How to avoid Root[] when using Solve[]

Question

I am trying to solve a system of algebraic equations using Solve. The solution Mathematica gives use Root[] as an intermediate:

m5 = {{k - m w^2, -k, 0, 0, 0}, {-k, 2 k - m w^2, -k, 0, 0}, {0, -k, 2 k - m w^2, -k, 0}, {0, 0, -k, 2 k - m w^2, -k}, {0, 0, 0, -k, k - m w^2}};

sol = Solve[Det[m5] == 0 , w, Reals]

The output looks like

{{w -> ConditionalExpression[0, (k > 0 && m > 0) || (k > 0 && m < 0) || (k < 0 && m > 0) || (k < 0 && m < 0)]}, 
 {w -> ConditionalExpression[Root[5 k^2 - 5 k m #1^2 + m^2 #1^4 &, 1], (k > 0 && m > 0) || (k < 0 && m < 0)]}, {w -> ConditionalExpression[ Root[5 k^2 - 5 k m #1^2 + m^2 #1^4 &, 2], (k > 0 && m > 0) || (k < 0 && m < 0)]}, 
 {w -> ConditionalExpression[Root[5 k^2 - 5 k m #1^2 + m^2 #1^4 &, 3], (k > 0 && m > 0) || (k < 0 && m < 0)]}, 
 {w -> ConditionalExpression[Root[5 k^2 - 5 k m #1^2 + m^2 #1^4 &, 4], (k > 0 && m > 0) || (k < 0 && m < 0)]},
 {w -> ConditionalExpression[Root[k^2 - 3 k m #1^2 + m^2 #1^4 &, 1], (k > 0 && m > 0) || (k < 0 && m < 0)]},
 {w -> ConditionalExpression[Root[k^2 - 3 k m #1^2 + m^2 #1^4 &, 2], (k > 0 && m > 0) || (k < 0 && m < 0)]},
 {w -> ConditionalExpression[Root[k^2 - 3 k m #1^2 + m^2 #1^4 &, 3], (k > 0 && m > 0) || (k < 0 && m < 0)]},
 {w -> ConditionalExpression[Root[k^2 - 3 k m #1^2 + m^2 #1^4 &, 4], (k > 0 && m > 0) || (k < 0 && m < 0)]}}

Even if I use Simplify[] to state the assumptions

sol = Simplify[sol, Assumptions -> {k > 0, m > 0}]

I still cannot get rid of Root[] in the solution

{{w -> 0}, {w -> Root[5 k^2 - 5 k m #1^2 + m^2 #1^4 &, 1]},
 {w -> Root[5 k^2 - 5 k m #1^2 + m^2 #1^4 &, 2]},
 {w -> Root[5 k^2 - 5 k m #1^2 + m^2 #1^4 &, 3]}, 
 {w -> Root[5 k^2 - 5 k m #1^2 + m^2 #1^4 &, 4]}, 
 {w -> Root[k^2 - 3 k m #1^2 + m^2 #1^4 &, 1]}, 
 {w -> Root[k^2 - 3 k m #1^2 + m^2 #1^4 &, 2]}, 
 {w -> Root[k^2 - 3 k m #1^2 + m^2 #1^4 &, 3]}, 
 {w -> Root[k^2 - 3 k m #1^2 + m^2 #1^4 &, 4]}}

This is annoying because the quartics are well solvable. Can anyone help please?

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score 4 · Accepted Answer · answered Oct 14 '16 at 00:57

4

From the documentation of ToRadicals:

ToRadicals[expr] attempts to express all Root objects in expr in terms of radicals.

Hence simply

sol = Solve[Det[m5] == 0 , w, Reals]
ToRadicals @ sol

Moreover, you can get rid of the ConditionalExpressions with Normal, which in this case assumes that all the conditions are met and only displays the formulae:

answered Oct 14 '16 at 00:57

corey979

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3

The OP was fortunate here that the expressions of interest were expressible in radicals. This is of course not the case in general. – J. M.'s missing motivation Oct 14 '16 at 03:35
I like this solution best. Intuitive and straightforward. Thanks. – Dan Zhang Oct 14 '16 at 21:14

score 3 · Answer 2 · answered Oct 14 '16 at 00:57

Does this work for you? You can try 5th order. Mathematica finds solution. Replace w^2 by z

ClearAll[k, m, w, s, s]
m5 = {{k - m w^2, -k, 0, 0, 0}, {-k, 2 k - m w^2, -k, 0, 0}, {0, -k, 
    2 k - m w^2, -k, 0}, {0, 0, -k, 2 k - m w^2, -k}, {0, 0, 0, -k, 
    k - m w^2}};
d = Det[m5]

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d = d /. w -> z^(1/2)

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s = Solve[d == 0, z]

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Now replace back and solve for w

s = Flatten[(s /. z -> w^2) /. Rule -> Equal]

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Solve[#, w] & /@ s

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How to avoid Root[] when using Solve[]

2 Answers2