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    How do you solve:
    X^3 + 2X^2 + X - 100000 = 0.

    There are no factors to take out.

    What's the best method of solving this? Thank you.

    Edit: This is a real one.

    Answer: 45.74923313.....
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    (Original post by Konnichiwa)
    How do you solve:
    X^3 + 2X^2 + X - 100000 = 0.

    There are no factors to take out.

    What's the best method of solving this? Thank you.
    Where did you get this from? Can you post the original question?
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    That's one you've created right? Because it doesn't have three real roots. Have you covered complex numbers yet?

    Easiest way I do it is using the following steps:

    1) By trial and error, you find a solution such that your function is 0. So let's say you have some function, f(x). You want to find a value for x, such that f(x) = 0.

    This is normally just done by guessing, but they tend to make the numbers easy for this. So you would really just try like x = ±1, ±2, ±3 and so on. You wouldn't need to go too far until you find your solution.

    2) Let's say that at some point 'a', you get f(a) = 0. What this means is that x = a is solution to your function (i.e a point where it crosses the 'x' axis) and so we can say that (x - a) = 0 is a root.

    3) Now you have two options. You can either use the long division way and divide f(x) by (x - a) and this will give you a quadratic. Either that or you can compare coefficients but i don't use this method so don't know how to explain it. You can then factorise this quadratic in the normal way.

    You will either be given a solution so you can skip that first step or, like I said, it won't take too long to find your first solution through trial and error.
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    Using Wolfram Alpha :P
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    Cubic formula!



    Don't try this one at home!
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    You need to find a value for which that equation is equal to zero. Usually you just do this by trial and error, so if this is from a textbook or something, it's unlikely that you will have to look further than about ±5. Then use synthetic division to turn this into a quadratic, and solve that. You then have values for x from the quadratic and the one you found by trial and error. Check with wolfram alpha
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    (Original post by claret_n_blue)
    That's one you've created right? Because it doesn't have three real roots. Have you covered complex numbers yet?

    Easiest way I do it is using the following steps:

    1) By trial and error, you find a solution such that your function is 0. So let's say you have some function, f(x). You want to find a value for x, such that f(x) = 0.

    This is normally just done by guessing, but they tend to make the numbers easy for this. So you would really just try like x = ±1, ±2, ±3 and so on. You wouldn't need to go too far until you find your solution.

    2) Let's say that at some point 'a', you get f(a) = 0. What this means is that x = a is solution to your function (i.e a point where it crosses the 'x' axis) and so we can say that (x - a) = 0 is a root.

    3) Now you have two options. You can either use the long division way and divide f(x) by (x - a) and this will give you a quadratic. Either that or you can compare coefficients but i don't use this method so don't know how to explain it. You can then factorise this quadratic in the normal way.

    You will either be given a solution so you can skip that first step or, like I said, it won't take too long to find your first solution through trial and error.
    It's a real one. Thanks.
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    (Original post by Konnichiwa)
    It's a real one. Thanks.
    What's the original question? You're not going to be expected to solve cubics like that except via numerical approximation!
 
 
 
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