Question

已知一元二次方程ax²+b=0有实根 则必须是

A a.b同号或b=0且a≠0 B a.b异号或b=0 且a≠0

C a ＞b 且a≠0 D a＜b 且a≠0

问下大家 怎么算的 谢谢解答

Answer 1

ax^2=-b
x^2=-b/a
所以-b/a>=0
A.同号一定是负数，肯定不对
B.对的，而且b=0,a不等于0的情况也考虑了
C.D这道题和a,b谁大谁小没关系，舍掉
所以答案是B

Answer 2

ax²+b=0，
ax² = -b，
x² = -b/a，
x² >0，
所以 -b/a >0，
b/a <0，
也就是说 a和b异号。

或者 b=0，那么解为 x=0

Answer 3

答案选B
首先一元二次方程需满足 a≠0
x²=-b/a>=0
所以：a.b异号或b=0
希望采纳，鼓励下，谢谢!

Answer 4

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