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thedarktiger

Member

Registered: 2014-01-10

Posts: 91

Hard algebra

Let a,b be real numbers, and let x_1, x_2 be the roots of the quadratic equation x^2+ax+b=0.

Prove that if x_1, x_2 are real and nonzero,

, and b>0, then |a+2|>2.

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Good. You can read.

Olinguito

Member

Registered: 2014-08-12

Posts: 649

Re: Hard algebra

[EDIT: Oops, mistake. Salvaging operation in progress.]

Last edited by Olinguito (2014-10-31 23:56:50)

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Bassaricyon neblina

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Hard algebra

hi thedarktiger,

So

but you are given

Oh.  I seem to have hit a glitch too.  Working on it. 

Suppose x1 = 2 and x2 = 3.

So a = -5 and b = 6

That's OK for both constraints because 1/2 + 1/3 < 1 and b = 6 > 0

a+2 > 2-b becomes -3 > -4 which is ok.  And |a+2| = 3 > 2 so that's ok too.

Hhmmm still

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

thedarktiger

Member

Registered: 2014-01-10

Posts: 91

Re: Hard algebra

thank you @Olinguito and @bob bundy

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Good. You can read.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Hard algebra

I've played about with this for ages and this is my best attempt.

Using results from my earlier post

b>0 => x1.x2 > 0 => both are negative or both are positive.

The negative case is easier.

Suppose x1,x2 < 0 then x1 + x2 < 0 => -a < 0 => a > 0 => a + 2 > 2

Now suppose x1,x2 > 0

If 0 < x1,x2 ≤ 1 then 1/x1 + 1/x2 ≥ 1 which is not allowed so

both x1 and x2 > 1

Now consider the function

This seems to have a maximum at x1 = x2 = 2.  (I still have hopes of proving this.  )

I'll assume it does for the moment.

At this maximum y = 1 and a = -4, giving |a+2| = 2.  When anywhere else than this maximum, a < -4 (also needs proving) and so |a+2| > 2

EDIT: Here's part of a 3D graph showing that the function rises continuously in both directions towards x1 and x2 becoming zero, so the values x1=x2=2 is the
best in the domain(s)

LATER EDIT: When x1 = x2 = 2 we have the sum = the product ie. x1 + x2 = x1.x2

What happens if I increase either by an amount e.

(x1 + e) + x2 = x1 + x2 + e = 2 + e

(X1+e)(x2 = X1.x2 + e.X2

Thus the sum gets bigger by e and the product gets bigger by e.x2.  But if x2 > 2 then e.x2 > e so the product gets bigger at a faster rate. 

Hence y = 1/x1 + 1/x2 = (x1+x2)/x1.x2 gets smaller.  This shows that I have the maximum value for y in the domain(s).

Also if I increase x1 or x2, then I increase a.  So a<-4 is also proved.

I think I'll stop there for the moment. 

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Shrimpy

Member

Registered: 2017-02-03

Posts: 4

Re: Hard algebra

In the positive case, where does the "less than or equal" to 1 come from?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Hard algebra

hi Shrimpy

I tried this a long time ago and none of it is still in my brain.  Please would you quote exactly the bit you are referring to and I'll have a look.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob