Find H

TheJason

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#1
There should be three answers (roots) (post edited).

\(\displaystyle 15 + 3\,h^{3} = 16 - 9\,h^{3}\)

\(\displaystyle 15 -15 + 3\,h^{3} = 16 - 15 - 9\,h^{3}\)

\(\displaystyle 3\,h^{3} = 1 - 9\,h^{3}\)

\(\displaystyle 3\,h^{3} + 9\,h^{3} = 1 - 9\,h^{3} + 9\,h^{3}\)

\(\displaystyle 12\,h^{3} = 1\)

\(\displaystyle 12(\dfrac{1}{12})\,h^{3} = 1(\dfrac{1}{12})\)

\(\displaystyle h^{3} = \dfrac{1}{12}\)

\(\displaystyle (h^{3})^{1/3} = (\dfrac{1}{12})^{1/3}\)

\(\displaystyle h = 0.4367902324\) (one root)

Check

\(\displaystyle 15 + 3\,(0.4367902324)^{3} = 16 - 9\,(0.4367902324)^{3}\)

\(\displaystyle 15.25 \approx 15.25\) (one root)

Yes

Your Turn

Find I. There should be three answers (roots).

\(\displaystyle 19 + 50\,i^{3} = 2 - \dfrac{4}{5}\,i^{3}\)
 
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MarkFL

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#2
There are likely a complex conjugate pair of roots as well. A cubic equation (degree 3) will have 3 roots. Also, I wouldn't use the equal sign when subbing in a decimal approximations, but rather, I would use the LaTeX code:

\approx
 

TheJason

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#3
There are likely a complex conjugate pair of roots as well. A cubic equation (degree 3) will have 3 roots. Also, I wouldn't use the equal sign when subbing in a decimal approximations, but rather, I would use the LaTeX code:

\approx
Post edited. OK, we got one root at least. What would be the other two?
 

MarkFL

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#4
You could write the equation as:

\(\displaystyle 12h^3-1=0\)

Or

\(\displaystyle \left(\sqrt[3]{12}h\right)^3-1^3=0\)

Now, factor as the difference of cubes, and equate the two factors to zero to find all 3 roots. :)
 
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TheJason

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#5
You could write the equation as:

\(\displaystyle 12h^3-1=0\)

Or

\(\displaystyle \left(\sqrt[3]{12}h\right)^3-1^3=0\)

Now, factor as the difference of cubes, and equate the two factors to zero to find all 3 roots. :)
So in all cases of the cubed stuff, do you always use the \(\displaystyle -1 = 0\,\) part? Why are those numbers used in this example?
 

MarkFL

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#6
Those numbers are the result of the equation you posted.