60x^37-63x^36+3=0 find x?

Question

answere is 1.03601. i want analytical answere.

answer is 1.03601. i want analytial ans..

Answer 1

A quick plot will tell you there are only 3 real solutions.

1 is clearly one.

You have one near -1 and one slightly higher than 1

If you want analytical answers well... ^o) im sure someone will find them... numerical answers work too

I used R to calculate the 3 real roots. (using newtons method)

-0.9032001053875137
1 is already a known solution
1.0360068534584774

The rest are imaginary

EDIT i have used matlab to calculate all the roots... here you go... enjoy

EDU» roots(p)

ans =

1.03600685345848
1.00000000000000
0.93941952208761 + 0.19413296609082i
0.93941952208761 - 0.19413296609082i
0.87713196799063 + 0.35030226542772i
0.87713196799063 - 0.35030226542772i
0.79548035780274 + 0.49207160521878i
0.79548035780274 - 0.49207160521878i
0.69386352920012 + 0.61750199119194i
0.69386352920012 - 0.61750199119194i
0.57416124248186 + 0.72374521999103i
0.57416124248186 - 0.72374521999103i
0.43932487170030 + 0.80812914340051i
0.43932487170030 - 0.80812914340051i
0.29296219803207 + 0.86845950067326i
0.29296219803207 - 0.86845950067326i
0.13911885212847 + 0.90315942761652i
0.13911885212847 - 0.90315942761652i
-0.01788677043796 + 0.91134742794900i
-0.01788677043796 - 0.91134742794900i
-0.17361132265423 + 0.89287889627887i
-0.17361132265423 - 0.89287889627887i
-0.32362688368333 + 0.84835919241695i
-0.32362688368333 - 0.84835919241695i
-0.46365551220520 + 0.77913155713409i
-0.46365551220520 - 0.77913155713409i
-0.90320010538751
-0.88991357725815 + 0.15488583645940i
-0.88991357725815 - 0.15488583645940i
-0.85043485442088 + 0.30527875115606i
-0.85043485442088 - 0.30527875115606i
-0.78589548164738 + 0.44681444071348i
-0.78589548164738 - 0.44681444071348i
-0.58969663919648 + 0.68724187131749i
-0.58969663919648 - 0.68724187131749i
-0.69814487395567 + 0.57538209321675i
-0.69814487395567 - 0.57538209321675i

Answer 2

rfamilymember
let t = x^35 and 20t^2 is 20x^37.
But 21t is *not* 21x^36.

Yeremy
Maybe it's a repeated root of order 37?

That's about the only thing that would save this one. If the exponents differed by 2 (instead of 1) then maybe a substitution would work to reduce it to a quadratic.

Maybe.

rondosxx
you can't subtract x^36 terms from x^37 terms.


Doug

Edit:

DAMN!!!!!! It must be nice to have access to matlab or an equivalent program.

As far as deriving an analytical solution...... Good Luck

Answer 3

you can get an approximated answer using the numerical analysis:
you suppose an initial answer:
x= -3
then you consider it as a solution to another equation with the same form:
60x^37-63x^36+3=0
the next solution you'll get is the answer of the next step...
until you get the solution you want.

Answer 4

1 is one of the 37 answers

Answer 5

60x^37 - 63x^36 +3 = 0
-3x + 3 = 0
x = -1
But there are a total of 37 possible values for x (though they could all be -1 (conceivably)); there could be a different value for each exponent of x.

Answer 6

Hi,

There are 3 real solutions and 34 complex ones.

The three real solutions are:

x = 1
x = -0.9032 (approx)
x = 1.03601 (approx)

The complex solutions are too long to write out here but consist of 17 pairs of complex conjugates, both real and complex parts between -1 and 1.

Answer 7

3000=<500-=0

Answer 8

two real values of x other than 1 are:
1.036006838
-0.903199792

17 complex conjugate pairs:
-0.323627022442891 +/- i0.848358562033605
-0.589696109769519 +/- i0.687241700692279
0.292961999148756 +/- i0.868459165534428
0.439323818709118 +/- i0.808128927583772
0.693863888141941 +/- i0.617502005640862
-0.785894759080031 +/- i0.446814946623903
0.7954804152233 +/- i0.492073036501474
-0.889913727856503 +/- i0.154886010136097
0.877131721834405 +/- i0.350301781237771
-1.78879792212231E-002 +/- i0.911346941827108
-0.850434272709277 +/- i0.305278990224512
0.939419136821416 +/- i0.194132782674957
-0.173609025819814 +/- i0.892877859433687
0.139118858832631 +/- i0.903159740490469
-0.698143625927589 +/- i0.57538178907986
0.574159946498502 +/- i0.723745394657546
-0.463655013320811 +/- i0.779131848529116

Answer 9

20X^37-21x^36+1=0
let x^35=t
20 t^2-21t+1=0
(20t-1)(t-1)=0
t=1/20 or 1
substituting back
x^35=1/20 x=(1/20)^1/35
x^35=1 so x=1