Discriminant and quadratic problem

[EF17xx]

Homework Statement

Homework Equations

Discriminant : b^2-4ac

When discriminant = 0 The function has two equal real roots
When discriminant < 0 The discriminant has NO real roots
When discriminant > 0 The function has 2 different real roots

The Attempt at a Solution

Why is it so that when the function is equal or less than 0 that the function has 2 equal roots or no roots.
I understand that when a quadratic is set equal to 0 that when you solve for x it will give you the roots. However when a quadratic is set to 0 the roots are not always necessarily equal they CAN be different. Could someone please explain why the statement is true?

 

[Mark44]

Homework Statement

View attachment 224102
View attachment 224101

Homework Equations

Discriminant : b^2-4ac

When discriminant = 0 The function has two equal real roots
When discriminant < 0 The discriminant has NO real roots
When discriminant > 0 The function has 2 different real roots

The Attempt at a Solution

Why is it so that when the function is equal or less than 0 that the function has 2 equal roots or no roots.
I understand that when a quadratic is set equal to 0 that when you solve for x it will give you the roots. However when a quadratic is set to 0 the roots are not always necessarily equal they CAN be different. Could someone please explain why the statement is true?

Note: Thread moved from Precalc section as it involves derivatives.

Your use of the word "function" is unclear as you seem to be confusing the derivative function (f') with the results of the discriminant.
You have $f'(x) = 3px^2 + 2px + q$, which is a quadratic function. The discriminant tells you that the equation $y = ax^2 + bx + c$ has two roots (discriminant > 0), a single root (discriminant = 0), or no real roots (discriminant < 0). Use this information to show that if $f'(x) \ge 0$, then f' has either two equal roots or no roots at all.

 

[BvU]

Hi,

Draw a bunch of parabolas that satisfy

and see that for more than one root the function has to 'go through' $f'= 0$ twice -- meaning it has to be $>0$ over some range.

What also helps is to work out $ax^2 + bx + c$ in the form $\ d(x+e)^2 - f^2\$: you see where this determinant comes from.

 

[epenguin]

Well just draw a picture of a function like that Complete with coordinates andsubject to the ≥ 0 condition and you may see.

 

[Ray Vickson]

Homework Statement

View attachment 224102
View attachment 224101

Homework Equations

Discriminant : b^2-4ac

When discriminant = 0 The function has two equal real roots
When discriminant < 0 The discriminant has NO real roots
When discriminant > 0 The function has 2 different real roots

The Attempt at a Solution

Why is it so that when the function is equal or less than 0 that the function has 2 equal roots or no roots.
I understand that when a quadratic is set equal to 0 that when you solve for x it will give you the roots. However when a quadratic is set to 0 the roots are not always necessarily equal they CAN be different. Could someone please explain why the statement is true?

I think the question is badly worded. It should either say $f'(x_0) \geq 0$ for some $x_0$ (in which case the stated result is not always true) or else it should say $f'(x) \geq 0$ for all $x$ (in which case the result is quite apparent, but I guess still needs a proof).

 

[BvU]

I think the question is badly worded

I agree: it should have read

if $\ f'(x) \ge 0 \$ then the equation $f'(x) = 0$ has two equal roots or no real roots​