okay the question is: find all the real roots of 5x^3-17^2-x+30=0
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okay the question is: find all the real roots of 5x^3-17^2-x+30=0
Hello,
Try this : Cubic function - Wikipedia, the free encyclopediaQuote:
Originally Posted by requal
Is there any other way of doing this ala factor theorem, or long divison? Because I have no clue of what a "cubic function is".(Doh) If there isnt, can sum1 explain it to me?(Talking)
PS- I just realised that this was in the wrong forum- feel free to move this to the high school section if u must
Do you have a typo in that? Should that be
If so, check out my cubic calculator that shows you the math:
http://www.mathcelebrity.com/cubic.php
Since the discriminant is less than 0, there are 3 real roots.
okayyyyyyyyyyy... I dont understand one bit of the theorem. I guess my teacher made a typo or somethin(Headbang). Thanks anyway(Clapping)
In solving quadratic equations that cannot be factored or hard to factor, the Quadratic Formula is very handy and useful and simple.
In soving cubic equations that cannot be factored, or are hard to factor, the Cubic Formula is hardly useful and "simple" is a term that is alien to its use.
For cubic equations then I usually resort to iteration or repeatition or trial and error. (The synthetic division is faster for integer values of x).
By iteration, I mean I try values for x that will make the given cubic equation true.
I always start with x=1. Then x=2, x=3, etc. Or x = -1, x = -2, x = -3, etc.
So, if x = 1,
5x^3 -17x^2 -x +30 = 0
5(1^3) -17(1^2) -1 +30 = 17 .......not zero.
If x = 2,
5(2^3) -17(2^20 -2 +30 = 0 ........hey, it is zero already!
Meaning, 2 is a root, and (x -2) is a factor.
So, use long division, or synthetic division, to find the qudratic remainder.
You shold find it as
5x^2 -7x -15 = 0
You cannot factor that anymore, so use the Quadratic Formula to find the other two roots.
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It does not always happen that an integer value of x will make the original cubic equation true. That is why watch the results as you test for integer values of x.
If x = 2 came out not zero, say it came out (- 3), then that means 2 is more than the correct value of x to make the cubic equation true or zero. That means x = 1.8 might be the correct root. So test x = 1.8 in the cubic equation. See if it will make the cubic equation to become zero.
Etc...
Iteration, repeatition, trial and error.
Later on , you'd come to established methods of iteration, like the Newton's method, to minimize trial and error.
By the way, if x = 2 resulted in, say, 4 for the above cubic, then try x = 3. You must go beyond zero, or to the negative, before you try non-integer values of x.
Quote:
Originally Posted by requal
requal,
I feel your pain, it is a LOT of math to go through. I designed the math section to be exactly like the Wikipedia link that Moo posted. That is the link I used to design the logic to solve the cubic.
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