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Tushar Watts Grade: 12
`        If the eqn ax2+bx+c=0  (0 1 and Z2, then show that |Z1| >1 , |Z2| >1.`
8 years ago

147 Points
```										If the eqn ax2+bx+c=0  (0 <a<b<c) has non real complex roots Z1 and Z2
then z1 & Z2 must be conjugate complex number.
and b2-4ac<0
and |Z1|= |Z2|
so i f  i will prove for one root then it is also true for other root
z1,z2={-b±i√(4ac-b2)}/2a
|z1|=√{b2 +(√(4ac-b2))2}/2a
|z1|=√{b2+4ac-b2}/2a
=√(c/a) >1
so |Z1|= |Z2| >1

```
8 years ago
mycroft holmes
271 Points
```										As in the previous thread we note that z1 and z2 are complex conjugates. Hence |z1| = |z2|.

Now z1 z2 = c/a and hence |z1 z2| = |z1|2 = |c/a|>1 since c>a>0
```
8 years ago
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