Question

Помогите, плиз!! Корни x1 и x2 уравнения x^2 +mx+27=0 удовлетворяют условию x1=3x2. Найти значение m и корни уравнения.

Answer 1

$\displaystyle\bf\\x^{2} +mx+27=0\\\\x_{1} =3x_{2} \\\\Teorema \ Vieta:\\\\x_{1} \cdot x_{2} =27\\\\3x_{2} \cdot x_{2} =27\\\\3x_{2} ^{2} =27\\\\x_{2} ^{2} =9\\\\x_{2} =-3 \ \ ili \ \ x_{2} =3\\\\x_{1} =3\cdot(-3)=-9 \ \ \ ili \ \ \ x_{1} =3\cdot 3=9\\\\Teorema \ Vieta:\\\\-m=x_{1} +x_{2} \\\\1) \ x_{1} =-9 \ \ , \ \ x_{2} =-3\\\\-m=-9+ (-3)=-12\\\\m_{1} =12\\\\2) \ x_{1} =9 \ \ , \ \ x_{2} =3\\\\-m=9+3=12\\\\m_{2} =-12$