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naseeam
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How to solve third order algebraic equation?

-8(4.2)^2 + 5(4.2)k + k^3 = -149

How to get started in solving this equation?  


I re-wrote equation as follows:

k( 5(4.2) + k^2) = -149 + 8( (4.2)^2) )

Am I on the right track?  If yes, what's the next step?

Math / Science

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phoffric

8/22/2022 - Mon
naseeam

ASKER
I re-wrote equation in standard form as follows:
k^3 + 21*k + 7.88 = 0
what's the next step?
ASKER CERTIFIED SOLUTION
phoffric

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naseeam

ASKER
It seems like there is not standard method to solve this equation.  Do we have to do trial and error?

Did they use trial and error to cancel two k^2 terms?  

Please explain steps that lead to real solution.
SOLUTION
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d-glitch

>> It seems like there is not standard method to solve this equation.

There is.  But, while there is a single method for all quadratic equations, there are 14 types of cubic.

The pro version of phoffric's Wolfram site can show you the derailed steps required for solving the problem.
Your help has saved me hundreds of hours of internet surfing.
fblack61
d-glitch

Cardano (another 16th century Italian) discovered something approaching a general method for solving cubics.  It is not for the faint of heart.

https://www.google.com/url?q=https://www.math.ucdavis.edu/~kkreith/tutorials/sample.lesson/cardano.html&sa=U&ved=2ahUKEwjn-tOx_b_zAhVAmHIEHaTKDFcQFnoECAAQAg&usg=AOvVaw1fFFtPKHgOXAQjihKkqWxE
phoffric

it is natural to read wikipedia's topic. You may even find the History section interesting, as well as the rest of the article.
https://en.m.wikipedia.org/wiki/Cubic_equation

If you form the cubic as y = f(x), where all the coefficients are real (but see the wiki for case where coefficients are rational, and therefore equivalently integers), then when y is 0, x has 3 values, not necessarily distinct. You can have 3 distinct real values, or one real and a conjugate pair of complex values, or 3 identical real values (e.g., y = f(x) = (x-2)^3, where x = 2 with multiplicity 3).
https://www.wolframalpha.com/input/?i=roots++%28x-2%29%5E3

Here is a case with 3 real roots.
https://www.wolframalpha.com/input/?i=roots+x%5E3+%2B+%2B2x%5E2+-+9x+%2B+2
In this case, you see two local extrema (one positive max and one negative min). Intuitively, you know that this has three real roots. Of course, if you have two distinct min/max local extrema having the same sign, then you only get one real value, and two complex conjugate pairs. The min/max extrema locations can be useful to determine where to start a guess using a numerical root finding method.

For the case where you have 3 real roots, to find out where these min/max values are, take the derivative which gives a quadratic equation, and solve for 0. Using previous example, here are the results:
https://www.wolframalpha.com/input/?i=roots+d%2Fdx%28x%5E3+%2B+%2B2x%5E2+-+9x+%2B+2%29