quadratic

[suyogya]

I have tried as:
Given: x² + bx + (cx + ca)(2x + b)
or (1 + 2c) x² + (b + bc + 2ca)x + abc
Discriminant D = (b + cb + 2 ca)² - 4abc (1 + 2c)
or D = (4c²)a² - (4bc²)a + b² (1+c)² . Now this needs to be > 0. How to do this?

 

[lev888]

View attachment 18290
I have tried as:
Given: x² + bx + (cx + ca)(2x + b)
or (1 + 2c) x² + (b + bc + 2ca)x + abc
Discriminant D = (b + cb + 2 ca)² - 4abc (1 + 2c)
or D = (4c²)a² - (4bc²)a + b² (1+c)² . Now this needs to be > 0. How to do this?

What about the first equitation? Doesn't it give you something?

 

[suyogya]

it of course will give me b² greater than or equal to 4c.

 

[lev888]

it of course will give me b² greater than or equal to 4c.

I would find the minimum of D(a). Maybe the fact above ensures that it's not negative.

 

[suyogya]

Thanks, I just done it.