Question

Questão 3 por favor alguém me ajuda

Answer 1

Boa tarde Nhatinha!

Solução!
$A=\begin{pmatrix} x & 1 \\ 6 & 3 \end{pmatrix} \Rightarrow \ Det _{A}= \begin{vmatrix} x & 1 \\ 6 & 3 \end{vmatrix}$

$B=\begin{pmatrix}10 & -x \\ 5 & 1 \end{pmatrix} \Rightarrow \ Det _{B}= \begin{vmatrix} 10 & -x \\ 5 & 1 \end{vmatrix}$

$det_{A} \times det_{B}=45x$

$\begin{vmatrix} x & 1 \\ 6 & 3 \end{vmatrix} \times \begin{vmatrix} 10 & -x \\ 5 & 1 \end{vmatrix}=45x$



Para calcular um determinante multiplica a diagonal principal e a diagonal secundária

$(3x-6)\times(5x+10)=45x\\\\\ 15 x^{2} +30x-60x-60=45x\\\\\ 15 x^{2} -45x-60=0$

Vamos dividir a equação por 15 para ficar mais fácil os cálculos.

$15 x^{2} -45x-60=0:(15) \\\\\\\\\\ x^{2} -3x-4=0 \\\\\\\\ x= \dfrac{-b\pm \sqrt{b^{2}-4.a.c } }{2.a} \\\\\\\ x= \dfrac{3\pm \sqrt{(-3)^{2}-4.1.(-4)} }{2.1} \\\\\\\ x= \dfrac{3\pm \sqrt{9+16} }{2} \\\\\\\\\ x= \dfrac{3\pm \sqrt{25} }{2} \\\\\\\ x= \dfrac{3\pm5 }{2} \\\\\\\ x_{1}= \dfrac{3+5}{2}= \dfrac{8}{2}=4\\\\\\\\\ x_{2}= \dfrac{3-5}{2}= \dfrac{-2}{2}=-1$


$\boxed{Resposta: S=\{-1,4\}~~Alternativa~~D}$

Boa tarde!
Bons estudos