设x1,x2是一元二次方程x2+5x-3=0的两根,且2x1(x22+6x2-3)+a=4,则a=______.
人气:382 ℃ 时间:2019-08-22 17:51:16
解答
∵x2是一元二次方程x2+5x-3=0的根,
∴x22+5x2-3=0,
∴x22+5x2=3,
∵2x1(x22+6x2-3)+a=4,
∴2x1•x2+a=4,
∵x1,x2是一元二次方程x2+5x-3=0的两根,
∴x1x2=-3,
∴2×(-3)+a=4,
∴a=10.
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