Question

tìm m để :   \(^{x^2}\)+ (m-2)x + m+5 = 0

Có 2 nghiệm x1 và x2 , sao cho \(x1^2\) + \(x2^2\) = 10

Answer 1

\(\Delta=\left(m-2\right)^2-4\left(m+5\right)=m^2-4m+4-4m-20=m^2-16\)

\(\Delta\ge0\Leftrightarrow m\ge4\)

theo hệ thức Vi - ét ta có \(\left\{{}\begin{matrix}x_1+x_2=2-m\\x_1x_2=m+5\end{matrix}\right.\)

 

\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\left(2-m\right)^2-2\left(m+5\right)=4-4m+m^2-2m-10=m^2-6m-6\)

\(\Delta_1'=\left(-1\right)^2-\left(-6\right)=7\Leftrightarrow\left\{{}\begin{matrix}m_1=1-\sqrt{7}\left(ktm\right)\\m_2=1+\sqrt{7}\left(tm\right)\end{matrix}\right.\)

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