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Find the positive root of $100^2=x^2+ left( frac100x100+x right)^2$

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4

1

$begingroup$

I was struggling with this problem:

$$100^2=x^2+ left( frac100x100+x right)^2$$

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

edited 5 mins ago

user21820

40.1k544162

asked 3 hours ago

shewlongshewlong

414

$endgroup$

• 
  
  
  

  
  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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

4

1

$begingroup$

I was struggling with this problem:

$$100^2=x^2+ left( frac100x100+x right)^2$$

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

edited 5 mins ago

user21820

40.1k544162

asked 3 hours ago

shewlongshewlong

414

$endgroup$

• 
  
  
  

  
  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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

I was struggling with this problem:

$$100^2=x^2+ left( frac100x100+x right)^2$$

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

edited 5 mins ago

user21820

40.1k544162

asked 3 hours ago

shewlongshewlong

414

$endgroup$

share|cite|improve this question

edited 5 mins ago

user21820

40.1k544162

asked 3 hours ago

shewlongshewlong

414

share|cite|improve this question

edited 5 mins ago

user21820

40.1k544162

asked 3 hours ago

shewlongshewlong

414

edited 5 mins ago

user21820

40.1k544162

edited 5 mins ago

user21820

40.1k544162

asked 3 hours ago

shewlongshewlong

414

asked 3 hours ago

shewlongshewlong

414

2 Answers
2

active

oldest

votes

3

$begingroup$

WA tell us that the positive root is
$$
50 (-1 + sqrt 2 + sqrt2 sqrt 2 - 1) approx 88.320
$$
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

edited 2 hours ago

answered 3 hours ago

lhflhf

167k11172404

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt2 + sqrt2sqrt2 -1)= 0$?
  $endgroup$
  – 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 $x in mathbbQ$.
  $endgroup$
  – David G. Stork
  2 hours ago
  

  
  
  
  

add a comment | 

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

answered 3 hours ago

Holding ArthurHolding Arthur

1,555417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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3

$begingroup$

WA tell us that the positive root is
$$
50 (-1 + sqrt 2 + sqrt2 sqrt 2 - 1) approx 88.320
$$
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

edited 2 hours ago

answered 3 hours ago

lhflhf

167k11172404

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt2 + sqrt2sqrt2 -1)= 0$?
  $endgroup$
  – 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 $x in mathbbQ$.
  $endgroup$
  – David G. Stork
  2 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

WA tell us that the positive root is
$$
50 (-1 + sqrt 2 + sqrt2 sqrt 2 - 1) approx 88.320
$$
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

edited 2 hours ago

answered 3 hours ago

lhflhf

167k11172404

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt2 + sqrt2sqrt2 -1)= 0$?
  $endgroup$
  – 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 $x in mathbbQ$.
  $endgroup$
  – David G. Stork
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

WA tell us that the positive root is
$$
50 (-1 + sqrt 2 + sqrt2 sqrt 2 - 1) approx 88.320
$$
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

edited 2 hours ago

answered 3 hours ago

lhflhf

167k11172404

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 3 hours ago

lhflhf

167k11172404

answered 3 hours ago

lhflhf

167k11172404

answered 3 hours ago

lhflhf

167k11172404

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

answered 3 hours ago

Holding ArthurHolding Arthur

1,555417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 3 hours ago

Holding ArthurHolding Arthur

1,555417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

answered 3 hours ago

Holding ArthurHolding Arthur

1,555417

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Holding ArthurHolding Arthur

1,555417

answered 3 hours ago

Holding ArthurHolding Arthur

1,555417

answered 3 hours ago

Holding ArthurHolding Arthur

1,555417