Question

sendo f(x)=[x²-3x] a solução da equação de modo que f(x)= 2 é:

Answer 1

Temos o valor de f(x). Agora é só troca na função, para encontrarmos o valor de x:

$f_{(x)}=2\\\\ 2=x^2-3x\\\\ x^2-3x-2=0\\\\\\ Termos\ da\ equa\c{c}\~ao:\ \ \ a=1\ \ \ b=-3\ \ \ c=-2\\\\ Resolvendo\ por\ Bh\'askara:\\\\ x=\dfrac{-b\pm\sqrt{b^2-4.a.c}}{2.a}\\\\ x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4.(1).(-2)}}{2.(1)}\\\\ x=\dfrac{3\pm\sqrt{9+8}}{2}\\\\ x=\dfrac{3\pm\sqrt{17}}{2}\\\\\\ Termos\ 2\ valores\ que\ solucionam:\\\\ x'=\dfrac{3+\sqrt{17}}{2}\ \ \ \ e\ \ \ \ x''=\dfrac{3-\sqrt{17}}{2}$



Tirando a prova real:

$f_{\left(\frac{3+\sqrt{17}}{2}\right)} = \left(\dfrac{3+\sqrt{17}}{2}\right)^2-3\left(\dfrac{3+\sqrt{17}}{2}\right)\\\\ f_{\left(\frac{3+\sqrt{17}}{2}\right)} = \dfrac{(3+\sqrt{17})^2}{2^2}-\dfrac{9+3\sqrt{17}}{2}\\\\ f_{\left(\frac{3+\sqrt{17}}{2}\right)} = \dfrac{3^2+2 \times 3 \times \sqrt{17}+\sqrt{17}^2}{4}-\dfrac{9+3\sqrt{17}}{2}\\\\ f_{\left(\frac{3+\sqrt{17}}{2}\right)} = \dfrac{9+6\sqrt{17}+\sqrt[\not2]{17}^{\not2}}{4}-\dfrac{9+3\sqrt{17}}{2}$


$f_{\left(\frac{3+\sqrt{17}}{2}\right)} = \dfrac{9+6\sqrt{17}+17}{4}-\dfrac{9+3\sqrt{17}}{2}\\\\ f_{\left(\frac{3+\sqrt{17}}{2}\right)} = \dfrac{26+6\sqrt{17}}{4}-\dfrac{9+3\sqrt{17}}{2}\\\\ f_{\left(\frac{3+\sqrt{17}}{2}\right)} = \dfrac{26+6\sqrt{17}-18-6\sqrt{17}}{4}\\\\ f_{\left(\frac{3+\sqrt{17}}{2}\right)} = \dfrac{8}{4}\\\\ \boxed{f_{\left(\frac{3+\sqrt{17}}{2}\right)} = 2}$


Espero ter ajudado.
Bons estudos!