Find the positive root of a 4-th degree polynomial equationSolve the equation $x^2+frac{9x^2}{(x+3)^2}=27$How...

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Find the positive root of a 4-th degree polynomial equation


Solve the equation $x^2+frac{9x^2}{(x+3)^2}=27$How many cups of sugar do I need for these 5th grade problems?Ecuation of Parabola from Point in vertex formGrouping and Factoring PolynomialsCoffee Table Trig (Finding angles when working with wider boards)How to solve this equation manually: $(x^2+100)^2=(x^3-100)^3$?Maths Puzzle!!!Largest integer less than $2013$ obtained by repeatedly doubling an integer $x$.Why does the definition of a square root for a number not include its solution's negative counterpart?Determining the number of non-real roots. Multiple choice strategy.Mixed percentage algebra problem













4












$begingroup$


I was struggling with this problem:



$$100^{2}=x^{2}+ left( frac{100x}{100+x} right)^{2}$$



It came up when i was developing a solution to a geometry problem. I've already checked in Mathematica and the solution is right, according to the answer. The answer needs to be a real positive number because it's a measure of a segment.



I've tried factoring, manipulating algebraically, but i couldn't solve the resulting 4th degree polynomial. I appreciate if someone could help me. Thanks!






algebra-precalculus polynomials contest-math real-numbers factoring







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What's wrong with just using the solution from Mathematica?
    $endgroup$
    – David G. Stork
    3 hours ago










  • $begingroup$
    It is a problem from a Olympiad contest. You can't use solvers there.
    $endgroup$
    – shewlong
    2 hours ago










  • $begingroup$
    math.stackexchange.com/questions/2020139/…
    $endgroup$
    – lab bhattacharjee
    1 hour ago
















4












$begingroup$


I was struggling with this problem:



$$100^{2}=x^{2}+ left( frac{100x}{100+x} right)^{2}$$



It came up when i was developing a solution to a geometry problem. I've already checked in Mathematica and the solution is right, according to the answer. The answer needs to be a real positive number because it's a measure of a segment.



I've tried factoring, manipulating algebraically, but i couldn't solve the resulting 4th degree polynomial. I appreciate if someone could help me. Thanks!






algebra-precalculus polynomials contest-math real-numbers factoring







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What's wrong with just using the solution from Mathematica?
    $endgroup$
    – David G. Stork
    3 hours ago










  • $begingroup$
    It is a problem from a Olympiad contest. You can't use solvers there.
    $endgroup$
    – shewlong
    2 hours ago










  • $begingroup$
    math.stackexchange.com/questions/2020139/…
    $endgroup$
    – lab bhattacharjee
    1 hour ago














4












4








4


1



$begingroup$


I was struggling with this problem:



$$100^{2}=x^{2}+ left( frac{100x}{100+x} right)^{2}$$



It came up when i was developing a solution to a geometry problem. I've already checked in Mathematica and the solution is right, according to the answer. The answer needs to be a real positive number because it's a measure of a segment.



I've tried factoring, manipulating algebraically, but i couldn't solve the resulting 4th degree polynomial. I appreciate if someone could help me. Thanks!






algebra-precalculus polynomials contest-math real-numbers factoring







share|cite|improve this question











$endgroup$




I was struggling with this problem:



$$100^{2}=x^{2}+ left( frac{100x}{100+x} right)^{2}$$



It came up when i was developing a solution to a geometry problem. I've already checked in Mathematica and the solution is right, according to the answer. The answer needs to be a real positive number because it's a measure of a segment.



