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#1 2014-11-04 03:16:46

PhuongMath
Member
Registered: 2014-10-25
Posts: 3

Find min of expression

1. Suppose that $x_{1}$ and $x_{2}$ different equation $ax^3+bx^2+cx+d=0$
Prove that: $x_{1}x_{2} \geq \frac{4ac-b^2} {4a^2}$
2. For the polynomial $f(x)=ax^2+bx+c$ knows that $f(x) \geq 0$ for all real numbers $x$ and $b>a$. Find min of expression: $P=\frac{a+b+c}{b-a}$

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#2 2014-11-04 06:12:04

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Find min of expression

Hi;

Please use:

instead of $$, also you have two questions that you have not replied to.


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.

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#3 2014-11-04 20:44:46

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Find min of expression

hi PhuongMath

Q!.  I've spent some time on this but not succeeded.  I notice that the required expression is close to b^2 - 4ac for a quadratic so I tried to make a quadratic from the cubic.

If you write the two cubics with x1 and x2 as the solutions and then subtract, you can factor out (x1-x2) and so end up with a quadratic with a, b and c.  But that's as far as I have got.  sad

Q2.  As f(x) is never negative, we know that a > 0 and so b >a > 0.  Also c  ≥ 0 else f(0) < 0.

Consider

So P = 1 is certainly a lower bound, as the fraction > 0.  So how small can I make the fraction ?

Is c=0 is allowable ? As f(x) cannot drop below zero, that would mean x = 0 would have to be the lowest point.  This point has coordinate -b/2a which would make b = 0 which contradicts b > 0.  So c> 0.

So P will never be 1.  If (b-a) > (2a+c) then we have a fraction < 1.  Can I make such a fraction ?

As f(x) has, at most, one solution (it never crosses the axis) => b^2 - 4ac ≤ 0

(b-a) > 2a + c => b > 3a + c

Then b^2 > (3a+c)^2 = 9a^2 + c^2 + 6ac

So we need  9a^2 + c^2 + 6ac < b^2 ≤ 4ac  This is impossible so I cannot make the fraction < 1.

=> The lower bound has moved up to P = 2

by trial, the best I have so far is P  ≈ 3.19 when a = 2  b = 9  c = 11

still working..................................

Bob


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 smile

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#4 2014-11-05 01:46:33

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Find min of expression

P = 3 when a = 2, b = c = 8

Bob


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 smile

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