若x1,x2是方程x²-5x-7=0的两根,那么x1²+x2²=____________,(x1-x2)²=________.
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若x1,x2是方程x²-5x-7=0的两根,那么x1²+x2²=____________,(x1-x2)²=________.
若x1,x2是方程x²-5x-7=0的两根,那么x1²+x2²=____________,(x1-x2)²=________.
若x1,x2是方程x²-5x-7=0的两根,那么x1²+x2²=____________,(x1-x2)²=________.
由韦达定理得:
①、x1+x2=5
②、x1×x2=-7
∴﹙x1﹚²+﹙x2﹚²
=﹙x1+x2﹚²-2x1×x2
=5²-2×﹙-7﹚
=39
∴﹙x1-x2﹚²
=﹙x1+x2﹚²-4x1×x2
=5²-4×﹙-7﹚
=53
由题,根据韦达定理得x1+x2=5,x1x2=-7
于是,x1²+x2²=(x1+x2)²-2x1x2=5²-2*(-7)=25+14=39
(x1-x2)²==(x1+x2)²-4x1x2=5²-4*(-7)=25+28=53
x1²+x2²=(x1+x2)^2-2x1x2=5^2-2*(-7)=25+14=39
(x1-x2)²=x1^2+x2^2-2x1x2=39-2*(-7)=39+14=53