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3 years ago

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Cho pt: x²+ (m-3)x-5=0Tìm m để pt có hai nghiệm x1 x2 thoả x1²+x2²-5x1x2+3x1+3x2=4

1 answer:

marin [14]3 years ago

6 0

Answer:

Step-by-step explanation:

delta=( m-3)^2+4*5

pt có 2 nghiệm x1 x2 khi delta>0<=>(m-3)^2+20>0(luôn đúng)

Viet: x1+x2=3-m

x1x2=-5

x1^{2} +x2^{2} -5x1x2+3x1+3x2=4

<=>(x1+x2)^{2}-7x1x2+3(x1+x2)=4

<=>(3-m)^2+35+3*(3-m)=4

<=>m^2-6m+9+35-3m+9-4=0

<=>m^2-9m+49=0

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