设方程x²+3x+1=0的两根为x1,x2,求下列各式的值:(1)x1²+x2²(2)(x1-3)(x2-3)
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设方程x²+3x+1=0的两根为x1,x2,求下列各式的值:(1)x1²+x2²(2)(x1-3)(x2-3)
设方程x²+3x+1=0的两根为x1,x2,求下列各式的值:
(1)x1²+x2²
(2)(x1-3)(x2-3)
设方程x²+3x+1=0的两根为x1,x2,求下列各式的值:(1)x1²+x2²(2)(x1-3)(x2-3)
解方程得两根分别为x1=-3+√5/2,x2=-3-√5/2
(1)x1²+x2²
=(-3-√5/2)²+(-3+√5/2)²
=(-3)²+(√5/2)²
=9+5/4
=10.25
(2)(x1-3)(x2-3)
=(-3-√5/2-3)(-3+√5/2-3)
=(-6-√5/2)(-6+√5/2)
=(-6)²-(√5/2)²
=36-5/4
=34.75
δ=3^2-4*1*1=5>0
韦达定理:
x1+x2=-3
x1x2=1
x1²+x2²=(x1+x2)^2-2x1x2=7
(x1-3)(x2-3)=x1x2-3(x1+x2)+9=19
x1²+x2²=(x1+x2)²-2x1x2=b²/4a²
(x1-3)(x2-3)=x1x2-3(x1+x2)+9=C/a+b/a+9
自己算。
(1)x1+x2=-3
x1*x2=1
x1²+x2²
=(x1+x2)²-2*x1*x2
= (-3)²-2*1
=7
(x1-3)(x2-3)=x1x2-3(x1+x2)+9=19