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












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












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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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Czorneboh

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Dieser Artikel erläutert den Berg Czorneboh, für die Gottheit siehe Czorneboh (Gottheit). Czorneboh Čornobóh Berggasthof mit Turm Höhe 555,7  m ü. NHN [1] Lage Freistaat Sachsen, Deutschland Gebirge Lausitzer Bergland Dominanz 15,4 km →  Kottmar Koordinaten 51° 7′ 11″  N , 14° 31′ 31″  O 51.119722222222 14.525194444444 555.7 Koordinaten: 51° 7′ 11″  N , 14° 31′ 31″  O Gestein Granodiorit Besonderheiten Aussichtsturm Czorneboh Der Czorneboh , obersorbisch Čornobóh , ist ein Berg zwischen Hochkirch und Cunewalde in der Oberlausitz. Er gehört zu der südöstlich von Bautzen gelegenen Czorneboh-Bergkette und ist mit einer Höhe von 555,7  m ü. NHN [1] der höchste Punkt dieses Ausläufers des Lausitzer Berglandes. Er leitet als Landschaftselement die wellige Granitlandschaft der Lausitzer Vorbergzone ein. Der Gipfel des Czorneboh liegt in der Gemarkung Meschwitz (Gemeinde Hochkirch), die Czornebohbaude jedoch
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