求证关于x的方程x2+2bx+1=0有两个负实根的充要条件是b≥1.

求证关于x的方程x2+2bx+1=0有两个负实根的充要条件是b≥1.

答案

证明:

充分性:∵b≥1,∴Δ=4b2-4≥0,

∴方程x2+2bx+1=0有实根并设为x1x2,

由韦达定理知x1<0,x2<0,即两根为负实数.

必要性:∵x2+2bx+1=0的两实根均为负,

x1·x2=1,

∴2b=-(x1+x2)=-(x1+)=(-x1)+

∵-x1>0,∴(-x1)+≥2,

即2b≥2,∴b≥1.

综上,方程x2+2bx+1有两负根的充要条件是b≥1.

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