I've tried factoring, manipulating algebraically, but i couldn't solve the resulting 4th degree polynomial. I appreciate if someone could help me. Thanks!






algebra-precalculus polynomials contest-math real-numbers factoring




algebra-precalculus polynomials contest-math real-numbers factoring






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









David G. Stork

12.1k41735




12.1k41735










asked 3 hours ago









shewlongshewlong

414




414








  • 1




    $begingroup$
    What's wrong with just using the solution from Mathematica?
    $endgroup$
    – David G. Stork
    3 hours ago










  • $begingroup$
    It is a problem from a Olympiad contest. You can't use solvers there.
    $endgroup$
    – shewlong
    2 hours ago










  • $begingroup$
    math.stackexchange.com/questions/2020139/…
    $endgroup$
    – lab bhattacharjee
    1 hour ago














  • 1




    $begingroup$
    What's wrong with just using the solution from Mathematica?
    $endgroup$
    – David G. Stork
    3 hours ago










  • $begingroup$
    It is a problem from a Olympiad contest. You can't use solvers there.
    $endgroup$
    – shewlong
    2 hours ago










  • $begingroup$
    math.stackexchange.com/questions/2020139/…
    $endgroup$
    – lab bhattacharjee
    1 hour ago








1




1




$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
3 hours ago




$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
3 hours ago












$begingroup$
It is a problem from a Olympiad contest. You can't use solvers there.
$endgroup$
– shewlong
2 hours ago




$begingroup$
It is a problem from a Olympiad contest. You can't use solvers there.
$endgroup$
– shewlong
2 hours ago












$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
1 hour ago




$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
1 hour ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

WA tell us that the positive root is
$$
50 (-1 + sqrt 2 + sqrt{2 sqrt 2 - 1}) approx 88.320
$$
WA also tells us that the minimal polynomial of this number over $mathbb Q$ is
$$
x^4 + 200 x^3 + 10000 x^2 - 2000000 x - 100000000
$$
and so there is no simpler answer.



On the other hand, substituting $x=50u$ in the minimal polynomial gives
$$
6250000 (u^4 + 4 u^3 + 4 u^2 - 16 u - 16)
$$
Now this can factored into two reasonably looking quadratics:
$$
u^4 + 4 u^3 + 4 u^2 - 16 u - 16 =
(u^2 + (2 + 2 sqrt 2) u + 4 sqrt 2 + 4)
(u^2 + (2 - 2 sqrt 2) u - 4 sqrt 2 + 4)
$$
But all this is in hindsight...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt{2} + sqrt{2sqrt{2} -1})= 0$?
    $endgroup$
    – David G. Stork
    3 hours ago










  • $begingroup$
    @DavidG.Stork, see my edited answer.
    $endgroup$
    – lhf
    3 hours ago










  • $begingroup$
    OK... I guess we can assume $x in mathbb{Q}$.
    $endgroup$
    – David G. Stork
    2 hours ago



















0












$begingroup$

Multiplying both sides by $(100+x)^2$, we get
$$
100^2(100+x)^2=x^2(100+x)^2+ 100^2x^2\
Rightarrow100^2[(100+x+x)(100+x-x)]=100^2x^2\
Rightarrow100(100+2x)=x^2\
Rightarrow x^2-200x-10000=0.
$$
You are then able to solve it. The positive root is $100(1+sqrt2)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you explain how you went from step 1 to step 2? I think you've made an error.
    $endgroup$
    – Clayton
    3 hours ago










  • $begingroup$
    I think there's an error in the solution. The answer is not that.
    $endgroup$
    – shewlong
    2 hours ago










  • $begingroup$
    The right side is (x(100+x))^2.
    $endgroup$
    – shewlong
    2 hours ago












Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

WA tell us that the positive root is
$$
50 (-1 + sqrt 2 + sqrt{2 sqrt 2 - 1}) approx 88.320
$$
WA also tells us that the minimal polynomial of this number over $mathbb Q$ is
$$
x^4 + 200 x^3 + 10000 x^2 - 2000000 x - 100000000
$$
and so there is no simpler answer.



