1. ## what's x's value?

could you tell me what is X's Value?

2. Originally Posted by neworld222
could you tell me what is X's Value?
Yikes!
$x^4 - x^3 - 4 = 0$
can be done analytically, but is much easier to approximate numerically. I get
x = -1.21733 or x = 1.7484
as the only real solutions.

-Dan

3. Originally Posted by topsquark
Yikes!
$x^4 - x^3 - 4 = 0$
can be done analytically, but is much easier to approximate numerically. I get
x = -1.21733 or x = 1.7484
as the only real solutions.

-Dan
those are correct, i used a utility on my pc to find them

how did you approximate it numerically (i was thinking about using Newton-Raphson approximation, but i couldn't bother)?

how can it be done analytically?

4. Solving a quartic can be done, but it is cumbersome. Here is a nice step-by-step approach to the process.

The "Quartic Formula"

5. Originally Posted by galactus
Solving a quartic can be done, but it is cumbersome. Here is a nice step-by-step approach to the process.

The "Quartic Formula"
thx,though the method is funny,but i think it's the best one if i don't want solve this subject by use advanced-algebra

6. Originally Posted by neworld222
thx,though the method is funny,but i think it's the best one if i don't want solve this subject by use a advanced-algebra
well, that's true i guess. Newton-Raphson approximation is a Calc 1 topic, so it's not really that advanced... in the way we would think of advanced.

the quartic formula, which is so cumbersome that i don't see my self using it anytime soon (if ever), is probably the method your advanced algebra professor would want. don't blame me if it takes up 6 pages (back and front) though

7. Originally Posted by Jhevon
well, that's true i guess. Newton-Raphson approximation is a Calc 1 topic, so it's not really that advanced... in the way we would think of advanced.

the quartic formula, which is so cumbersome that i don't see my self using it anytime soon (if ever), is probably the method your advanced algebra professor would want. don't blame me if it takes up 6 pages (back and front) though
i should'nt blame you,if i'll have a headache when i calculate by 'advanced method',to a certainty blame Mr.Newton

8. Originally Posted by galactus
Solving a quartic can be done, but it is cumbersome. Here is a nice step-by-step approach to the process.

The "Quartic Formula"
Thank you! This is the answer to a post I made a while ago. (I was looking for the logic behind the quartic formula.)

-Dan

9. Thanks for the thanks. How serendipitous

10. Certain Computer Algebra Systems can solve certain equations, for example, Mathematica.

This is what I got (not exactly sure what it means because it says "too long, it will appear as a clipboard").

Code:
MATRIX([[1/4+1/12*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)
+576)/(54+6*3153^(1/2))^(1/3))^(1/2)+1/12*6^(1/2)*((3*(54+6*3153^(1/2))^(1/3)*
((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^
(1/3))^(1/2)+2*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+
6*3153^(1/2))^(1/3))^(1/2)*(54+6*3153^(1/2))^(2/3)-96*((9*(54+6*3153^(1/2))^(
1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)+9*(54+6*
3153^(1/2))^(1/3))/(54+6*3153^(1/2))^(1/3)/((9*(54+6*3153^(1/2))^(1/3)-12*(54+6
*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2))^(1/2)], [1/4+1/12*((9*(54
+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))
^(1/2)-1/12*6^(1/2)*((3*(54+6*3153^(1/2))^(1/3)*((9*(54+6*3153^(1/2))^(1/3)-12*(
54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)+2*((9*(54+6*3153^(
1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)*(54+
6*3153^(1/2))^(2/3)-96*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+
576)/(54+6*3153^(1/2))^(1/3))^(1/2)+9*(54+6*3153^(1/2))^(1/3))/(54+6*3153^(
1/2))^(1/3)/((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*
3153^(1/2))^(1/3))^(1/2))^(1/2)], [1/4-1/12*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*
3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)+1/12*I*6^(1/2)*((-3*(54+6
*3153^(1/2))^(1/3)*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/
(54+6*3153^(1/2))^(1/3))^(1/2)-2*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(
1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)*(54+6*3153^(1/2))^(2/3)+96*((9*
(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(
1/3))^(1/2)+9*(54+6*3153^(1/2))^(1/3))/(54+6*3153^(1/2))^(1/3)/((9*(54+6*3153^(
1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2))^(
1/2)], [1/4-1/12*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54
+6*3153^(1/2))^(1/3))^(1/2)-1/12*I*6^(1/2)*((-3*(54+6*3153^(1/2))^(1/3)*((9*(54+
6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(
1/2)-2*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^
(1/2))^(1/3))^(1/2)*(54+6*3153^(1/2))^(2/3)+96*((9*(54+6*3153^(1/2))^(1/3)-12*(
54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)+9*(54+6*3153^(1/2))
^(1/3))/(54+6*3153^(1/2))^(1/3)/((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(
1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2))^(1/2)]])

11. Originally Posted by galactus
Solving a quartic can be done, but it is cumbersome. Here is a nice step-by-step approach to the process.

The "Quartic Formula"
Personally, I prefer The "TI-89 Formula"

12. Originally Posted by ThePerfectHacker
Certain Computer Algebra Systems can solve certain equations, for example, Mathematica.

This is what I got (not exactly sure what it means because it says "too long, it will appear as a clipboard").
In this case Ma is making a meal of it. Looking only for the real roots derive gives what is in the attachment:

RonL