# solving equation

• Aug 1st 2007, 02:17 PM
WolfMV
solving equation
Greetings all,
I have a test tomorrow and i am having difficulties with one section of the study guide. If anyone could help me with solutions that would be really appreciated.

Basically he wants us to find all solutions and factor the following problems.

1. 54x + 2x^3=0

2. -11 + 2x - 3x^2=0

3. 3x^4 + 18x^2 - 48=0

4. x^3-8=0

Once again thank you!! :)
• Aug 1st 2007, 02:26 PM
tukeywilliams
1. $x(54+2x^2) = 0$ so $x = 0$ or $x^2 = -26 \implies x = \pm i \sqrt{26}$.

2. $-3x^2 +2x - 11$. Use quadratic formula.

3.
• Aug 1st 2007, 02:47 PM
galactus
Quote:

3. $3x^{4} + 18x^{2} - 48=0$
Divide by 3 and get $x^{4}+6x^{2}-16=0$

Let $u=x^{2}$

$u^{2}+6u-16=0$

$(u-2)(u+8)=0$

$3(x^{2}-2)(x^{2}+8)=0$

Quote:

4. $x^{3}-8=0$
This is just the difference of two cubes

Once again thank you!! :)[/quote]
• Aug 1st 2007, 06:31 PM
topsquark
Quote:

Originally Posted by WolfMV
4. x^3-8=0

Let me expand a little on galactus' comment, since most who haven't seen how to solve this would likely miss two solutions.

You need to solve
$x^3 - 8 = (x - 2)(x^2 + 2x + 4) = 0$

-Dan
• Aug 1st 2007, 07:52 PM
Ideasman
Quote:

Originally Posted by topsquark
Let me expand a little on galactus' comment, since most who haven't seen how to solve this would likely miss two solutions.

You need to solve
$x^3 - 8 = (x - 2)(x^2 + 2x + 4) = 0$

-Dan

How's this complicated?

x^3 = 8, and then it's easy to see x = 2.

I see that x^3 - 8 factors into (x - 2)(x^2 + 2x + 4); this is equal to 0 when x = 2 for the (x - 2) but for the (x^2 + 2x + 4) factor, this will never equal 0.

Did I miss something?
• Aug 1st 2007, 08:18 PM
ThePerfectHacker
Quote:

Originally Posted by Ideasman
How's this complicated?

x^3 = 8, and then it's easy to see x = 2.

I see that x^3 - 8 factors into (x - 2)(x^2 + 2x + 4); this is equal to 0 when x = 2 for the (x - 2) but for the (x^2 + 2x + 4) factor, this will never equal 0.

Did I miss something?

The second factor is never equal to zero when its is real. It can still have complex solutions.
• Aug 2nd 2007, 04:15 AM
topsquark
Quote:

Originally Posted by Ideasman
How's this complicated?

x^3 = 8, and then it's easy to see x = 2.

I see that x^3 - 8 factors into (x - 2)(x^2 + 2x + 4); this is equal to 0 when x = 2 for the (x - 2) but for the (x^2 + 2x + 4) factor, this will never equal 0.

Did I miss something?