Question

5. Resolva a equação
$(2 x 2 )\\( 1 \: 1 \: x) \: \: = - 3 \\ (1 \: 1 \: 6)$
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Answer 1

Resposta:

x = 4 - √7 ou x = 4 + √7

Explicação passo a passo:

$\displaystyle \left[\begin{array}{ccc}2&x&2\\1&1&x\\1&1&6\end{array}\right] =3\\\\\\(2).(1).(6) + (x).(x).(1)+(1).(1).(2) - (1).(1).(2)-(1).(x).(6)-(1).(x).(2)=3\\12+x^2+2-2-6x-2x=3\\x^2-8x+12-3=0\\x^2-8x+9=0$

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}-8x+9=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-8~e~c=9\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-8)^{2}-4(1)(9)=64-(36)=28\\\sqrt{\Delta} = \sqrt{28} =\sqrt{2^2.7} =\sqrt{2^2} .\sqrt{7} =2\sqrt{7}$$\displaystyle C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-8)-\sqrt{28}}{2(1)}=\frac{8-2\sqrt{7}}{2}=4-\sqrt{7} \\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-8)+\sqrt{28}}{2(1)}=\frac{8+2\sqrt{7}}{2}=4+\sqrt{7} \\\\S=\{4-\sqrt{7} ,~4+\sqrt{7} \}$