\(x^2-2\left(m+1\right)x+m^2+3=0\)
\(\Delta'=\left(m+1\right)^2-m^2-3=m^2+2m+1-m^2-3=2m-2\)
Để phương trình có 2 nghiệm\(\Leftrightarrow2m-2\ge0\Leftrightarrow m\ge1\)
Theo Vi ét có:
\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=m^2+3\end{matrix}\right.\)
Có \(\left|x_1\right|+\left|x_2\right|=10\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=10\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=10\)
\(\Leftrightarrow\left(2m+2\right)^2-2\left(m^2+3\right)+2\left|m^2+3\right|=10\)
Có \(m^2+3>0\forall m\Rightarrow\left|m^2+3\right|=m^2+3\)
\(\Rightarrow4m^2+8m+4-2m^2-6+2m^2+6=10\)
\(\Leftrightarrow4m^2+8m-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=\frac{-2+\sqrt{10}}{2}\\m=\frac{-2-\sqrt{10}}{2}\end{matrix}\right.\) (loại)
\(x^2-2\left(m+1\right)x+m^2+3=0\)
\(\Delta'\)\(=\left[-\left(m+1\right)\right]^2-\left(m^2+3\right)\)
\(=m^2+2m+1-m^2-3\)\(=2m-2\)
Để pt có 2 nghiệm ⇔ Δ' ≥ 0 ⇔ \(2m-2\ge0\)
\(\Leftrightarrow m\ge1\)
Ta có: \(\left|x_1\right|+\left|x_2\right|=10\)
\(\Leftrightarrow\left(\left|x_1\right|+\left|x_2\right|\right)^2=100\)
\(\Leftrightarrow x_1^2+x_2^2+2\left(\left|x_1x_2\right|\right)=100\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=100\) (1)
Áp dụng định lí Vi-ét, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=m^2+3\end{matrix}\right.\)
⇒(1) ⇔\(\left(2m+2\right)^2-2\left(m^2+3\right)+2\left|m^2+3\right|=100\)
\(\Leftrightarrow4m^2+8m+4-2m^2-6+2m^2+6=100\)
\(\Leftrightarrow4m^2+8m-96=0\)
\(\Leftrightarrow4m^2+24m-16m-96=0\)
\(\Leftrightarrow\left(m+6\right)\left(4m-16\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=-6\left(KTM\right)\\m=4\left(TM\right)\end{matrix}\right.\)
Vậy m=4 là gt cần tìm
