Find the positive root of $100^{2}=x^{2}+ left( frac{100x}{100+x} right)^{2}$












3












$begingroup$


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
    10 hours ago






  • 1




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










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
















3












$begingroup$


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
    10 hours ago






  • 1




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










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














3












3








3


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 7 hours ago









user21820

40.1k544162




40.1k544162










asked 11 hours ago









shewlongshewlong

364




364








  • 1




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






  • 1




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










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














  • 1




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






  • 1




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










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








1




1




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




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




1




1




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




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












$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
8 hours ago




$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
8 hours 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
    10 hours ago










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










  • $begingroup$
    OK... I guess we can assume $x in mathbb{Q}$.
    $endgroup$
    – David G. Stork
    9 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
    10 hours ago










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










  • $begingroup$
    The right side is (x(100+x))^2.
    $endgroup$
    – shewlong
    10 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
    10 hours ago










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










  • $begingroup$
    OK... I guess we can assume $x in mathbb{Q}$.
    $endgroup$
    – David G. Stork
    9 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
    10 hours ago










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










  • $begingroup$
    OK... I guess we can assume $x in mathbb{Q}$.
    $endgroup$
    – David G. Stork
    9 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 9 hours ago

























answered 10 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
    10 hours ago










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










  • $begingroup$
    OK... I guess we can assume $x in mathbb{Q}$.
    $endgroup$
    – David G. Stork
    9 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
    10 hours ago










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










  • $begingroup$
    OK... I guess we can assume $x in mathbb{Q}$.
    $endgroup$
    – David G. Stork
    9 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
10 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
10 hours ago












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




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












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




$begingroup$
OK... I guess we can assume $x in mathbb{Q}$.
$endgroup$
– David G. Stork
9 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
    10 hours ago










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










  • $begingroup$
    The right side is (x(100+x))^2.
    $endgroup$
    – shewlong
    10 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
    10 hours ago










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










  • $begingroup$
    The right side is (x(100+x))^2.
    $endgroup$
    – shewlong
    10 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 10 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
    10 hours ago










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










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


















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










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










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
















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




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












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




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












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




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


















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