Question

d) Resolva a seguinte inequação:

$x^{2} - x - 160 \leq 0$

Answer 1

Resposta correta:

$\space\space{\because\space\space\boxed{\boxed{\begin{array}{lr}\red{\space S = \{ \frac{ - \sqrt{641} + 1 }{2} \leqslant x \leqslant \frac{ \sqrt{641} + 1 }{2} \}\space}\end{array}}}}$

$\bf\large\gray{\underline{\qquad \qquad\qquad \qquad \qquad \qquad \qquad \quad }}$

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$\space\space\space\space\space\huge\mid{\boxed{\bf{\blue{Matem\acute{a}tica}}}\mid}$

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Solução passo a passo

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$\large\underline{ \overline{\boxed{\begin{array}{clr}\\ \displaystyle{ ~ x^{2} - x - 160 \leqslant 0 ~ }\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \end{array}}}}$

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Por equação quadrática $\sf { ax^{2} + bx + c = 0 }$

$\large\gray{\space\space\space\space\downarrow}\space\space\space\space\displaystyle\begin{aligned} x^{2} - x - 160 = 0  \end{aligned}$

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Utilizando a fórmula de Bhaskara: $\large\sf {x = \frac{ - b \pm \sqrt{b^{2} - 4 \: \cdot \: a \: \cdot \: c} }{2 \: \cdot \: a}}$

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$\large{\gray{\space\space\space\downarrow}\space\space\space\space\displaystyle\text{ \begin{aligned} x = \frac{1 \pm \sqrt{( -1 )^{2} - 4 \: \cdot \: 1( - 160)} }{1\: \cdot \:2}  \end{aligned} } }$

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Resolvendo

$\large\gray{\space\space\space\space\downarrow}\space\space\space\space\displaystyle\text{ \begin{aligned} x = \frac{ 1 \: \pm \: \sqrt{641} }{2}  \end{aligned} }\left\{\begin{array}{ll}\displaystyle{ x = \frac{1 \pm \sqrt{( -1 )^{2} - 4 \: \cdot \: 1( - 160)} }{1\: \cdot \:2} } \\ \\ \displaystyle{ x = \frac{1 \pm \sqrt{( -1 )^{2} - 4 \: \cdot \: 1( - 160)} }{2}}\\ \\ \displaystyle{ x = \frac{1 \pm \sqrt{( -1 )^{2} - 4 \: ( - 160)} }{2}} \\ \\ \displaystyle{ x = \frac{1 \pm \sqrt{1 - 4 \: ( - 160)} }{2}} \\ \\ \displaystyle{ x = \frac{1 \pm \sqrt{1 + 640} }{2}}\\ \\~~\displaystyle{= x = \frac{ 1 \: \pm \: \sqrt{641} }{2} ~~} \\ \end{array}\right.$

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$\large{\gray{\space\space\space\space\downarrow}\space\space\space\space\displaystyle\text{ \begin{aligned} = \frac{ - \sqrt{641} + 1 }{2} \leqslant x \leqslant \frac{ \sqrt{641} + 1 }{2}  \end{aligned} } }$

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Notas:

• $\boxed{\displaystyle\text{ \begin{aligned} \: \: \sf{ ax^{2} + bx + c = 0 \: \rightarrow \: x = \frac{ - b \: \pm \: \sqrt{b^{2} - 4ac}}{2a} } \: \:\end{aligned} }}$

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Solução:

$~~{\therefore~~\boxed{\boxed{\begin{array}{lr}\red{~S = \{ \frac{ - \sqrt{641} + 1 }{2} \leqslant x \leqslant \frac{ \sqrt{641} + 1 }{2} \} ~}\end{array}}}}$

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$\: \: \: \text{\underline{Att.}}$

${\huge\boxed { {\bf{M}}}\boxed { \red {\bf{y}}} \boxed { \blue {\bf{t}}} \boxed { \gray{\bf{h}}} \boxed { \red {\bf{}}} \boxed { \orange {\bf{P}}} \boxed {\bf{i}}}$

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${~~\vdots~~~\large\boxed {\boxed{17:42h}~22.10.21}}$

$\bf\large\gray{\underline{\qquad \qquad\qquad \qquad \qquad \qquad \qquad \quad }}$

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