Find the positive root of $100^2=x^2+ left( frac100x100+x right)^2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Solve the equation $x^2+frac9x^2(x+3)^2=27$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
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Find the positive root of $100^2=x^2+ left( frac100x100+x right)^2$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Solve the equation $x^2+frac9x^2(x+3)^2=27$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
$begingroup$
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
edited Apr 9 at 6:26
user21820
40.3k544163
asked Apr 9 at 2:35
shewlongshewlong
715
$endgroup$
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
edited Apr 9 at 6:26
user21820
40.3k544163
asked Apr 9 at 2:35
shewlongshewlong
715
$endgroup$
1
$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
Apr 9 at 2:44
2
$begingroup$
It is a problem from a Olympiad contest. You can't use solvers there.
$endgroup$
– shewlong
Apr 9 at 3:26
$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
Apr 9 at 5:01
$begingroup$
Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
$endgroup$
– shewlong
Apr 9 at 15:08
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
edited Apr 9 at 6:26
user21820
40.3k544163
asked Apr 9 at 2:35
shewlongshewlong
715
$endgroup$
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
algebra-precalculus polynomials contest-math real-numbers factoring
edited Apr 9 at 6:26
user21820
40.3k544163
asked Apr 9 at 2:35
shewlongshewlong
715
edited Apr 9 at 6:26
user21820
40.3k544163
asked Apr 9 at 2:35
shewlongshewlong
715
edited Apr 9 at 6:26
user21820
40.3k544163
edited Apr 9 at 6:26
user21820
40.3k544163
edited Apr 9 at 6:26
user21820
40.3k544163
40.3k544163
asked Apr 9 at 2:35
shewlongshewlong
715
asked Apr 9 at 2:35
shewlongshewlong
715
asked Apr 9 at 2:35
shewlongshewlong
715
715
1
$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
Apr 9 at 2:44
2
$begingroup$
It is a problem from a Olympiad contest. You can't use solvers there.
$endgroup$
– shewlong
Apr 9 at 3:26
$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
Apr 9 at 5:01
$begingroup$
Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
$endgroup$
– shewlong
Apr 9 at 15:08
add a comment |
1
$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
Apr 9 at 2:44
2
$begingroup$
It is a problem from a Olympiad contest. You can't use solvers there.
$endgroup$
– shewlong
Apr 9 at 3:26
$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
Apr 9 at 5:01
$begingroup$
Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
$endgroup$
– shewlong
Apr 9 at 15:08
1
1
$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
Apr 9 at 2:44
$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
Apr 9 at 2:44
2
2
$begingroup$
It is a problem from a Olympiad contest. You can't use solvers there.
$endgroup$
– shewlong
Apr 9 at 3:26
$begingroup$
It is a problem from a Olympiad contest. You can't use solvers there.
$endgroup$
– shewlong
Apr 9 at 3:26
$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
Apr 9 at 5:01
$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
Apr 9 at 5:01
$begingroup$
Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
$endgroup$
– shewlong
Apr 9 at 15:08
$begingroup$
Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
$endgroup$
– shewlong
Apr 9 at 15:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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...
answered Apr 9 at 2:54
lhflhf
168k11172404
$endgroup$
$begingroup$
How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt2 + sqrt2sqrt2 -1)= 0$?
$endgroup$
– David G. Stork
Apr 9 at 3:12
$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
Apr 9 at 3:13
$begingroup$
OK... I guess we can assume $x in mathbbQ$.
$endgroup$
– David G. Stork
Apr 9 at 3:52
add a comment |
$begingroup$
Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac9x^2(x+3)^2=27$ shows that, if we have an equation:
$$a^2+b^2=k$$
Then, calling $c=fracaba-b$, we can manipulate algebraically the expression above:
$$(a-b)^2+2c(a-b)-k=0$$
For our problem, let $a=x$, $b=frac100x100+x$ and $k=100^2$. It turns out that $c=100$ and $a-b=fracx^2100+x$. So, if we put $u=fracx^2100+x$, we will have:
$$u^2+200u-100^2=0$$
Which the only positive root is $100(sqrt2-1)$. Solving the equation:
$$fracx^2100+x=100(sqrt2-1)$$
The only positive root is $50(sqrt2-1+sqrt-1+2sqrt2)$.
