Math Is Fun Forum

juantheron

Member

Registered: 2011-10-19

Posts: 312

quadratic equation

if the equation

does not have

distinct real roots and

then prove that

.

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: quadratic equation

As far as I can tell this isn't true.  For example, try a = -1, b = 1, c = -1.  -x^2 + x - 1 does not have 2 distinct real roots, a + b > c, but f(x) < 0 for all real values of x.

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Wrap it in bacon

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: quadratic equation

i think here

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: quadratic equation

Hi;

Are you sure about your constraints?

When

a = -2
b = -2
c = 1
x = -2

This has no real roots and f(-2) < 0.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: quadratic equation

That has 2 real roots bobby, at

The new constraint fixes it.  First I'll assume that 'a' is not allowed to be 0, since otherwise the result is wrong.  In order for a quadratic to have fewer than 2 real roots you need

Since b^2 is always >= 0, 4ac > 0 (it can equal 0 if b and c equal 0, but that's an uninteresting exception I'll address at the end) that means a and c must either both be positive or both be negative.  If a > 0 then f(x) >= 0 for all x, if a < 0 then f(x) <= 0 for all x, so we need to prove that a > 0 under the constraints.  Do this by contradiction and assume a, and also c, are less than 0:

This is a contradiction so our assumption that a < 0 must be wrong.  Since a > 0 that means f(x) >= 0 for all x.

(Note that if b = c = 0 we're left with a + 0 > 0 ==> a > 0 by our constraint, so the proof is immediately clear).

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Wrap it in bacon

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: quadratic equation

Your right but those are not the roots of that quadratic. But it does not matter for your proof.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: quadratic equation

Your right but those are not the roots of that quadratic. But it does not matter for your proof.

Hah, I can do a proof concerning quadratic equations but can't solve a simple one

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Wrap it in bacon

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: quadratic equation

I can not complain. Obviously it is a tricky one, I got it wrong too! Good proof though!

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: quadratic equation

thaks the dude got it

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: quadratic equation

i have to complain because f(x) was nowhere defined,so you should fix that.say f(x)=ax^2+bx+c.

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