Hi

 

I'm trying to solve this :

assume:  a*x^2+b*x+c>=0 for all x with a≠0 then
have: b^2-4*a*c<=0

but I couldn't. I proved it when the quadratic equation is greater than zero

a*x^2+b*x+c>0 for all x with a≠0

but not for greater then or equal to zero. So any one help in solving this please.

Edited June 8, 20169 yr by O.J

If the quadratic = 0 for some x and is never < 0, then it has a double root at that point and the discriminant=0.

This is typically a statement that is best proven by contraposition. Assume that the discriminant is positive and then prove that your polynomial cannot be non negative in all points.

 

When you have a positive discriminant, your polynomial has exactly two roots and depending on the sign of a, it is either positive between these roots or negative. If it is negative, you're done. If not then it must be positive between the largest of the two roots and infinity and between minus infinity and the smallest of your two roots.

 

I guess you could also prove it directly, but then you'd probably have to waste time with who knows how many different cases.

 

Hope it is clear enough.

Edited June 28, 20169 yr by Keen

Hi

 

I'm trying to solve this :

assume:  a*x^2+b*x+c>=0 for all x with a≠0 then
have: b^2-4*a*c<=0

but I couldn't. I proved it when the quadratic equation is greater than zero

a*x^2+b*x+c>0 for all x with a≠0

but not for greater then or equal to zero. So any one help in solving this please.

 

 

you seem like you are missing something or we are unsure about something.

 

look please this example :

 

x² + 4 = ax² +bx +c , there a = 1 ( > 0 ) ( b= 0 no problem) and c= 4 , ∆ < 0 ( ∆ = - 16) (as you would) then look at the solution

 

x₁= (-b +√ ∆)/2 x₂ = (-b -√ ∆)/2

 

x₁ = 2i x₂ = - 2i both x₁ & x₂ ϵ ₵

 

and how did you compare with zero ?

 

( when you say x₁ ,x₂ are >0 ??)