if ay²+by+c=0 so find zeros of ax²+bx+c
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If it has no real zeros, then
b2−4ac<0b2−4ac<0
b2<4acb2<4ac
so the sign of a and c are the same, either both positive or both negative.
If both are positive then 0 < a+c < -b because a+b+c <0
0<(a+c)2<b2<4ac0<(a+c)2<b2<4ac
(a+c)2<4ac(a+c)2<4ac
a2+2ac+c2<4aca2+2ac+c2<4ac
a2−2ac+c2<0a2−2ac+c2<0
(a−c)2<0(a−c)2<0
This is clearly not possible, so a and c can not be both positive.
This means they can only be both negative, and indeed such equations exist, for example:
f(x)=−x2−1
b2−4ac<0b2−4ac<0
b2<4acb2<4ac
so the sign of a and c are the same, either both positive or both negative.
If both are positive then 0 < a+c < -b because a+b+c <0
0<(a+c)2<b2<4ac0<(a+c)2<b2<4ac
(a+c)2<4ac(a+c)2<4ac
a2+2ac+c2<4aca2+2ac+c2<4ac
a2−2ac+c2<0a2−2ac+c2<0
(a−c)2<0(a−c)2<0
This is clearly not possible, so a and c can not be both positive.
This means they can only be both negative, and indeed such equations exist, for example:
f(x)=−x2−1
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