x=x1,x=x2是方程x²-(2m-1)x+m²+3m+4=0的兩實(shí)根.由韋達(dá)定理得
x1+x2=2m-1
x1x2=m²+3m+4
代入(x1+x2)²-2x1x2=5
(2m-1)²-2(m²+3m+4)=5
整理,得
m²-5m-6=0
(m-6)(m+1)=0
m=6或m=-1
m=6時(shí),方程變?yōu)閤²-11x+58=0,方程無解.
m=-1時(shí),方程變?yōu)閤²+3x+2=0
(x+1)(x+2)=0
x=-1或x=-2點(diǎn)A坐標(biāo)(-1,0),點(diǎn)B坐標(biāo)(-2,0)
函數(shù)解析式為y=x²+3x+2
令x=0,y=0+0+2=2點(diǎn)C坐標(biāo)(0,2)
y=x²+3x+2=(x +3/2)² -1/4,頂點(diǎn)坐標(biāo)M(-3/2,-1/4)
S四邊形AMBC=S△ABM+S△ABC
=|(-1)-(-2)|×|-1/4|/2 +|(-1)-(-2)|×2/2
=9/8