solve this.... (1 Viewer)

[onebytwo]

x^5 = (x-3)^5, find possible values for x

i know that they are the same type of line but (x-3)^5 is shifted to the right three units, so they have no points of intersection and thus no real solution.
i tried using mod-arg form but couldnt get anywhere

any help appreciated thanks

 

[sasquatch]

maybe take the 5th-root of both sides. Since the power (5) is odd, the is only a positive solution for both sides. Then upon solving,

x = x - 3
0 = -3
.;. no solution.

I dunno if thats correct though.. but it should be..

------

I think its ok, cuz if you consider this example,

x² = x² - 2x + 1

there is a solution for x = (1/2)

so writing the statement as

x² = (x - 1)²
|x| = |x-1| (as the square-root is an even power)

x = x - 1 or x = -(x-1)
0 = -1, no solution

x = -x + 1
2x = 1
x = (1/2)

hence x = 1/2


Yeah but i yeah i still cant be 100% sure of the 5-th root.

 

[Riviet]

Well if you expand the RHS, you can get rid of the x⁵ term, leaving you a polynomial of 4th degree. I would expect two sets of complex conjugate roots since we know there are no real roots. Hope that helps.

 

[who_loves_maths]

x^5 = (x-3)^5, find possible values for x

i know that they are the same type of line but (x-3)^5 is shifted to the right three units, so they have no points of intersection and thus no real solution.
i tried using mod-arg form but couldnt get anywhere

any help appreciated thanks

Dear onebytwo, hoping this isn't too late to help.

First, begin with the following identity: (from memory, I did this last year so you probably would know it already but just in case you didn't I have laid out the proof in the appendix at the end of this post)

1 - Cis@ = (2(1- Cos@))^1/2 .Cis[(@ - pi)/2] ____________________ (1)
where @ here denotes an angle in radians.

Knowing that, we proceed as follows:

1) x = 0 does not satisfy the equation x⁵ = (x-3)⁵

2) So for all x not 0, we divide the equation throughout by x⁵ :

---> 1 = (x-3)⁵/x⁵ = [(x-3)/x]⁵ = (1 - 3/x)⁵

LHS = 1 = Cis(0 + 2k.pi) = Cis(2k.pi) , where k = {1, -1, 2, -2} gives all the solution. k cannot be 0 as the roots of the equation occur in conjugates since we already know no real roots exist by virtue of previous authors sasquatch & Riviet.

So we have, (1 - 3/x)⁵ = Cis(2k.pi)

---> 1 - 3/x = Cis(2k.pi/5) ____________________ By DeMoivre's Theorem

Then, by identity (1):
3/x = 1 - Cis(2k.pi/5) = (2(1 - Cos(2k.pi/5)))^1/2.Cis[(2k.pi - 5pi)/10]

---> x = 3/(2(1 - Cos(2k.pi/5)))^1/2.Cis[(2k.pi - 5pi)/10]]

But 1/Cis@ = Cis(-@)

---> x = (3Cis[(5pi - 2k.pi)/10])/(2(1 - Cos(2k.pi/5)))^1/2

And finally, for k = {1, -1, 2, -2} we have the solutions:

x = {(3Cis(3pi/10))/(2(1 - Cos(2pi/5)))^1/2, (3Cis(7pi/10))/(2(1 - Cos(2pi/5)))^1/2, (3Cis(pi/10))/(2(1 - Cos(4pi/5)))^1/2, (3Cis(9pi/10))/(2(1 - Cos(4pi/5)))^1/2}

- are the solutions of the quartic equation: x⁵ = (x-3)⁵

Hope that helps .


APPENDIX
Prove identity (1): 1 - Cis@ = (2(1- Cos@))^1/2 .Cis[(@ - pi)/2]

1) Modulus:
|1 - Cis@| = |(1 - Cos@) + i(-Sin@)| = [(1 - Cos@)² + Sin²@]^1/2 = (1 - 2Cos@ + Cos²@ + Sin²@)^1/2
= (2(1- Cos@))^1/2

2) Argument:
Arg{1 - Cis@} = Arg{(1 - Cos@) + i(-Sin@)} = Tan^-1(Sin@/(Cos@ - 1))

Sin@ = 2Sin(@/2).Cos(@/2)
Cos@ = 1 - 2Sin²(@/2) ---> Cos@ - 1 = -2Sin²(@/2)

---> Sin@/(Cos@ - 1) = [2Sin(@/2).Cos(@/2)]/-2Sin²(@/2) = -Cot(@/2)

---> Arg{1 - Cis@} = Tan^-1(-Cot(@/2)) = -Tan^-1(Cot(@/2))

But there exist the identity: pi/2 = Tan^-1(#) + Tan^-1(1/#)

Hence: Arg{1 - Cis@} = -(pi/2 - Tan^-1(1/Cot(@/2))) = -(pi/2 - Tan^-1(Tan(@/2))) = @/2 - pi/2
= (@ - pi)/2

Finally, putting 1) and 2) together yields:
1 - Cis@ = (2(1- Cos@))^1/2 .Cis[(@ - pi)/2]

 

[bboyelement]

is this in the syllabus because i would have no idea if it was in the test ...

