Question

cho pt: x² - 2(m - 1)x + m² -2m - 3 = 0 (*)

tìm m để x₁ + 4 = \(\sqrt{x_2}\)

Answer 1

\(\Delta'=\left(m-1\right)^2-\left(m^2-2m-3\right)=4>0;\forall m\)

\(\Rightarrow\left[{}\begin{matrix}x=m-3\\x=m+1\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x_1=m-3\\x_2=m+1\end{matrix}\right.\)

Để \(\sqrt{x_2}\) xác định \(\Leftrightarrow m\ge-1\)

\(\Rightarrow m+1=\sqrt{m+1}\Leftrightarrow\left[{}\begin{matrix}m+1=0\\m+1=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=-1\\m=0\end{matrix}\right.\) (thỏa mãn)

TH2: \(\left\{{}\begin{matrix}x_1=m+1\\x_2=m-3\end{matrix}\right.\) \(\left(m\ge3\right)\)

\(m+5=\sqrt{m-3}\)

\(\Leftrightarrow m^2+10m+25=m-3\)

\(\Leftrightarrow m^2+9m+28=0\) (ptvn)

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