Real Number System

**fdrhs** · February 1st 2007, 02:58 PM

For both questions below, solve each equation in the real number system.

[1] x^3 + (3x^2/2) + 3x - 2 = 0

[2] 2x^4 + x^3 - 24x^2 + 20x + 16 = 0

NOTE: I do not understand the wording "...solve each equation in the real number system."

**ThePerfectHacker** · February 1st 2007, 04:31 PM

Originally Posted by fdrhs

NOTE: I do not understand the wording "...solve each equation in the real number system."

Meaning if you get any complex numbers (imaginary) then disregard* them.

*)Pretend they are rules of the forum, that is, ignore them as you did.

**AlvinCY** · February 1st 2007, 05:30 PM

$x^3 + (3x^2/2) + 3x - 2 = 0$

By trial and error (or the way I did it was to plot the graph), you'll find that $x = \frac{1}{2}$ is a solution.

Indeed, by polynomial division, we see that:

$x^3 + (3x^2/2) + 3x - 2 = (x - \frac{1}{2})(x^2 + 2x + 4)$ and we know that $x^2 + 2x + 4$ cannot be factorised over the Real field as its determinant is less than 0.

$2x^4 + x^3 - 24x^2 + 20x + 16 = 0$

Once again, by trial and error (or plot the graph), you'll find that $x = 2$ is a solution.

Factorising this:

$2x^4 + x^3 - 24x^2 + 20x + 16$
$= (x - 2)(2x^3 + 5x^2 - 14x - 8)$
$= (x - 2)(x + \frac{1}{2})(2x^2 + 4x - 16)$ , for neater form:
$= (x - 2)(2x + 1)(x^2 + 2x - 8)$

We know we can't simplify this further as $(x^2 + 2x - 8)$ has a determinant which is less than 0, so this second question has $x = 2$ and $x = -\frac{1}{2}$

**CaptainBlack** · February 1st 2007, 10:39 PM

Originally Posted by AlvinCY

[
We know we can't simplify this further as $(x^2 + 2x - 8)$ has a determinant which is less than 0, so this second question has $x = 2$ and $x = -\frac{1}{2}$

discriminant not determinant

RonL

**AlvinCY** · February 2nd 2007, 02:40 AM

Originally Posted by CaptainBlack

discriminant not determinant

RonL

LoL... I've been doing way too much physics and advanced algebra at uni, and ahve been swapping the terms around way too often.

**topsquark** · February 2nd 2007, 04:06 AM

Originally Posted by fdrhs

For both questions below, solve each equation in the real number system.

[1] x^3 + (3x^2/2) + 3x - 2 = 0

[2] 2x^4 + x^3 - 24x^2 + 20x + 16 = 0

NOTE: I do not understand the wording "...solve each equation in the real number system."

In the event you don't know a way to start finding solutions to these (AlvinCY used, presumably logical, guesswork) you can use the "Rational Root Theorem" to come up with some possibilities.

Given $px^n + ax^{n-1} + ... + bx + q = 0$ , any rational roots of this polynomial equation will be of the form: $\pm \frac{\text{factor of q}}{\text{factor of p}}$ . There may, of course, be other real solutions to the equation, but this will give you a list to start with, anyway.

-Dan

Originally Posted by AlvinCY

LoL... I've been doing way too much physics and advanced algebra at uni, and ahve been swapping the terms around way too often.

Can't EVER do too much Physics!

-Dan

**Soroban** · February 2nd 2007, 06:22 AM

Hello, fdrhs!

The second one factors . . . if you hammer at it long enough.

$[2]\;\;2x^4 + x^3 - 24x^2 + 20x + 16 \:= \:0$

Factor "by grouping" . . .

$x^3(2x + 1) - 4(x^2 + 5x - 4) \:=\:0$

$x^3\underbrace{(2x + 1)} - 4\underbrace{(2x + 1)}(3x - 4) \:=\:0$

$(2x + 1)(x^3 - 4[3x-4]) \:=\:0\quad\Rightarrow\quad (2x + 1)(x^3 - 12x + 16)\:=\:0$

We find that $x = 2$ is a zero of cubic. .Hence, $(x - 2)$ is a factor.

And we have: . $(2x + 1)(x^3-12x + 16)\:=\:0$

. . . . . Then: . $(2x + 1)\overbrace{(x - 2)(x^2 + 2x - 8)}\:=\:0$

. . . . . And: . $(2x + 1)(x - 2)\overbrace{(x - 2)(x + 4)}\:=\:0$

Therefore, the four roots are: . $x\;=\;\text{-}\frac{1}{2},\:2,\:2,\:\text{-}4$

**earboth** · February 2nd 2007, 07:54 AM

Originally Posted by AlvinCY

...
We know we can't simplify this further as $(x^2 + 2x - 8)$ has a determinant which is less than 0, so this second question has $x = 2$ and $x = -\frac{1}{2}$

Hello,

there must be a mistake: To calculate the dicrimininant you have to subtract the (-8) thus you'll get a positive number.

EB

**AlvinCY** · February 2nd 2007, 01:47 PM

I do apologise, I was doing it at late hours of night while chatting on MSN with others about their weekends.

Thanks for picking it up guys, and yes, I do mean discriminants... and no it wasn't negative, therefore you CAN factorise it.

I certainly look like a little bit of a fool at the moment

Thanks guys.

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