 

[who_loves_maths]

...... we have the solutions:

x = {(3Cis(3pi/10))/(2(1 - Cos(2pi/5)))^1/2, (3Cis(7pi/10))/(2(1 - Cos(2pi/5)))^1/2, (3Cis(pi/10))/(2(1 - Cos(4pi/5)))^1/2, (3Cis(9pi/10))/(2(1 - Cos(4pi/5)))^1/2}

- are the solutions of the quartic equation: x⁵ = (x-3)⁵

......

But there exist the identity: pi/2 = Tan^-1(#) + Tan^-1(1/#)

......

Grave apologies people... made a terrible mistake in the bolded lines above ^ . Thankfully no-one has complained of it yet.

The quoted line:
"" pi/2 = Tan^-1(#) + Tan^-1(1/#) ""

should in fact read:
pi/2⁺_- = Tan^-1(#) + Tan^-1(1/#)

where I've used the notation x⁺_- to denote "plus or minus x".

This correction is due to the fact that '#' in the above equation can, of course, be both positive or negative, and the ArcTan function is an odd function.

Whether LHS = +pi/2 or -pi/2 is directly related to the polarity of '#'.

Thus, for k={1, -1, 2, -2}, we have:
Cot(2k.pi/5) = {Cot(2pi/5), Cot(-2pi/5), Cot(4pi/5), Cot(-4pi/5)} = {Cot(2pi/5), Cot(-2pi/5), Cot(pi/5), Cot(-pi/5)}

And:
i) # = {Cot(2pi/5), Cot(pi/5)} > 0 -----> corresponding to +pi/2
ii) # = {Cot(-2pi/5), Cot(-pi/5)} < 0 -----> corresponding to -pi/2

Finally, the major implication of this correction is a re-shuffle of the roots of the quartic equation (after working through some facile algebra which I won't show here):

x = {(3Cis(3pi/10))/(2(1 - Cos(2pi/5)))^1/2, (3Cis(-3pi/10))/(2(1 - Cos(2pi/5)))^1/2, (3Cis(4pi/10))/(2(1 - Cos(4pi/5)))^1/2, (3Cis(-4pi/10))/(2(1 - Cos(4pi/5)))^1/2}

- are in fact the actual solutions of the equation as opposed to what I'd put up in the last post.

Of course, the simple observation is that these roots are now finally conjugates pairs, as they must be, as opposed to before!

Sorry about this.

 

[who_loves_maths]

is this in the syllabus because i would have no idea if it was in the test ...

Yes to the best of my knowledge, this is in fact a standard 4u question.
It has been examined before in HSC exams as well as holding presence in exercises in many standard 4u textbooks.

 

[icycloud]

z^5 = (z-3)^5
(z/(z-3))^5 = 1

Let rcis@ = z/(z-3) = 1 + 3/(z-3)
r^5 cis5@ = cis0
r = 1, @ = 2k(pi)/5

Thus, 1+3/(z-3) = cis(2k(pi)/5) for k=0,1,2,3
z = 3/(cis[2k(pi)/5]-1) + 3

Let c = cos[k(pi)/5], s = sin[k(pi)/5]

Thus, z = 3/(2is[c+is]) + 3
= -3i/2 cosec(k(pi)/5) cis(-k(pi)/5) + 3
#

 

[who_loves_maths]

z^5 = (z-3)^5
(z/(z-3))^5 = 1

Let rcis@ = z/(z-3) = 1 + 3/(z-3)
r^5 cis5@ = cis0
r = 1, @ = 2k(pi)/5

Thus, 1+3/(z-3) = cis(2k(pi)/5) for k=0,1,2,3
z = 3/(cis[2k(pi)/5]-1) + 3

Let c = cos[k(pi)/5], s = sin[k(pi)/5]

Thus, z = 3/(2is[c+is]) + 3
= -3i/2 cosec(k(pi)/5) cis(-k(pi)/5) + 3
#

^ Although it's essentially the same method, using z/(z-3) is really taking the long way of doing it instead of (1 - 3/z), particularly in the solution of the roots where one would usually be required to express them in mod-arg form instead of an inverted complex numbers added to a real constant.