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Question 384042: I need to find the x-intercepts for the function f(x)= 2-x+2x^2-x^3 x^2 means x squared
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 
From the graph, it is clear that the x intercept is around x = 2.
When x = 2, the equation becomes 2 - 2 + 8 - 8 = 0
The answer is x = 2.
That is the only time that the graph crosses the x-axis so that is the only x-intercept of the equation.
I don't quite know how to derive that algebraically.
I looked on the web for a clue.
The bottom line on the search is there is no formula, similar to the quadratic formula, that can find the roots of the equation easily.
There are formulas, but they are much more complex to use than the quadratic formula is for quadratic equations.
A method used is to determine if there are any integral roots to the cubic equation.
If you look at the cubic equation, you will see that it ends in 2.
This indicates some possible roots are multiples of 2, like 1, 2, -1, 2.
Try each one of those values to see if any of them are a root.
It turns out that 2 is a root.
The equation becomes 0 when x = 2.
We just found one of the roots to the equation.
Are there others?
There are, but they are not real.
There is a handy little calculator that does the grunt work for you.
That calculator can be found at the following website address:
http://www.cubicsolver.com/
You just enter the coefficients and constants of the respective terms in the cubic equation after you have set it to standard form (higher order exponents to the left of lower order exponents and set the equation equal to 0) and it does the rest.
The answer shows a real part and an imaginary part.
If the imaginary part is 0, then the root is real.
Otherwise, the root is complex (real part and imaginary part).
I used it for your problem, and it gave the following result.
x = 2
x = +/- i
The following website provides you with some general methods for finding the roots of a cubic equation.
http://everything2.com/title/Cubic+equation
In your problem, I cheated by creating a graph of the equation and eye balling where the roots would be, if any.
I saw it was around 2 and tried to substitute 2 in the equation to see if I got a root.
It appeared that I did at x = 2 because the equation became equal to 0 when i replaced x with 2.
To find the other roots, I needed to divide the cubic equation by x-2.
Doing that I got the quadratic equation of -x^2-1 with a remainder of 0.
To find the roots of this quadratic equation, I needed to set it equal to 0 and solve.
I got:
-x^2-1 = 0
multiply both sides of this equation by -1 to get:
x^2 + 1 = 0
subtract 1 from both sides of this equation to get:
x^2 = -1
take square root of both sides of this equation to get:
x = +/- sqrt(-1) which is equivalent to x = +/- i
Those are the roots of this equation.
They are imaginary because the real part is 0.
If the real part was not zero and the imaginary part was not zero, then the roots would be complex (contain a real part and an imaginary part).
example of a real root = 2
example of an imaginary root = i
example of a complex root = 2 + i
If you do not have a real root to the equation, or an integer root, then there are ways to find the root but they are quite complex.
One way is by iteration.
If x = 1 yields a positive solution and x = 2 yields a negative solution, then the root exists between 1 and 2. In that case you may be able to find it by iterating different values of x between 1 and 2 until you land on the solution.
Computer programs are well suited to this task. Humans would find it a lot more labor intensive.
That second website alludes to the methods for finding the roots.
There is another website that gives similar information.
That website can be found here:
http://www.sosmath.com/algebra/factor/fac11/fac11.html
Another website can be found here:
http://www.math.vanderbilt.edu/~schectex/courses/cubic/
Bottom Line is that there is a general solution to a cubic equation but it is much more complex than the solution for the roots of a quadratic equation.
Fortunately there are calculators available that do the work for you.
I can't think anybody would want to do it manually, at least not over and over and over again.
graphing, as I did, is also a good method to narrow the solution down.
There are also methods to determine how many real roots there will be.
I believe it has to do with Descartes' Rule.
The following website discusses Descartes' Rule, and also the Rational Root Test, both of which are used.
http://www.purplemath.com/modules/drofsign.htm
Bottom Line is, if you have access to a graphing calculator, narrowing down the selection to what the real roots might be is so much easier.
In your case, the only real root is x = 2 which I narrowed down by graphing and then solved by substitution into the original equation.
More help on the web is found by doing a search on "finding the roots of a cubic equation" or some other related search string.
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