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

**nathan02079** · January 28th 2009, 04:19 PM

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

).

1.

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

2.

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

3.

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

Thank you sooo much!

Nate

**Constatine11** · January 28th 2009, 08:54 PM

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

).

1.

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

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

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

which you expand and simplify and draw conclusions.

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

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

which you expand and simplify and draw conclusions.

.

**Constatine11** · January 28th 2009, 08:56 PM

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

).

3.

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

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

.

**Constatine11** · January 28th 2009, 08:58 PM

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

).

2.

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

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

.

**nathan02079** · January 28th 2009, 09:45 PM

Originally Posted by Constatine11

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

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

which you expand and simplify and draw conclusions.

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

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

which you expand and simplify and draw conclusions.

.

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

and

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

I have yet to get to the answer

Originally Posted by Constatine11

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

.

Sorry, what does this mean exactly?

Originally Posted by Constatine11

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

.

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

Is it?

**Constatine11** · January 28th 2009, 10:16 PM

Originally Posted by nathan02079

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

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

which is impossible for real $x$

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

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

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

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

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

.

**Constatine11** · January 28th 2009, 10:22 PM

Originally Posted by nathan02079

Originally Posted by Constatine11

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

.

Sorry, what does this mean exactly?

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

$<br /> y^2-9y+8<br />$

and so:

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

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

.

**Constatine11** · January 28th 2009, 10:29 PM

Originally Posted by nathan02079

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

Is it?

Consider:

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

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

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

into, and draw any conclusions.

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

.

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