# inequality

• May 31st 2010, 02:49 PM
alexandros
inequality
Challenge problem:

Given:

1) $\displaystyle x^2-2ax+b=0$ with $\displaystyle b\neq 0$

2) $\displaystyle |x_{2}|< |x_{1}|$,where $\displaystyle x_{1},x_{2}$ are the roots of the above equation

Prove:

$\displaystyle 2(2a^2-b)-|b|< x_{1}^2<2(2a^2-b)$

Moderator approved, CB
• Jun 2nd 2010, 04:45 AM
Defunkt
Quote:

Originally Posted by alexandros
Given:

1) $\displaystyle x^2-2ax+b=0$ with $\displaystyle b\neq 0$

2) $\displaystyle |x_{2}|< |x_{1}|$,where $\displaystyle x_{1},x_{2}$ are the roots of the above equation

Prove:

$\displaystyle 2(2a^2-b)-|b|< x_{1}^2<2(2a^2-b)$

Moderator approved, CB

In order to even consider the $\displaystyle <$ relation on $\displaystyle x_1^2$, shouldn't we require that $\displaystyle a^2 \geq b$? (That is, assuming $\displaystyle a,b \in \mathbb{R}$)
• Jun 2nd 2010, 07:26 AM
alexandros
Quote:

Originally Posted by Defunkt
In order to even consider the $\displaystyle <$ relation on $\displaystyle x_1^2$, shouldn't we require that $\displaystyle a^2 \geq b$? (That is, assuming $\displaystyle a,b \in \mathbb{R}$)

in fact $\displaystyle 2a^2>b$
• Jun 7th 2010, 12:49 PM
Defunkt
Quote:

Originally Posted by alexandros
in fact $\displaystyle 2a^2>b$

This still doesn't give us that the solutions are real. Take $\displaystyle a=2, ~ b=3$ to get the equation $\displaystyle x^2-4x+6=0$ which has no real solutions, although $\displaystyle 2a^2=2 \cdot 2^2 = 8 > 6 = b$... So are you requiring that $\displaystyle a^2>b$ or not?
• Jun 7th 2010, 04:29 PM
alexandros
Quote:

Originally Posted by Defunkt
This still doesn't give us that the solutions are real. Take $\displaystyle a=2, ~ b=3$ to get the equation $\displaystyle x^2-4x+6=0$ which has no real solutions, although $\displaystyle 2a^2=2 \cdot 2^2 = 8 > 6 = b$... So are you requiring that $\displaystyle a^2>b$ or not?

No the problem does require $\displaystyle 2a^2>b$

In your example : $\displaystyle |x_{1}| = |x_{2}|$ ,in our problem we have :

$\displaystyle |x_{2}|<|x_{1}|$
• Jul 21st 2010, 08:30 AM
melese
Here is what I did.
By Vieta's formulas, we have $\displaystyle x_1+x_2=2a$, and $\displaystyle x_1\cdot x_2=b$. Then $\displaystyle {x_1}^2+{x_2}^2+2x_1x_2={x_1}^2+{x_2}^2+2b=4a^2$. And then, $\displaystyle {x_1}^2+{x_2}^2=4a^2-2a=2(2a^2+b)$.

Now, we have $\displaystyle |x_1\cdot x_2|=|x_2|\cdot|x_1|=|b|$, and since $\displaystyle |x_{2}|< |x_{1}|$ we arrive at (*)$\displaystyle {x_2}^2<|b|$.

Eventually, $\displaystyle {x_1}^2=2(2a^2+b)-{x_2}^2$. Notice that $\displaystyle x_1, x_2 \neq 0$ because $\displaystyle 0^2-2a0+b=b\neq 0$.
It follows that $\displaystyle {x_1}^2=2(2a^2+b)-{x_2}^2<2(2a^2+b)$ ($\displaystyle {x_2}^2$ is positive).
Also from (*), $\displaystyle -{x_2}^2>-|b|$ and it follows that $\displaystyle {x_1}^2=2(2a^2+b)-{x_2}^2>2(2a^2+b)-|b|$.

From all this we get the inequality: $\displaystyle 2(2a^2-b)-|b|< x_{1}^2<2(2a^2-b)$.
• Jul 21st 2010, 01:12 PM
There are a few typos in there...

(1) $\displaystyle x^2-2ax+b=0,\ b\ne0$

Roots of f(x)=0 are $\displaystyle x_1,\ x_2$

(2) $\displaystyle |x_2|<|x_1|$

Prove the following...

$\displaystyle 2\left(2a^2-b\right)-|b|<\left(x_1\right)^2<2\left(2a^2-b\right)$

Proof

$\displaystyle \left(x-x_1\right)\left(x-x_2\right)=x^2-2\left(x_1+x_2\right)+x_1x_2$

or

$\displaystyle x^2-2ax+b=x^2-(sum\ of\ roots)x+product\ of\ roots$

Therefore

$\displaystyle x_1+x_2=2a,\ x_1x_2=b$

$\displaystyle \left(x_1+x_2\right)^2=\left(2a\right)^2\ \Rightarrow\ \left(x_1\right)^2+2x_1x_2+\left(x_2\right)^2=4a^2 =\left(x_1\right)^2+\left(x_2\right)^2+2b$

$\displaystyle \left(x_1\right)^2+\left(x_2\right)^2=4a^2-2b=2\left(2a^2-b\right)$

Hence...since $\displaystyle |x_2|<|x_1|,$

$\displaystyle |x_1x_2|>\left(x_2\right)^2\ \Rightarrow\ \left(x_1\right)^2+\left(x_2\right)^2-|x_1x_2|=\left(x_1\right)^2-|k|$

This is $\displaystyle <\left(x_1\right)^2$

it's already been established that $\displaystyle 2\left(2a^2-b\right)=\left(x_1\right)^2+\left(x_2\right)^2$

which is $\displaystyle >\left(x_1\right)^2$
• Jul 25th 2010, 06:11 PM
alexandros
Here is an easy double implication proof:

$\displaystyle 2(2a^2-b)-|b|<(x_{1})^2< 2(2a^2-b)\Longleftrightarrow$ $\displaystyle -|b|<(x_{1})^2-4a^2+2b<0\Longleftrightarrow$ $\displaystyle -|x_{1}x_{2}|<(x_{1})^2-(x_{1}+x_{2})^2+2x_{1}x_{2}<0\Longleftrightarrow$

...................(since $\displaystyle x_{1}+x_{2}=2a$ and $\displaystyle x_{1}x_{2}=b$)

$\displaystyle \Longleftrightarrow |x_{1}x_{2}|>|x_{2}|^2>0\Longleftrightarrow$ $\displaystyle |x_{2}|<|x_{1}|$

............(since $\displaystyle b\neq 0\Longrightarrow |x_{1}|\neq 0\wedge|x_{2}|\neq 0$)