0
$\begingroup$

Find $k= constant$ so that $\lceil$ https://www.wolframalpha.com/input/?i=discriminant%5By%5E6%2By%5E2z%5E6%2Bz%5E2-y%5E2z%5E2(1%2By%5E2%2Bz%5E2)-k(1%2Bz%5E2)z%5E2(y%2Bz)(y-z)(z-1)(z%2B1)%5D,z%5D $\rfloor$ $$disc\!\left [\!disc[\!z^{2}(\!z^{2}- 1\!)^{2}(\!z^{2}+ 1\!)+ (\!y^{2}- z^{2}\!)( \!y^{4}+ z^{6}- z^{4}- z^{2}\!)- k(\!z^{4}- 1\!)z^{2}( \!y^{2}- z^{2}\!)\!,\!y\!]\!,\!z\!\right ]\!=\!0$$

$\endgroup$

closed as off-topic by bbgodfrey, m_goldberg, MarcoB, Daniel Lichtblau, Michael E2 Feb 23 at 18:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "The question is out of scope for this site. The answer to this question requires either advice from Wolfram support or the services of a professional consultant." – bbgodfrey, m_goldberg, MarcoB
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful $\endgroup$ – Michael E2 Feb 23 at 17:00
  • 1
    $\begingroup$ This site is about using Mathematica. Questions about W|A are off topic. $\endgroup$ – Michael E2 Feb 23 at 17:03
  • $\begingroup$ My view is it would have made for a reasonable question had there been copy-and-pasteable input provided. As for W|A query length limitations, might want to try it on the Wolfram Cloud instead. $\endgroup$ – Daniel Lichtblau Feb 24 at 16:54
2
$\begingroup$

Your nested discriminants are zero for any real k.

Discriminant[
  z^2 (z^2 - 1)^2 (z^2 + 1) + (y^2 - z^2) (y^4 + z^6 - z^4 - z^2) - 
    k (z^4 - 1) z^2 (y^2 - z^2), y], z]

0

This is correct because

zpoly = 
  Discriminant[
    z^2 (z^2 - 1)^2 (z^2 + 1) + (y^2 - z^2) (y^4 + z^6 - z^4 - z^2) - 
      k (z^4 - 1) z^2 (y^2 - z^2), y]

gives

zpoly

As you see zpoly has a multiple root at zero, so its discriminant must be zero.

$\endgroup$
1
$\begingroup$

There are line length limits in WolframAlpha and I think that may be some of the reason you were not able to get a solution in a single step from WolframAlpha. So we use a trick that sometimes works to do somewhat more complicated calculations within WolframAlpha.

I did your inner discriminant as a separate WolframAlpha calculation

Discriminant[y^6+y^2z^6+z^2-y^2z^2(1+y^2+z^2)-k(1+z^2)z^2(y+z)(y-z)(z-1)(z+1),y]

and scraped the result

-64 z^8 (z^2 - 1)^4 ((z^2 - 1) z^4 k(z^2 + 1) + z^2) (4 z^2 k(z^2 + 1) (-k(z^2 + 1)*
((z^2 - 1) k(z^2 + 1) - 3 z^4 + z^2 + 3) - 3 z^6 + 8 z^4 + 5 z^2 - 6) + (z^2 + 1)*
(4 z^8 - 9 z^6 - 3 z^4 + 5 z^2 + 27))^2

and then tried to rearrange and simplify that and eliminate as many spaces as I could, without confusing WolframAlpha, to get the shortest possible expression to paste into this second separate WolframAlpha calculation of the outer overall discriminant.

Discriminant[-64*z^8*(z^2-1)^4*(1+z^2)^2*(27+(5+4*k*((k-3)*k-6))*z^2+(4*k*(5+
(k-4)*k)-3)*z^4+(-9-4*(k-4)*k*(2+k))*z^6-4*(k-1)^3*z^8)^2*(z^2*k*z^4+k*z^8),z]

or you can reach that via this

[WolframAlphalink][1]

And the result of that is

0

So I think the result is zero for all values of k.

Please check all this very carefully to make certain there are no mistakes.

$\endgroup$
  • $\begingroup$ $\text{k}= 1$ is true but I don$'$t think the result is for all values of $\text{k}$ $.$ The square can be equal to $0$ not $> 0$ $.$ $\endgroup$ – user61956 Feb 23 at 12:18