inequality

**alexandros** · May 31st 2010, 03:49 PM

Challenge problem:

Given:

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

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

Prove:

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

Moderator approved, CB

**Defunkt** · June 2nd 2010, 05:45 AM

Originally Posted by alexandros

Given:

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

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

Prove:

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

Moderator approved, CB

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

**alexandros** · June 2nd 2010, 08:26 AM

Originally Posted by Defunkt

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

in fact $2a^2>b$

**Defunkt** · June 7th 2010, 01:49 PM

Originally Posted by alexandros

in fact $2a^2>b$

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

**alexandros** · June 7th 2010, 05:29 PM

Originally Posted by Defunkt

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

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

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

$|x_{2}|<|x_{1}|$

**melese** · July 21st 2010, 09:30 AM

Here is what I did.
By Vieta's formulas, we have $x_1+x_2=2a$ , and $x_1\cdot x_2=b$ . Then ${x_1}^2+{x_2}^2+2x_1x_2={x_1}^2+{x_2}^2+2b=4a^2$ . And then, ${x_1}^2+{x_2}^2=4a^2-2a=2(2a^2+b)$ .

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

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

From all this we get the inequality: $2(2a^2-b)-|b|< x_{1}^2<2(2a^2-b)$ .

**Archie Meade** · July 21st 2010, 02:12 PM

There are a few typos in there...

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

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

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

Prove the following...

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

Proof

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

or

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

Therefore

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

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

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

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

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

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

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

which is $>\left(x_1\right)^2$

**alexandros** · July 25th 2010, 07:11 PM

Here is an easy double implication proof:

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

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

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

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

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