{2} = \bf{1} \\ \\ x_{2} = \frac{ - 4 - 6}{2} = - \frac{10}{2} = \bf{ - 5}$

Espero ter ajudado! :)

Answer 2

$\large\boxed{\begin{array}{l} \rm{a)}\rm \: x {}^{2} - 5x + 6 = 0 \\ \\ \rm\Delta = b {}^{2} - 4 \cdot{a} \cdot{c } \\\Delta = ( - 5) {}^{2} - 4 \cdot1 \cdot6 \\ \Delta = 25 - 24 \\ \Delta = 1 \\ \\ \rm \: x = \dfrac{ - b \pm \sqrt{\Delta} }{2 \cdot{a}} \\ \\ \rm \: x = \dfrac{ - ( - 5) \pm \sqrt{1} }{2 \cdot1} \\ \\ \rm \: x = \dfrac{5 \pm1}{2} \begin{cases} \rm \: x _{1} = \dfrac{5 + 1}{2} = \dfrac{6}{2} = 3 \\ \\ \rm \: x_{2} = \dfrac{5 - 1}{2} = \dfrac{4}{2} = 2\end{cases} \\ \\ \rm \: S = \{3,2 \}\end{array}}$

$\large\boxed{\begin{array}{l} \rm{b)} \rm \: 2x {}^{2} + 12x - 14 = 0 \\ \\ \rm\Delta = b {}^{2} - 4 \cdot{a} \cdot{c} \\\Delta = 12 {}^{2} - 4 \cdot2 \cdot( - 14) \\ \Delta = 144 + 112 \\ \Delta = 256 \\ \\ \rm \: x = \dfrac{ - b \pm \sqrt{\Delta} }{2 \cdot{a}} \\ \\ \rm \: x = \dfrac{ - 12 \pm \sqrt{256} }{2 \cdot2} \\ \\ \rm \: x = \dfrac{ - 12 \pm16}{4} \begin{cases} \rm \: x _1 = \dfrac{ - 12 + 16}{4} = \dfrac{4}{4} = 1 \\ \\ \rm \: x _2 = \dfrac{ - 12 - 16}{4} = \dfrac{ - 28}{4} = - 7\end{cases} \\ \\ \rm \: S = \{1, - 7 \}\end{array}}$

$\large\boxed{\begin{array}{l} \rm{c)} \rm \: - x {}^{2} - 4x + 5 = 0 \\ \\ \rm\Delta = b {}^{2} - 4 \cdot{a} \cdot{c} \\ \Delta = ( - 4) {}^{2} - 4 \cdot{ (- 1)} \cdot5 \\\Delta = 16 + 20 \\\Delta = 36 \\ \\ \rm \: x = \dfrac{ - b \pm \sqrt{\Delta} }{2 \cdot{a}} \\ \\ \rm \: x = \dfrac{ - ( - 4) \pm \sqrt{36} }{2 \cdot{( - 1)}} \\ \\ \rm \: x = \dfrac{4 \pm6}{ - 2} \begin{cases} \rm \: x _1 = \dfrac{4 + 6}{ - 2} = \dfrac{10}{ - 2} = - 5 \\ \\ \rm \: x _{2} = \dfrac{4 - 6}{ - 2} = \dfrac{ - 2}{ - 2} = 1\end{cases} \\ \\ \rm \: S = \{1, - 5 \}\end{array}}$