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### Solving a Quartic Polynomial using a Trig Substitution

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Date: 12/27/2001 at 15:23:23
From: Kevin Sobieski
Subject: Algebra

I need help solving this system:

y/sqrt(y^2+64) = (x+y)/25

x/sqrt(x^2+64) = (x+y)/20

I tried solving for y in terms of x, but after 1.5 pages, I still
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```
Date: 12/28/2001 at 04:28:29
From: Doctor Pete
Subject: Re: Algebra

Hi,

This is a somewhat challenging and tedious system of equations to
solve. The solution set for x is a subset of the roots of a related
degree-4 polynomial.

Let us write

[1] y/Sqrt[y^2+64] = (x+y)/25
[2] x/Sqrt[x^2+64] = (x+y)/20.

Clearly, we can see there is the trivial solution (x,y) = (0,0). So we
may assume x and y are nonzero. If we were to use a straightforward
approach, we would begin by finding y in terms of x in [2], and then
substituting this result into [1]. After a great deal of computation
you would eventually arrive at a degree-8 polynomial for x,

[3] 0 = 196000874496 - 612204544x^2 - 19609344x^4 - 44064x^6 + 81x^8.

Obviously this is really a quartic equation, since the powers of x
are all even. We could even use a substitution of the form
x = Sqrt[(z+408)/3] to obtain a monic degree-4 polynomial in z with
rational coefficients, and no term of degree 3. But ultimately this
method is not inspiring, and as you observed, it is very tedious.
(In fact, I obtained the above using a computer.)

So, we want another method.  Consider the trigonometric identity

Tan[t]^2 + 1 = Sec[t]^2,

for any angle t.  We are therefore motivated to make the substitutions

x = 8*Tan[s],
y = 8*Tan[t],

and we immediately see that [1,2] become

[4] Sin[t] = (8/25)(Tan[s]+Tan[t])
[5] Sin[s] = (2/5)(Tan[s]+Tan[t]).

While this looks easy (or easier) to solve, it is actually no better
or worse than our original equations. If we multiply [5] by 4/5 and
subtract the result from [4], we obtain the relation

Sin[t] = (4/5)Sin[s],

from which it follows that

[6] Sin[s] = (2/5)(Tan[s]+(4/5)Sin[s]/Sqrt[1-(16/25)Sin[s]^2])
= (2/5)Tan[s] + (8/5)Sin[s]/Sqrt[25-16*Sin[s]^2],

and since Sin[s] is nonzero, we have

[7] 5/2 = 1/Cos[s] + 4/Sqrt[25-16*Sin[s]^2].

If we then let z = 4*Cos[s], we obtain

[8] 5/2 = 4/z + 4/Sqrt[9+z^2],

or equivalently,

[9] (5z-8)Sqrt[9+z^2] = 8z.

We readily obtain a quartic in z,

[10] 576 - 720z + 225z^2 - 80z^3 + 25z^4 = 0.

Since z = 4*Cos[s], and x = 8*Tan[s], it follows that

x = 32*Sin[s]/z = 8 Sqrt[16-z^2]/z
= 8 Sqrt[(4/z)^2 - 1].

So, we have greatly simplified the algebra by using a trigonometric
substitution. But that's not all....

Alternatively, we may proceed from [8] as follows: let z = 3Tan[w];
hence

[11] 5/2 = (4/3)Cot[w] + (4/3)Cos[w],

or

[12] 15/8 = (Cos[w])(1+Sin[w])/Sin[w].

Now let v = Sin[w] to get

[13] (15/8)v = Sqrt[1-v^2](1+v)

and we now have a new polynomial,

[14] -1 - 2v + (225/64)v^2 + 2v^3 + v^4 = 0,

where z = 3*Sin[w]/Cos[w] = 3v/Sqrt[1-v^2], so

x = (8/3)Sqrt[(4/v)^2 - 25].

That's about the simplest you can get.  Of course it is possible to
solve all the quartics [3,10,14] in terms of radicals, but this is far
too computational to be of much use. Numerically, there are two real
solutions for v in [14], approximately

v = {-0.3344924259407920018155742, 0.6637031578486982907001578}.

Only the second value yields a correct value for x, the other being an
extraneous solution. We find approximately,

(x,y) = +/- (8.972929477351965431166023, 5.955361629272169366821429).

where +/- indicates x and y are both positive or both negative.

All said, I'm afraid we are not much better off for our efforts, as
ultimately we are still required to solve a quartic polynomial.
However, by using some clever trigonometric substitutions, we can at
least do two things: (a) Simplify the computation to obtain a quartic,
and (b) obtain a simpler quartic polynomial (with smaller
coefficients.)

- Doctor Pete, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
College Trigonometry
High School Polynomials
High School Trigonometry

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