2 replies to this topic

[Rob_Gandy]

I'd say 'Can you find all the roots?,' but I know this will get solved.


This problem was posted a few years ago in the math department of the college I attended. I got a math club t-shirt and a certificate for getting the correct answer.

Find all the roots of the following equation without the use of technology.:

x⁶sin(x)+240-14x⁵sin(x)-308x+69x⁴sin(x)+22x²-126x³sin(x)+126x³-22x²sin(x)-69x⁴+308xsin(x)+14x⁵-240sin(x)-x⁶=0

Spoiler for Small hint.

There are four consecutive integer roots.

[mmiguel]

I'd say 'Can you find all the roots?,' but I know this will get solved.


This problem was posted a few years ago in the math department of the college I attended. I got a math club t-shirt and a certificate for getting the correct answer.

Find all the roots of the following equation without the use of technology.:

x⁶sin(x)+240-14x⁵sin(x)-308x+69x⁴sin(x)+22x²-126x³sin(x)+126x³-22x²sin(x)-69x⁴+308xsin(x)+14x⁵-240sin(x)-x⁶=0

Spoiler for Small hint.

There are four consecutive integer roots.

Spoiler for fun

put all the non-sin(x) terms on other side of equation to get

sin(x)*(x^6-14x^5+69x^4-126x^3-22x^2+308x-240) = (x^6-14x^5+69x^4-126x^3-22x^2+308x-240)

Let y = (x^6-14x^5+69x^4-126x^3-22x^2+308x-240)

sin(x)*y = y

If y != 0, then we have sin(x) = 1, which is true for x = pi/2 + 2*pi*k for any integer k.

if y == 0, then we need to find the roots of the polynomial
y = (x^6-14x^5+69x^4-126x^3-22x^2+308x-240)

Looking at the prime decomposition of 240,
we get 240 = 2^4*3*5

This makes it seem likely that (x-5), (x-3), and (x-2) are factors.
Polynomial long division to get
y/((x-5)*(x-3)*(x-2)) = x^3-4x^2-2x+8

Prime decomp of 8 is 2^3,

(x-2) doesn't work, try (x-4) [which agrees with hint]:

success:

y/((x-5)*(x-4)*(x-3)*(x-2)) = x^2-2
x^2-2 = 0 if x = plus/minus sqrt 2


Therefore......

roots are:

2,3,4,5, +sqrt(2), -sqrt(2), [pi/2 + 2*pi*k for any integer k]

Edited by mmiguel, 19 September 2012 - 07:51 AM.

[Rob_Gandy]

Spoiler for fun

put all the non-sin(x) terms on other side of equation to get

sin(x)*(x^6-14x^5+69x^4-126x^3-22x^2+308x-240) = (x^6-14x^5+69x^4-126x^3-22x^2+308x-240)

Let y = (x^6-14x^5+69x^4-126x^3-22x^2+308x-240)

sin(x)*y = y

If y != 0, then we have sin(x) = 1, which is true for x = pi/2 + 2*pi*k for any integer k.

if y == 0, then we need to find the roots of the polynomial
y = (x^6-14x^5+69x^4-126x^3-22x^2+308x-240)

Looking at the prime decomposition of 240,
we get 240 = 2^4*3*5

This makes it seem likely that (x-5), (x-3), and (x-2) are factors.
Polynomial long division to get
y/((x-5)*(x-3)*(x-2)) = x^3-4x^2-2x+8

Prime decomp of 8 is 2^3,

(x-2) doesn't work, try (x-4) [which agrees with hint]:

success:

y/((x-5)*(x-4)*(x-3)*(x-2)) = x^2-2
x^2-2 = 0 if x = plus/minus sqrt 2


Therefore......

roots are:

2,3,4,5, +sqrt(2), -sqrt(2), [pi/2 + 2*pi*k for any integer k]

Spoiler for The only difference in my method was

I started by writing it as (sin(x)-1)(polynomial)=0. The rest is basically the same.