Thanks!
answered Apr 9 at 18:52
shewlongshewlong
715
$endgroup$
$begingroup$
Excellent! @shewlong
$endgroup$
– Dr. Mathva
Apr 9 at 19:09
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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...
answered Apr 9 at 2:54
lhflhf
168k11172404
$endgroup$
$begingroup$
How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt2 + sqrt2sqrt2 -1)= 0$?
$endgroup$
– David G. Stork
Apr 9 at 3:12
$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
Apr 9 at 3:13
$begingroup$
OK... I guess we can assume $x in mathbbQ$.
$endgroup$
– David G. Stork
Apr 9 at 3:52
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...
answered Apr 9 at 2:54
lhflhf
168k11172404
$endgroup$
$begingroup$
How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt2 + sqrt2sqrt2 -1)= 0$?
$endgroup$
– David G. Stork
Apr 9 at 3:12
$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
Apr 9 at 3:13
$begingroup$
OK... I guess we can assume $x in mathbbQ$.
$endgroup$
– David G. Stork
Apr 9 at 3:52
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...
answered Apr 9 at 2:54
lhflhf
168k11172404
$endgroup$
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...
answered Apr 9 at 2:54
lhflhf
168k11172404
edited Apr 9 at 3:52
answered Apr 9 at 2:54
lhflhf
168k11172404
answered Apr 9 at 2:54
lhflhf
168k11172404
answered Apr 9 at 2:54
lhflhf
168k11172404
168k11172404
$begingroup$
How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt2 + sqrt2sqrt2 -1)= 0$?
$endgroup$
– David G. Stork
Apr 9 at 3:12
$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
Apr 9 at 3:13
$begingroup$
OK... I guess we can assume $x in mathbbQ$.
$endgroup$
– David G. Stork
Apr 9 at 3:52
add a comment |
$begingroup$
How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt2 + sqrt2sqrt2 -1)= 0$?
$endgroup$
– David G. Stork
Apr 9 at 3:12
$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
Apr 9 at 3:13
$begingroup$
OK... I guess we can assume $x in mathbbQ$.
$endgroup$
– David G. Stork
Apr 9 at 3:52
$begingroup$
How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt2 + sqrt2sqrt2 -1)= 0$?
$endgroup$
– David G. Stork
Apr 9 at 3:12
$begingroup$
How do we know this is the minimal polynomial? What about $x - 50(-1+sqrt2 + sqrt2sqrt2 -1)= 0$?
$endgroup$
– David G. Stork
Apr 9 at 3:12
$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
Apr 9 at 3:13
$begingroup$
@DavidG.Stork, see my edited answer.
$endgroup$
– lhf
Apr 9 at 3:13
$begingroup$
OK... I guess we can assume $x in mathbbQ$.
$endgroup$
– David G. Stork
Apr 9 at 3:52
$begingroup$
OK... I guess we can assume $x in mathbbQ$.
$endgroup$
– David G. Stork
Apr 9 at 3:52
add a comment |
$begingroup$
Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac9x^2(x+3)^2=27$ shows that, if we have an equation:
$$a^2+b^2=k$$
Then, calling $c=fracaba-b$, we can manipulate algebraically the expression above:
$$(a-b)^2+2c(a-b)-k=0$$
For our problem, let $a=x$, $b=frac100x100+x$ and $k=100^2$. It turns out that $c=100$ and $a-b=fracx^2100+x$. So, if we put $u=fracx^2100+x$, we will have:
$$u^2+200u-100^2=0$$
Which the only positive root is $100(sqrt2-1)$. Solving the equation:
$$fracx^2100+x=100(sqrt2-1)$$
The only positive root is $50(sqrt2-1+sqrt-1+2sqrt2)$.