On the other hand, substituting $x=50u$ in the minimal polynomial gives
$$
6250000 (u^4 + 4 u^3 + 4 u^2 - 16 u - 16)
$$
Now this can factored into two reasonably looking quadratics:
$$
u^4 + 4 u^3 + 4 u^2 - 16 u - 16 =
(u^2 + (2 + 2 sqrt 2) u + 4 sqrt 2 + 4)
(u^2 + (2 - 2 sqrt 2) u - 4 sqrt 2 + 4)
$$
But all this is in hindsight...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt{2} + sqrt{2sqrt{2} -1})= 0$?
    $endgroup$
    – David G. Stork
    3 hours ago










  • $begingroup$
    @DavidG.Stork, see my edited answer.
    $endgroup$
    – lhf
    3 hours ago










  • $begingroup$
    OK... I guess we can assume $x in mathbb{Q}$.
    $endgroup$
    – David G. Stork
    2 hours ago
















3












$begingroup$

WA tell us that the positive root is
$$
50 (-1 + sqrt 2 + sqrt{2 sqrt 2 - 1}) approx 88.320
$$
WA also tells us that the minimal polynomial of this number over $mathbb Q$ is
$$
x^4 + 200 x^3 + 10000 x^2 - 2000000 x - 100000000
$$
and so there is no simpler answer.



On the other hand, substituting $x=50u$ in the minimal polynomial gives
$$
6250000 (u^4 + 4 u^3 + 4 u^2 - 16 u - 16)
$$
Now this can factored into two reasonably looking quadratics:
$$
u^4 + 4 u^3 + 4 u^2 - 16 u - 16 =
(u^2 + (2 + 2 sqrt 2) u + 4 sqrt 2 + 4)
(u^2 + (2 - 2 sqrt 2) u - 4 sqrt 2 + 4)
$$
But all this is in hindsight...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt{2} + sqrt{2sqrt{2} -1})= 0$?
    $endgroup$
    – David G. Stork
    3 hours ago










  • $begingroup$
    @DavidG.Stork, see my edited answer.
    $endgroup$
    – lhf
    3 hours ago










  • $begingroup$
    OK... I guess we can assume $x in mathbb{Q}$.
    $endgroup$
    – David G. Stork
    2 hours ago














3












3








3





$begingroup$

WA tell us that the positive root is
$$
50 (-1 + sqrt 2 + sqrt{2 sqrt 2 - 1}) approx 88.320
$$
WA also tells us that the minimal polynomial of this number over $mathbb Q$ is
$$
x^4 + 200 x^3 + 10000 x^2 - 2000000 x - 100000000
$$
and so there is no simpler answer.



On the other hand, substituting $x=50u$ in the minimal polynomial gives
$$
6250000 (u^4 + 4 u^3 + 4 u^2 - 16 u - 16)
$$
Now this can factored into two reasonably looking quadratics:
$$
u^4 + 4 u^3 + 4 u^2 - 16 u - 16 =
(u^2 + (2 + 2 sqrt 2) u + 4 sqrt 2 + 4)
(u^2 + (2 - 2 sqrt 2) u - 4 sqrt 2 + 4)
$$
But all this is in hindsight...






share|cite|improve this answer











$endgroup$



WA tell us that the positive root is
$$
50 (-1 + sqrt 2 + sqrt{2 sqrt 2 - 1}) approx 88.320
$$
WA also tells us that the minimal polynomial of this number over $mathbb Q$ is
$$
x^4 + 200 x^3 + 10000 x^2 - 2000000 x - 100000000
$$
and so there is no simpler answer.



On the other hand, substituting $x=50u$ in the minimal polynomial gives
$$
6250000 (u^4 + 4 u^3 + 4 u^2 - 16 u - 16)
$$
Now this can factored into two reasonably looking quadratics:
$$
u^4 + 4 u^3 + 4 u^2 - 16 u - 16 =
(u^2 + (2 + 2 sqrt 2) u + 4 sqrt 2 + 4)
(u^2 + (2 - 2 sqrt 2) u - 4 sqrt 2 + 4)
$$
But all this is in hindsight...







