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

Multi tool use
Multi tool use












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
    yesterday






  • 2




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










  • $begingroup$
    math.stackexchange.com/questions/2020139/…
    $endgroup$
    – lab bhattacharjee
    yesterday










  • $begingroup$
    Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
    $endgroup$
    – shewlong
    13 hours 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
    yesterday






  • 2




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










  • $begingroup$
    math.stackexchange.com/questions/2020139/…
    $endgroup$
    – lab bhattacharjee
    yesterday










  • $begingroup$
    Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
    $endgroup$
    – shewlong
    13 hours 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 22 hours ago









user21820

40.1k544162




40.1k544162










asked yesterday









shewlongshewlong

715




715








  • 1




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






  • 2




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










  • $begingroup$
    math.stackexchange.com/questions/2020139/…
    $endgroup$
    – lab bhattacharjee
    yesterday










  • $begingroup$
    Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
    $endgroup$
    – shewlong
    13 hours ago














  • 1




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






  • 2




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










  • $begingroup$
    math.stackexchange.com/questions/2020139/…
    $endgroup$
    – lab bhattacharjee
    yesterday










  • $begingroup$
    Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
    $endgroup$
    – shewlong
    13 hours ago








1




1




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




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




2




2




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




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












$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
yesterday




$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
yesterday












$begingroup$
Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
$endgroup$
– shewlong
13 hours ago




$begingroup$
Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
$endgroup$
– shewlong
13 hours ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

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
    yesterday










  • $begingroup$
    @DavidG.Stork, see my edited answer.
    $endgroup$
    – lhf
    yesterday










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



















3












$begingroup$

Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac{9x^2}{(x+3)^2}=27$ shows that, if we have an equation:



$$a^{2}+b^{2}=k$$



Then, calling $c=frac{ab}{a-b}$, we can manipulate algebraically the expression above:



$$(a-b)^{2}+2c(a-b)-k=0$$



For our problem, let $a=x$, $b=frac{100x}{100+x}$ and $k=100^{2}$. It turns out that $c=100$ and $a-b=frac{x^{2}}{100+x}$. So, if we put $u=frac{x^{2}}{100+x}$, we will have:



$$u^{2}+200u-100^{2}=0$$



Which the only positive root is $100(sqrt{2}-1)$. Solving the equation:



$$frac{x^{2}}{100+x}=100(sqrt{2}-1)$$



The only positive root is $50(sqrt{2}-1+sqrt{-1+2sqrt{2}})$.



Thanks!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Excellent! @shewlong
    $endgroup$
    – Dr. Mathva
    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
    yesterday










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










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












Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3180476%2ffind-the-positive-root-of-1002-x2-left-frac100x100x-right2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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
    yesterday










  • $begingroup$
    @DavidG.Stork, see my edited answer.
    $endgroup$
    – lhf
    yesterday










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
















4












$begingroup$

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
    yesterday










  • $begingroup$
    @DavidG.Stork, see my edited answer.
    $endgroup$
    – lhf
    yesterday










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














4












4








4





$begingroup$

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 yesterday

























answered yesterday









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
    yesterday










  • $begingroup$
    @DavidG.Stork, see my edited answer.
    $endgroup$
    – lhf
    yesterday










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


















  • $begingroup$
    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
    yesterday










  • $begingroup$
    @DavidG.Stork, see my edited answer.
    $endgroup$
    – lhf
    yesterday










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
















$begingroup$
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
yesterday




$begingroup$
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
yesterday












$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
yesterday




$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
yesterday












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




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











3












$begingroup$

Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac{9x^2}{(x+3)^2}=27$ shows that, if we have an equation:



$$a^{2}+b^{2}=k$$



Then, calling $c=frac{ab}{a-b}$, we can manipulate algebraically the expression above:



$$(a-b)^{2}+2c(a-b)-k=0$$



For our problem, let $a=x$, $b=frac{100x}{100+x}$ and $k=100^{2}$. It turns out that $c=100$ and $a-b=frac{x^{2}}{100+x}$. So, if we put $u=frac{x^{2}}{100+x}$, we will have:



$$u^{2}+200u-100^{2}=0$$



Which the only positive root is $100(sqrt{2}-1)$. Solving the equation:



$$frac{x^{2}}{100+x}=100(sqrt{2}-1)$$



The only positive root is $50(sqrt{2}-1+sqrt{-1+2sqrt{2}})$.



