Question

cho pt x - 2mx + m² -m-6 = 0. Tìm m để pt có 2 nghiệm x1, x2 thỏa mãn \(\left|x1\right|\) + \(\left|x2\right|\) = 8

Answer 1

Δ= 4m^2 - 4m^2 + 4m + 24 = 4m + 24

để pt có 2 nghiệm thì Δ ≥ 0 => 4m + 24 ≥ 0 <=> m ≥ -6

viet: \(\left\{{}\begin{matrix}x1+x2=2m\\x1\cdot x2=m^2-m+6\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left(x1+x2\right)^2=4m^2\\2x1\cdot x2=2m^2-2m+12\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x1^2+x2^2=4m^2-2x1\cdot x2\\2x1\cdot x2=2m^2-2m+12\end{matrix}\right.\)

|x1| + |x2| = 8

<=> (|x1| + |x2|)^2 = 64

<=> x1^2 + x2^2 + 2|x1|*|x2| = 64

<=> 4m^2 - 2m^2+2m-12 + 2m^2-2m+12 = 64

<=> 4m^2 = 64

<=> m = -4; m = 4