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 3 hours ago









lhflhf

167k11172404




167k11172404












  • $begingroup$
    How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt{2} + sqrt{2sqrt{2} -1})= 0$?
    $endgroup$
    – David G. Stork
    3 hours ago










  • $begingroup$
    @DavidG.Stork, see my edited answer.
    $endgroup$
    – lhf
    3 hours ago










  • $begingroup$
    OK... I guess we can assume $x in mathbb{Q}$.
    $endgroup$
    – David G. Stork
    2 hours ago


















  • $begingroup$
    How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt{2} + sqrt{2sqrt{2} -1})= 0$?
    $endgroup$
    – David G. Stork
    3 hours ago










  • $begingroup$
    @DavidG.Stork, see my edited answer.
    $endgroup$
    – lhf
    3 hours ago










  • $begingroup$
    OK... I guess we can assume $x in mathbb{Q}$.
    $endgroup$
    – David G. Stork
    2 hours ago
















$begingroup$
How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt{2} + sqrt{2sqrt{2} -1})= 0$?
$endgroup$
– David G. Stork
3 hours ago




$begingroup$
How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt{2} + sqrt{2sqrt{2} -1})= 0$?
$endgroup$
– David G. Stork
3 hours ago












$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
3 hours ago




$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
3 hours ago












$begingroup$
OK... I guess we can assume $x in mathbb{Q}$.
$endgroup$
– David G. Stork
2 hours ago




$begingroup$
OK... I guess we can assume $x in mathbb{Q}$.
$endgroup$
– David G. Stork
2 hours ago











0












$begingroup$

Multiplying both sides by $(100+x)^2$, we get
$$
100^2(100+x)^2=x^2(100+x)^2+ 100^2x^2\
Rightarrow100^2[(100+x+x)(100+x-x)]=100^2x^2\
Rightarrow100(100+2x)=x^2\
Rightarrow x^2-200x-10000=0.
$$
You are then able to solve it. The positive root is $100(1+sqrt2)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you explain how you went from step 1 to step 2? I think you've made an error.
    $endgroup$
    – Clayton
    3 hours ago










  • $begingroup$
    I think there's an error in the solution. The answer is not that.
    $endgroup$
    – shewlong
    2 hours ago










  • $begingroup$
    The right side is (x(100+x))^2.
    $endgroup$
    – shewlong
    2 hours ago
















0












$begingroup$

Multiplying both sides by $(100+x)^2$, we get
$$
100^2(100+x)^2=x^2(100+x)^2+ 100^2x^2\
Rightarrow100^2[(100+x+x)(100+x-x)]=100^2x^2\
Rightarrow100(100+2x)=x^2\
Rightarrow x^2-200x-10000=0.
$$
You are then able to solve it. The positive root is $100(1+sqrt2)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you explain how you went from step 1 to step 2? I think you've made an error.
    $endgroup$
    – Clayton
    3 hours ago










  • $begingroup$
    I think there's an error in the solution. The answer is not that.
    $endgroup$
    – shewlong
    2 hours ago










  • $begingroup$
    The right side is (x(100+x))^2.
    $endgroup$
    – shewlong
    2 hours ago














0












0








0





$begingroup$

Multiplying both sides by $(100+x)^2$, we get
$$
100^2(100+x)^2=x^2(100+x)^2+ 100^2x^2\
Rightarrow100^2[(100+x+x)(100+x-x)]=100^2x^2\
Rightarrow100(100+2x)=x^2\
Rightarrow x^2-200x-10000=0.
$$
You are then able to solve it. The positive root is $100(1+sqrt2)$.






share|cite|improve this answer









$endgroup$



Multiplying both sides by $(100+x)^2$, we get
$$
100^2(100+x)^2=x^2(100+x)^2+ 100^2x^2\
Rightarrow100^2[(100+x+x)(100+x-x)]=100^2x^2\
Rightarrow100(100+2x)=x^2\
Rightarrow x^2-200x-10000=0.
$$
You are then able to solve it. The positive root is $100(1+sqrt2)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









Holding ArthurHolding Arthur

1,555417




1,555417












  • $begingroup$
    Can you explain how you went from step 1 to step 2? I think you've made an error.
    $endgroup$
    – Clayton
    3 hours ago










