Thread: 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!!

1. so or .

2. . Use quadratic formula.

3.

3.

Divide by 3 and get

Let

4.

This is just the difference of two cubes


Once again thank you!! [/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


-Dan

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


-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?

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.

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?

Use the quadratic formula on the quadratic factor.

-Dan