Find the positive root of a 4-th degree polynomial equationSolve the equation $x^2+frac{9x^2}{(x+3)^2}=27$How...
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Find the positive root of a 4-th degree polynomial equation
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$begingroup$
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
edited 3 hours ago
David G. Stork
12.1k41735
asked 3 hours ago
shewlongshewlong
414
$endgroup$
add a comment |
$begingroup$
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
edited 3 hours ago
David G. Stork
12.1k41735
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
2 hours ago
$begingroup$
math.stackexchange.com/questions/2020139/…
$endgroup$
– lab bhattacharjee
1 hour ago
add a comment |
$begingroup$
I was struggling with this problem:
$$100^{2}=x^{2}+ left( frac{100x}{100+x} right)^{2}$$
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
edited 3 hours ago
David G. Stork
12.1k41735
asked 3 hours ago
shewlongshewlong
414
$endgroup$
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
edited 3 hours ago
David G. Stork
12.1k41735
asked 3 hours ago
shewlongshewlong
414
edited 3 hours ago
David G. Stork
12.1k41735
asked 3 hours ago
shewlongshewlong
414
edited 3 hours ago
David G. Stork
12.1k41735
edited 3 hours ago
David G. Stork
12.1k41735
edited 3 hours ago
David G. Stork
12.1k41735
12.1k41735
asked 3 hours ago
shewlongshewlong
414
asked 3 hours ago
shewlongshewlong
414
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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...
answered 3 hours ago
lhflhf
167k11172404
$endgroup$
$begingroup$
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
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 $100(1+sqrt2)$.
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
2 hours ago
$begingroup$
The right side is (x(100+x))^2.
$endgroup$
– shewlong
2 hours ago
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
$$
50 (-1 + sqrt 2 + sqrt{2 sqrt 2 - 1}) approx 88.320
$$
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...
answered 3 hours ago
lhflhf
167k11172404
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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...
answered 3 hours ago
lhflhf
167k11172404
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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...
answered 3 hours ago
lhflhf
167k11172404
$endgroup$
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...
answered 3 hours ago
lhflhf
167k11172404
edited 2 hours ago
answered 3 hours ago
lhflhf
167k11172404
answered 3 hours ago
lhflhf
167k11172404
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
add a comment |
$begingroup$
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
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 $100(1+sqrt2)$.
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
2 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 $100(1+sqrt2)$.
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
2 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 $100(1+sqrt2)$.
answered 3 hours ago
Holding ArthurHolding Arthur
1,555417
$endgroup$
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 $100(1+sqrt2)$.
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
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
add a comment |
$begingroup$
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
add a comment |
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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