Thanks!
answered Apr 9 at 18:52
shewlongshewlong
715
$endgroup$
$begingroup$
Excellent! @shewlong
$endgroup$
– Dr. Mathva
Apr 9 at 19:09
add a comment |
$begingroup$
Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac9x^2(x+3)^2=27$ shows that, if we have an equation:
$$a^2+b^2=k$$
Then, calling $c=fracaba-b$, we can manipulate algebraically the expression above:
$$(a-b)^2+2c(a-b)-k=0$$
For our problem, let $a=x$, $b=frac100x100+x$ and $k=100^2$. It turns out that $c=100$ and $a-b=fracx^2100+x$. So, if we put $u=fracx^2100+x$, we will have:
$$u^2+200u-100^2=0$$
Which the only positive root is $100(sqrt2-1)$. Solving the equation:
$$fracx^2100+x=100(sqrt2-1)$$
The only positive root is $50(sqrt2-1+sqrt-1+2sqrt2)$.
Thanks!
answered Apr 9 at 18:52
shewlongshewlong
715
$endgroup$
$begingroup$
Excellent! @shewlong
$endgroup$
– Dr. Mathva
Apr 9 at 19:09
add a comment |
$begingroup$
Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac9x^2(x+3)^2=27$ shows that, if we have an equation:
$$a^2+b^2=k$$
Then, calling $c=fracaba-b$, we can manipulate algebraically the expression above:
$$(a-b)^2+2c(a-b)-k=0$$
For our problem, let $a=x$, $b=frac100x100+x$ and $k=100^2$. It turns out that $c=100$ and $a-b=fracx^2100+x$. So, if we put $u=fracx^2100+x$, we will have:
$$u^2+200u-100^2=0$$
Which the only positive root is $100(sqrt2-1)$. Solving the equation:
$$fracx^2100+x=100(sqrt2-1)$$
The only positive root is $50(sqrt2-1+sqrt-1+2sqrt2)$.
Thanks!
answered Apr 9 at 18:52
shewlongshewlong
715
$endgroup$
Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac9x^2(x+3)^2=27$ shows that, if we have an equation:
$$a^2+b^2=k$$
Then, calling $c=fracaba-b$, we can manipulate algebraically the expression above:
$$(a-b)^2+2c(a-b)-k=0$$
For our problem, let $a=x$, $b=frac100x100+x$ and $k=100^2$. It turns out that $c=100$ and $a-b=fracx^2100+x$. So, if we put $u=fracx^2100+x$, we will have:
$$u^2+200u-100^2=0$$
Which the only positive root is $100(sqrt2-1)$. Solving the equation:
$$fracx^2100+x=100(sqrt2-1)$$
The only positive root is $50(sqrt2-1+sqrt-1+2sqrt2)$.
Thanks!
answered Apr 9 at 18:52
shewlongshewlong
715
answered Apr 9 at 18:52
shewlongshewlong
715
answered Apr 9 at 18:52
shewlongshewlong
715
answered Apr 9 at 18:52
shewlongshewlong
715
715
$begingroup$
Excellent! @shewlong
$endgroup$
– Dr. Mathva
Apr 9 at 19:09
add a comment |
$begingroup$
Excellent! @shewlong
$endgroup$
– Dr. Mathva
Apr 9 at 19:09
$begingroup$
Excellent! @shewlong
$endgroup$
– Dr. Mathva
Apr 9 at 19:09
$begingroup$
Excellent! @shewlong
$endgroup$
– Dr. Mathva
Apr 9 at 19:09
add a comment |
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1
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What's wrong with just using the solution from Mathematica?
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– David G. Stork
Apr 9 at 2:44
2
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It is a problem from a Olympiad contest. You can't use solvers there.
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– shewlong
Apr 9 at 3:26
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math.stackexchange.com/questions/2020139/…
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– lab bhattacharjee
Apr 9 at 5:01
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Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later!
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– shewlong
Apr 9 at 15:08