# Thread: to prove that...

1. ## to prove that...

I have a problem I can not solve this:

Let a, b, c are real numbers and f(x) = x^2 + ax + b.
if the equation f(x) = 0 has two distinct real solutions,
and the equation (x^2-2x + c)^2 + a (x^2-2x + c) + b = 0
has no real solutions, to prove that f(c) > 1

2. ## Re: to prove that...

1. For f(x)=0 to have 2 distinct real solutions it must be D>0 where D stands for discriminant
2. Since D>0 it has to be (a^2)>4b in the first equation {solve from here}
3. For f(x^2-2x + c)=0 to have no real solutions it must be D<0 {solve from here}

Hint: For all quadratics, f(x) is greater or equal to 0.

3. ## Re: to prove that...

Originally Posted by alexagak
1. For f(x)=0 to have 2 distinct real solutions it must be D>0 where D stands for discriminant
2. Since D>0 it has to be (a^2)>4b in the first equation {solve from here}

Hint: For all quadratics, f(x) is greater or equal to 0.
This version of f(x) had better not be greater than or equal to zero! Else it will not have two distinct real solutions.

For the last line, yes.

-Dan