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A quartic that is biquadratic once depressed is solved - #2314
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SolveQuartic depresses a x^4 + b x^3 + c x^2 + d x + e by y = x + b/(4a), and where the
depressed form has no cubic term it solves the quadratic in y^2 and shifts back. The shift
was written -b / 4 * a, which is -a b / 4, where it is -b / (4 a), as the general branch
below it has it. With a leading coefficient of 1 or -1 the two agree, so it went unseen;
with any other every candidate was off by a multiple of b, failed its check against the
equation, and the empty set that was left claimed there was no root.
31x^4 - 62x^3 + 33x^2 - 2x - 1 = 0 has four real roots, 1/2 +- sqrt((27/62 +- 8 sqrt(2)/31)/2),
and was answered { }.
An equation in a square root whose squared form is such a quartic was answered { } with it:
x^2 - x - 1 = 4x sqrt(2x^2 - 4x + 2), which has the root -0.1326, and
(2x^2 + 4x - 2)/sqrt(-2x - x^2) = -2, which has four complex ones. Both were found by a
cross-check of equations in roots against a Newton search, and 2.6.0 answers them the same.
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Codecov Report✅ All modified and coverable lines are covered by tests. Additional details and impacted files@@ Coverage Diff @@
## master #2314 +/- ##
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Coverage 86.76% 86.76%
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Files 311 311
Lines 68199 68199
Branches 19821 19823 +2
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+ Hits 59175 59176 +1
+ Misses 4881 4880 -1
Partials 4143 4143
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Every root a solver finds is put back into the equation as written, and the ones that fail
are dropped. Where all of them failed, the empty set was left, and the empty set claims
there is no root. For a polynomial of degree one or more, read over the complex numbers,
that is never true, so what failed was the solver: (x + 1)^3 = 2 came back { } while
Cardano's C was 0 for a shifted cube (#2315), and 31x^4 - 62x^3 + 33x^2 - 2x - 1 = 0 while a
biquadratic quartic was shifted by -b a / 4 (#2314). Such a polynomial is now left as the
set of the x that satisfy it, which claims only that its roots were not found.
Only a polynomial with numbers for coefficients and over the complex numbers: an equation
that is not a polynomial can have no root, as sqrt(x) = -1 has none, and so can a
polynomial over the reals.
With those two fixed, no equation known reaches this, so nothing answered changes.
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
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SolveQuarticdepressesa x^4 + b x^3 + …byy = x + b/(4a). Where the depressed form has no cubic term, it solves the quadratic iny^2and shifts back. The shift was written-b / 4 * a, which is-a b / 4, where it is-b / (4 a), as in the general branch just below. With a leading coefficient of 1 or -1 the two agree, so it went unseen. With any other, every candidate was off, failed its check against the equation, and the empty set that was left claimed there was no root.Solve("x")31x^4 - 62x^3 + 33x^2 - 2x - 1 = 0{ }1/2 ± sqrt((27/62 ± 8/31 * sqrt(2)) / 2), four real roots2x^4 - 8x^3 + 10x^2 - 4x - 1 = 0{ }1 ± sqrt((1 ± sqrt(3)) / 2)x^2 - x - 1 = 4x * sqrt(2x^2 - 4x + 2){ }{ 1/2 - sqrt((27/62 + 8/31 * sqrt(2)) / 2) }, which is -0.1326(2x^2 + 4x - 2) / sqrt(-2x - x^2) = -2{ }-1 ± sqrt((3 ± sqrt(-3)) / 2){ }there, and each squares to such a quartic. Both quartics are irreducible over the rationals, so no factoring answers them first.BREAKING-CHANGES.mdhas an entry. No issue was filed.🤖 Generated with Claude Code