# (Algebraic Inequalities) Can somebody show me the first step to these 3 equations?

• Jan 28th 2009, 04:19 PM
nathan02079
(Algebraic Inequalities) Can somebody show me the first step to these 3 equations?
Hello, I'm absolutely stuck. :S My teacher gave me these 'challenge' problems and I have no idea how to get started. Can somebody explain how to start each equation? (Not solving it would be okay. I would love to get the answer...just don't know how to start it(Worried)(Worried)).

1.

$\displaystyle {{(x-1)(x-2)(x-3)}\over{(x+1)(x+2)(x+3)}}>1$

2.

$\displaystyle a^4+a^3-a-1<0$

3.

$\displaystyle x^6-9x^3+8>0$

Thank you sooo much!

Nate
• Jan 28th 2009, 08:54 PM
Constatine11
Quote:

Originally Posted by nathan02079
Hello, I'm absolutely stuck. :S My teacher gave me these 'challenge' problems and I have no idea how to get started. Can somebody explain how to start each equation? (Not solving it would be okay. I would love to get the answer...just don't know how to start it(Worried)(Worried)).

1.

$\displaystyle {{(x-1)(x-2)(x-3)}\over{(x+1)(x+2)(x+3)}}>1$

If $\displaystyle (x+1)(x+2)(x+3)>0$ multiplying through by it gives:

$\displaystyle (x-1)(x-2)(x-3)>(x+1)(x+2)(x+3)$

which you expand and simplify and draw conclusions.

But if $\displaystyle (x+1)(x+2)(x+3)<0$ we get instead:

$\displaystyle (x-1)(x-2)(x-3)<(x+1)(x+2)(x+3)$

which you expand and simplify and draw conclusions.

.
• Jan 28th 2009, 08:56 PM
Constatine11
Quote:

Originally Posted by nathan02079
Hello, I'm absolutely stuck. :S My teacher gave me these 'challenge' problems and I have no idea how to get started. Can somebody explain how to start each equation? (Not solving it would be okay. I would love to get the answer...just don't know how to start it(Worried)(Worried)).

3.

$\displaystyle x^6-9x^3+8>0$

$\displaystyle x^6-9x^3+8$ is a quadratic in $\displaystyle y=x^3$

.
• Jan 28th 2009, 08:58 PM
Constatine11
Quote:

Originally Posted by nathan02079
Hello, I'm absolutely stuck. :S My teacher gave me these 'challenge' problems and I have no idea how to get started. Can somebody explain how to start each equation? (Not solving it would be okay. I would love to get the answer...just don't know how to start it(Worried)(Worried)).

2.

$\displaystyle a^4+a^3-a-1<0$

Does the left hand side have any obvious roots or factors?

.
• Jan 28th 2009, 09:45 PM
nathan02079
Quote:

Originally Posted by Constatine11
If $\displaystyle (x+1)(x+2)(x+3)>0$ multiplying through by it gives:

$\displaystyle (x-1)(x-2)(x-3)>(x+1)(x+2)(x+3)$

which you expand and simplify and draw conclusions.

But if $\displaystyle (x+1)(x+2)(x+3)<0$ we get instead:

$\displaystyle (x-1)(x-2)(x-3)<(x+1)(x+2)(x+3)$

which you expand and simplify and draw conclusions.

.

$\displaystyle (x-1)(x-2)(x-3)>(x+1)(x+2)(x+3)$
$\displaystyle X^3-3x^2+2x-3x^2+9x-6>x^3+3x^2+2x+3x^2+9x+6$
$\displaystyle -12x^2-12>0$
$\displaystyle -12(x+1)(x-1)>0$

and

$\displaystyle (x-1)(x-2)(x-3)<(x+1)(x+2)(x+3)$
$\displaystyle X^3-3x^2+2x-3x^2+9x-6<x^3+3x^2+2x+3x^2+9x+6$
$\displaystyle -12x^2-12<0$
$\displaystyle -12(x+1)(x-1)<0$

I have yet to get to the answer

Quote:

Originally Posted by Constatine11
$\displaystyle x^6-9x^3+8$ is a quadratic in $\displaystyle y=x^3$

.

Sorry, what does this mean exactly?

Quote:

Originally Posted by Constatine11
Does the left hand side have any obvious roots or factors?

.

$\displaystyle a(a^3+a^2-1)-1<0$

Is it?
• Jan 28th 2009, 10:16 PM
Constatine11
Quote:

Originally Posted by nathan02079
$\displaystyle (x-1)(x-2)(x-3)>(x+1)(x+2)(x+3)$
$\displaystyle X^3-3x^2+2x-3x^2+9x-6>x^3+3x^2+2x+3x^2+9x+6$
$\displaystyle -12x^2-12>0$
$\displaystyle -12(x+1)(x-1)>0$

$\displaystyle -12x^2-12>0 \implies x^2<-1$

which is impossible for real $\displaystyle x$

Quote:

$\displaystyle (x-1)(x-2)(x-3)<(x+1)(x+2)(x+3)$
$\displaystyle X^3-3x^2+2x-3x^2+9x-6<x^3+3x^2+2x+3x^2+9x+6$
$\displaystyle -12x^2-12<0$
$\displaystyle -12(x+1)(x-1)<0$
$\displaystyle -12x^2-12<0 \implies x^2>-1$

which is always true for real $\displaystyle x$. So we conclude that:

$\displaystyle (x+1)(x+2)(x+3)<0$

or that: $\displaystyle x \in (-\infty,-3) \cup (-2,-1)$

.
• Jan 28th 2009, 10:22 PM
Constatine11
Quote:

Originally Posted by nathan02079
Quote:

Originally Posted by Constatine11
$\displaystyle x^6-9x^3+8$ is a quadratic in $\displaystyle y=x^3$

.

Sorry, what does this mean exactly?

It means that if you make the substitution $\displaystyle y=x^3$ you get the quadratic:

$\displaystyle y^2-9y+8$

and so:

$\displaystyle y^2-9y+8>0$

is satisfied on $\displaystyle (-\infty, y_1) \cup (y_2, \infty)$ , where $\displaystyle y_1 \le y_2$ are the roots of $\displaystyle y^2-9y+8=0$, which now needs to be translated into something to do with $\displaystyle x$

.
• Jan 28th 2009, 10:29 PM
Constatine11
Quote:

Originally Posted by nathan02079
$\displaystyle a(a^3+a^2-1)-1<0$

Is it?

Consider:

$\displaystyle a^4+a^3-a-1=0$

The rational root theorem suggests that if this has rational roots then they are either $\displaystyle a=1$ or $\displaystyle a=-1$, test these. If one or other of these are roots then $\displaystyle (a-1)$ and/or $\displaystyle (a+1)$ are factors of $\displaystyle a^4+a^3-a-1$, so take out the factors and see what that transforms:

$\displaystyle a^4+a^3-a-1<0$

into, and draw any conclusions.

(what this should do, if you get it right, is lead you to find all the real roots of $\displaystyle a^4+a^3-a-1$ then the inequality will be satisfied between appropriate consecutive pairs of these roots)

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