1. ## Real Number System

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."

2. 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.

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

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

Indeed, by polynomial division, we see that:

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

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

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

Factorising this:

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

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

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

RonL

5. 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.

6. 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 $\displaystyle px^n + ax^{n-1} + ... + bx + q = 0$, any rational roots of this polynomial equation will be of the form: $\displaystyle \pm \frac{\text{factor of q}}{\text{factor of p}}$. 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

7. Hello, fdrhs!

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

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

Factor "by grouping" . . .

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

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

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

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

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

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

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

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

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

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

EB

9. 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.