Thanks!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Excellent! @shewlong
    $endgroup$
    – Dr. Mathva
    9 hours ago
















3












$begingroup$

Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac{9x^2}{(x+3)^2}=27$ shows that, if we have an equation:



$$a^{2}+b^{2}=k$$



Then, calling $c=frac{ab}{a-b}$, we can manipulate algebraically the expression above:



$$(a-b)^{2}+2c(a-b)-k=0$$



For our problem, let $a=x$, $b=frac{100x}{100+x}$ and $k=100^{2}$. It turns out that $c=100$ and $a-b=frac{x^{2}}{100+x}$. So, if we put $u=frac{x^{2}}{100+x}$, we will have:



$$u^{2}+200u-100^{2}=0$$



Which the only positive root is $100(sqrt{2}-1)$. Solving the equation:



$$frac{x^{2}}{100+x}=100(sqrt{2}-1)$$



The only positive root is $50(sqrt{2}-1+sqrt{-1+2sqrt{2}})$.



Thanks!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Excellent! @shewlong
    $endgroup$
    – Dr. Mathva
    9 hours ago














3












3








3





$begingroup$

Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac{9x^2}{(x+3)^2}=27$ shows that, if we have an equation:



$$a^{2}+b^{2}=k$$



Then, calling $c=frac{ab}{a-b}$, we can manipulate algebraically the expression above:



$$(a-b)^{2}+2c(a-b)-k=0$$



For our problem, let $a=x$, $b=frac{100x}{100+x}$ and $k=100^{2}$. It turns out that $c=100$ and $a-b=frac{x^{2}}{100+x}$. So, if we put $u=frac{x^{2}}{100+x}$, we will have:



$$u^{2}+200u-100^{2}=0$$



Which the only positive root is $100(sqrt{2}-1)$. Solving the equation:



$$frac{x^{2}}{100+x}=100(sqrt{2}-1)$$



The only positive root is $50(sqrt{2}-1+sqrt{-1+2sqrt{2}})$.



Thanks!






share|cite|improve this answer









$endgroup$



Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac{9x^2}{(x+3)^2}=27$ shows that, if we have an equation:



$$a^{2}+b^{2}=k$$



Then, calling $c=frac{ab}{a-b}$, we can manipulate algebraically the expression above:



$$(a-b)^{2}+2c(a-b)-k=0$$



For our problem, let $a=x$, $b=frac{100x}{100+x}$ and $k=100^{2}$. It turns out that $c=100$ and $a-b=frac{x^{2}}{100+x}$. So, if we put $u=frac{x^{2}}{100+x}$, we will have:



$$u^{2}+200u-100^{2}=0$$



Which the only positive root is $100(sqrt{2}-1)$. Solving the equation:



$$frac{x^{2}}{100+x}=100(sqrt{2}-1)$$



The only positive root is $50(sqrt{2}-1+sqrt{-1+2sqrt{2}})$.



Thanks!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 10 hours ago









shewlongshewlong

715




715












  • $begingroup$
    Excellent! @shewlong
    $endgroup$
    – Dr. Mathva
    9 hours ago


















  • $begingroup$
    Excellent! @shewlong
    $endgroup$
    – Dr. Mathva
    9 hours ago
















$begingroup$
Excellent! @shewlong
$endgroup$
– Dr. Mathva
9 hours ago




$begingroup$
Excellent! @shewlong
$endgroup$
– Dr. Mathva
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
    yesterday










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










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
















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
    yesterday










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










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














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 yesterday









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
    yesterday










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










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


















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










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










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
















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




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












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




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












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




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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3180476%2ffind-the-positive-root-of-1002-x2-left-frac100x100x-right2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







RVEzoc9hfEm T iI4njKBq5nPUHy6dzZqJ OZ,tTTgd2NYXkpQX,EHMHtf,4 Kb8MlgOoXLQc
-
1FrljfZ yczDMDuUldgFZjSDRZkFTq55MsSypwNXBuvz2pWw B7y y

Popular posts from this blog

Tabula Rosettana

Image
Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
Read more

How to label and detect the document text images

Image
1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
Read more

LSTM sequence prediction: 3d input to 2d output

Image
1 $begingroup$ I have this LSTM model model = Sequential() model.add(Masking(mask_value=0, input_shape=(timesteps, features))) model.add(LSTM(100, dropout=0.2, recurrent_dropout=0.2, return_sequences=False)) model.add(Dense(features, activation='softmax')) model.compile(loss='categorical_crossentropy', optimizer='adam', metrics=['accuracy']) and shapes X_train (21, 11, 5), y_train (21, 5) . Each timestep is represented by 5 features and return_sequences is set to False because I want to predict one 5D array (the next timestep) for each input sequence of 11 timesteps. I get the error ValueError: y_true and y_pred have different number of output (5!=1) If I reshape the data as X_train (21, 11, 5), y_train (21, 1, 5) instead I get the error ValueError: Inva...
Read more