Question

\(x^2+5x+3m-1=0\left(1\right)\)

tìm m để (1) có 2 nghiệm x1; x2 thỏa mãn \(x_1^3+x_2^3+3x_1x_2=-35\)

Answer 1

\(\Delta=25-4\left(3m-1\right)=29-12m\ge0\Rightarrow m\le\dfrac{29}{12}\)

Theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=-5\\x_1x_2=3m-1\end{matrix}\right.\)

\(x_1^3+x_2^3+3x_1x_2=-35\)

\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+3x_1x_2=-35\)

\(\Leftrightarrow\left(-5\right)^3+15\left(3m-1\right)+3\left(3m-1\right)=-35\)

\(\Leftrightarrow18\left(3m-1\right)=90\)

\(\Rightarrow m=2\) (thỏa mãn)

Answer 2

\(\text{Δ}=5^2-4\cdot1\cdot\left(3m-1\right)\)

\(=25-4\left(3m-1\right)\)

\(=25-12m+4=-12m+29\)

Để phương trình (1) có hai nghiệm thì Δ>=0

=>-12m+29>=0

=>-12m>=-29

=>\(m< =\dfrac{29}{12}\)

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-5}{1}=-5\\x_1x_2=\dfrac{c}{a}=\dfrac{3m-1}{1}=3m-1\end{matrix}\right.\)

\(x_1^3+x_2^3+3x_1x_2=-35\)

=>\(\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+3x_1x_2=-35\)

=>\(\left(-5\right)^3-3\cdot\left(3m-1\right)\cdot\left(-5\right)+3\cdot\left(3m-1\right)=-35\)

=>\(-125+15\left(3m-1\right)+9m-3=-35\)

=>\(-125+45m-15+9m-3=-35\)

=>54m-143=-35

=>54m=108

=>m=2(nhận)