Question

Boa tarde pessoa, alguem pode me ajudar ?

Determine M, de modo que uma raiz seja o triplo da outra
$x^{2} + (3m + 2)x + m^2 + 2m + 4 = 0$

Answer 1

Olá Rafael.


Vamos usar as relações de Girard para resolver essa questão.


Sabendo que a soma e o produto das raízes de uma equação do segundo grau é dado por:

$\mathsf{ax^2+bx+c=0~~~~~}\qquad onde\qquad\qquad\mathsf{a\neq 0}\\\\\\\mathsf{x_1+x_2=\dfrac{-b}{a}}\qquad\qquad ~~e\qquad\qquad\qquad\mathsf{x_1\cdot x_2=\dfrac{c}{a}}$


Temos:


$\mathsf{x^2+(3m+2)x+m^2+2m+4=0}\\\\\mathsf{a=1}\\\mathsf{b=2m+2}\\\mathsf{c=m^2+2m+4}\\\\\\\mathsf{x_1=3x_2}\\\\\\\mathsf{3x_2+x_2=\dfrac{-(3m+2)}{1}~~\Rightarrow~~4x_2=-3m-2~~\Rightarrow~~x_2=\dfrac{-3m-2}{4}~(i)}\\\\\\\mathsf{3x_2\cdot x_2=\dfrac{m^2+2m+4}{1}~\Rightarrow ~3x_2^2=m^2+2m+4~(ii)}\\\\\\\\\mathsf{3\cdot\Big(\dfrac{-3m-2}{4}\Big)^2=m^2+2m+4}\\\\=\\\\\mathsf{~3\cdot\Big(\dfrac{9m^2+12m+4}{16}\Big)=m^2+2m+4~\cdot(16)}\\\\=\\\\\mathsf{3\cdot(9m^2+12m+4)=16m^2+32m+64}\\\\=$


$\mathsf{27m^2+36m+12=16m^2+32m+64}\\\\=\\\\\mathsf{11m^2+4m-52=0}\\\\\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=4^2-4\cdot11\cdot(-52)}\\\mathsf{\Delta=16-2.288}\\\mathsf{\Delta=2.304}\\\\\\\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\\mathsf{m^+=\dfrac{-4+\sqrt{2.304}}{2\cdot11}\qquad\qquad\qquad\qquad m^-=\dfrac{-4-\sqrt{2.304}}{2\cdot11}}\\\\\\\mathsf{m^+=\dfrac{-4+48}{22}\qquad\qquad\qquad\qquad\qquad m^-=\dfrac{-4-48}{22}}$


$\mathsf{m^+=\dfrac{44}{22}\qquad\qquad\qquad\qquad\qquad\qquad~~m^-=\dfrac{-52}{22}}\\\\\\~~\boxed{\boxed{\mathsf{m^+=2}}}\mathsf{\qquad\qquad\qquad\qquad\qquad~~\boxed{\boxed{\mathsf{m^-=-\dfrac{26}{11}}}}}$



Dúvidas? comente.