A quadratic function judgment problem! If b > A + C, then ax & sup2; + BX + C = 0 has two unequal real roots How can this sentence be proved right or wrong? Online kneeling for the emperor of mathematics! It is better to give the process of proof

A quadratic function judgment problem! If b > A + C, then ax & sup2; + BX + C = 0 has two unequal real roots How can this sentence be proved right or wrong? Online kneeling for the emperor of mathematics! It is better to give the process of proof

Conclusion: uncertain (strictly speaking)
Proof: according to the discriminant rule of the root of quadratic equation of one variable, the
(delta) ⊿ = B ^ 2 (the square of B) - 4ac
If ⊿ 0, there are two different real roots
1. If b > A + C > 0, B and a + C are both positive numbers
There must be
B ^ 2 > (a + C) ^ 2 = a ^ 2 + 2Ac + C ^ 2 (1). In addition, because there must be (A-C) ^ 2 ≥ 0, it is simplified
A ^ 2 + C ^ 2 ≥ 2Ac, and then substitute it into (1)
⊿ = B ^ 2 (the square of B) - 4ac > 0, there are two different real roots
2. But if B and a + C are not both zero, it's hard to say, because
B ^ 2 > (a + C) ^ 2 is not necessarily true, the previous conclusion is not applicable. For example, B = - 1, a = - 2, C = - 3, of course
B > A + C, but
⊿ = B ^ 2 (the square of B) - 4ac = 1-24 = - 23