  • $begingroup$
    I think there's an error in the solution. The answer is not that.
    $endgroup$
    – shewlong
    2 hours ago










  • $begingroup$
    The right side is (x(100+x))^2.
    $endgroup$
    – shewlong
    2 hours ago


















  • $begingroup$
    Can you explain how you went from step 1 to step 2? I think you've made an error.
    $endgroup$
    – Clayton
    3 hours ago










  • $begingroup$
    I think there's an error in the solution. The answer is not that.
    $endgroup$
    – shewlong
    2 hours ago










  • $begingroup$
    The right side is (x(100+x))^2.
    $endgroup$
    – shewlong
    2 hours ago
















$begingroup$
Can you explain how you went from step 1 to step 2? I think you've made an error.
$endgroup$
– Clayton
3 hours ago




$begingroup$
Can you explain how you went from step 1 to step 2? I think you've made an error.
$endgroup$
– Clayton
3 hours ago












$begingroup$
I think there's an error in the solution. The answer is not that.
$endgroup$
– shewlong
2 hours ago




$begingroup$
I think there's an error in the solution. The answer is not that.
$endgroup$
– shewlong
2 hours ago












$begingroup$
The right side is (x(100+x))^2.
$endgroup$
– shewlong
2 hours ago




$begingroup$
The right side is (x(100+x))^2.
$endgroup$
– shewlong
2 hours ago


















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Gersau Kjelder | Navigasjonsmeny46°59′0″N 8°31′0″E46°59′0″N...

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Tidlegare republikkarSkipingar i 1433Kommunar i SchwyzDistrikt i SchwyzNedleggingar i 1798Stader ved Vierwaldstättersee kommunekantonenSchwyzSveitskm²VierwaldstätterseeLauerz Gersau Kommunevåpenet til Gersau Plassering Styresmakter Land   Sveits Kanton kantonen Schwyz Distrikt Gersau Språk tysk Geografi Flatevidd  - kommune 23,70 km² Innbyggjarar  - kommune (desember 2010)    - folketettleik 2 094   88,4 /km² Koordinatar 46°59′0″N 8°31′0″E  Koordinatar: 46°59′0″N 8°31′0″E  Høgd over havet 435 moh Diverse annan informasjon Postnummer 6442 Nettstad: www.gersau.ch Freistaat und Republik Gersau Fristaten og republikken Gersau By-republikk og alliert protektorat i Det gamle sveitsiske eidsforundet ←   ← 1433–1798 1814–1818 →   → Hovudstad Gersau Styreform Republikk Historisk periode Tidleg moderne tid  - Kjøpte fridom frå ...
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Hestehale Innhaldsliste Hestehale på kvinner | Hestehale på menn | Galleri | Sjå òg |...

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Frisyrar frisyrehårhårstrikkhårklemmerøyrethesthovudverkeuropeisk motejobbtreningfletterparykkarungdomsopprørmotkulturmusikararmotorsykkelentusiastarpolitietpolitiskforretningverksemdfiskeflette Hestehale sett bakfrå. Hestehale på ein mannleg indianarleiar. Hestehale er ein frisyre der ein samlar det meste eller alt av langt hår [1] i bakhovudet med ein hårstrikk eller liknande. Avhengig av mote kan det òg vere vanleg å sette opp hestehalen med hårklemmer eller å ha hestehalen hengande på den eine sida av hovudet ved å samle håret ved øyret. Hestehale er ein vanleg frisyre for jenter og for gutar med langt hår. Han har fått namnet sitt etter likskapen med halen på ein hest. For stramme hestehaler kan føre til hårtap, [2] i nokre høve også hovudverk. [3] Innhaldsliste 1 Hestehale på kvinner 2 Hestehale på menn 3 Galleri 4 Sjå òg 5 Bakgrunnsstoff 6 Kjelder Hestehale på kvinner | Ungjente med hestehale som held håret unna ansiktet...
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