# Cute elementary algebra problem

Discussion in 'Undergraduate Math' started by Stephen J. Herschkorn, Sep 15, 2007.

1. ### Stephen J. HerschkornGuest

Came across this problem in the first section of the first chapter of
the intro calculus text from which I am teaching this semester. I
thought it was kind of cute.

Solve x^4 - 4x = 1 exactly.

Attributed to Murray Klamkin and published in The Mathematics Student
Journal (1980).

Stephen J. Herschkorn, Sep 15, 2007

2. ### CompendiumGuest

x=-sqrt(2)/2-iÂ·sqrt(sqrt(2)+1/2) or
x=-sqrt(2)/2+iÂ·sqrt(sqrt(2)+1/2) or
x=sqrt(2)/2-sqrt(sqrt(2)-1/2) or
x=sqrt(sqrt(2)-1/2)+sqrt(2)/2

Compendium, Sep 15, 2007

3. ### Brian VanPeltGuest

http://www.msc.uky.edu/ken/ma110/text/poly.htm

My 2nd grade teacher beat this technique into us!

Brian

Brian VanPelt, Sep 15, 2007
4. ### Jeremy WattsGuest

if by 'exactly' you mean solving it in terms of radicals then the only
method i know of to do that is 'ferrari's method' , which you'd have to be
out of your mind to attempt as its hideously complicated.

is there a solution printed for this?

Jeremy Watts, Sep 15, 2007
5. ### ªºª rrockGuest

It's somewhere (not exactly) around x ~ -0.24903837639842
but it seems to me it's probably irrational.

ªºª rrock, Sep 15, 2007
6. ### Jeremy WattsGuest

all the roots are irrational, two are real, two complex

Jeremy Watts, Sep 15, 2007
7. ### DanaGuest

Here's cheating...

x /. Solve[x^4 - 4*x == 1, x][[ 1 ;; 2]]

{-((Sqrt[-1 + 2*Sqrt[2]] - 1)/Sqrt[2]),
(Sqrt[-1 + 2*Sqrt[2]] + 1)/Sqrt[2]}

N[%]

{x->-0.2490383763983743},
{x-> 1.663251938771469}

Dana, Sep 15, 2007
8. ### Stephen J. HerschkornGuest

It is possible to do this with only elementary algebra one would learn
in high school (at the latest) in the U.S. I will post a solution in a
few days if no one else does.

Stephen J. Herschkorn, Sep 15, 2007
9. ### Brian M. ScottGuest

On Fri, 14 Sep 2007 23:09:24 -0400, "Stephen J. Herschkorn"
Let s = sqrt(2) and write x^4 - 4x - 1 as the difference of
two squares:

x^4 - 4x - 1 =
(x^4 + 2x^2 + 1) - (2x^2 + 4x + 2) =
(x^2 + 1)^2 - 2(x + 1)^2 =
(x^2 + 1 + s(x + 1))(x^2 + 1 - s(x + 1)) =
(x^2 + sx + (1 + s)) (x^2 - sx + (1 - s))

Clearly either x^2 + sx + (1+s) = 0 or x^2 - sx + (1-s) = 0.
But s^2 - 4(1+s) = 2 - 4 - 4s < 0, so only the second factor
produces real roots. They are (1/2)(s +/- sqrt(-2 + 4s)) =
(s/2)(1 +/- sqrt(-1 + 2s)). (Barring algebraic errors at
some point, of course.)

I actually got this after the fact, by trying to write
x^4 - 4x - 1 as a product of two monic quadratics.
Specifically, I divided x^4 - 4x - 1 by x^2 - ax - b; this
yields a quotient of x^2 + ax + (a^2 + b), where a and b
must satisfy a^3 + 2ab - 4 = 0 and (a^2 + b)b = 1 in order
to avoid having a remainder. Note that without loss of
generality a >= 0.

Consider the second of these as a quadratic in b and solve
to get

b = (1/2)[-a^2 +/- sqrt(a^4 + 4)].

Substitute this into the first condition:

a^3 - a^3 +/- a sqrt(a^4 + 4) - 4 = 0,
+/- sqrt(a^4 + 4) = 4/a,
a^4 + 4 = 16/a^2,
a^6 + 4a^2 - 16 = 0.

Let c = a^2, making this c^3 + 4c - 16 = 0; by inspection
(or the rational root test) this has a solution c = 2.
Dividing out c - 2 leaves c^2 + 2c + 8, which has no real
roots, so we take c = 2 and a = sqrt(2) = s; the condition
a^3 + 2ab - 4 = 0 then becomes 2s + 2sb - 4 = 0, b + 1 = 2/s
= s, and b = -1 + s. Thus, the quadratic factors are
x^2 - sx + (1 - s) and x^2 + sx + (1 + s). I then noticed
that these can be written (x^2 + 1) +/- s(x + 1), which led
to the slick solution given above.

[...]

Brian

Brian M. Scott, Sep 16, 2007
10. ### Stephen J. HerschkornGuest

Why do you exclude the complex roots, easily found from your derivation?

I got there this way, which is pretty close to Brian's route.

x^4 - 4x = 1
x^4 = 4x + 1
x^4 + 2x^2 + 1 = 2 x^2 + 4x + 2
(x^2 + 1)^2 = 2 (x + 1)^2
x^2 + 1 = +/- sqrt(2) (x+1)

We now have two quadratic equations to solve.

Any idea if we can express sqrt(sqrt(2) - 1) as a rational linear
combination of roots of integers? I suspect not.

Stephen J. Herschkorn, Sep 16, 2007
11. ### Brian M. ScottGuest

On Sat, 15 Sep 2007 21:15:57 -0400, "Stephen J. Herschkorn"
Because of the context that you gave and the later comment
that the solution can be found using only elementary high
school algebra: both suggest that real roots are expected,
though as you say the others are easy enough to find if one
It does seem unlikely, but it's been a *long* time since I
did anything with field extensions.

Brian

Brian M. Scott, Sep 16, 2007
12. ### Stan BrownGuest

Sat, 15 Sep 2007 21:15:57 -0400 from Stephen J. Herschkorn
I used the method at
http://mathforum.org/library/drmath/view/52657.html
and unless I made an algebraic error there is no expression
sqrt(a) ± sqrt(b) for integer a,b that equals sqrt(sqrt(2)-1).

Stan Brown, Sep 16, 2007
13. ### Brian VanPeltGuest

Yeah, if

sqrt(sqrt(2) - 1) = p sqrt(a) + q sqrt(b)

for p and q rational numbers, a and b non-square naturals, then

sqrt(2) - 1 = p^2 a + q^2 b + 2 pq sqrt(ab)

It seems like -1 would have to equal p^2 a + q^2 b.

Brian

Brian VanPelt, Sep 16, 2007
14. ### Ken PledgerGuest

No it isn't in cases like this where the reducing cubic has an
easy factor.

Only by inspiration. (I've sometimes told students that the best
method of integrating functions is "integration by inspiration," where
you can guess the answer and then just check it by differentiating.)

Ferrari's method leads to the difference of squares

(x^2 + 1)^2 - 2(x + 1)^2 = 0

which is easy to factorize. But finding that expression without
Ferrari's help would involve more inspiration than most of us could hope
for.

Ken Pledger.

Ken Pledger, Sep 17, 2007
15. ### JillBonesGuest

I get (x^2 + 2*sqrt(x)) * (x^2 - 2*sqrt(x)) = 1 by simple
factoring. Doesn't solve the problem but raises the question;

"Does this preclude negative values for x?"

Bill J

JillBones, Sep 19, 2007
16. ### ProginoskesGuest

That doesn't get you anywhere, since you don't have a 0 on the right-
hand side.
Not necessarily; if we're working with complex numbers, square roots
of negative numbers are okay.

--- Christopher Heckman

Proginoskes, Sep 19, 2007