Find the positive root of $100^2=x^2+ left( frac100x100+x right)^2$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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New order #4: World
Find the positive root of $100^2=x^2+ left( frac100x100+x right)^2$
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 28 mins ago
user21820
40.1k544162
asked 4 hours ago
shewlongshewlong
364
$endgroup$
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 28 mins ago
user21820
40.1k544162
asked 4 hours ago
shewlongshewlong
364
$endgroup$
1
$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
4 hours ago
1
$begingroup$
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
edited 28 mins ago
user21820
40.1k544162
asked 4 hours ago
shewlongshewlong
364
$endgroup$
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 28 mins ago
user21820
40.1k544162
asked 4 hours ago
shewlongshewlong
364
edited 28 mins ago
user21820
40.1k544162
asked 4 hours ago
shewlongshewlong
364
edited 28 mins ago
user21820
40.1k544162
edited 28 mins ago
user21820
40.1k544162
edited 28 mins ago
user21820
40.1k544162
40.1k544162
asked 4 hours ago
shewlongshewlong
364
asked 4 hours ago
shewlongshewlong
364
asked 4 hours ago
shewlongshewlong
364
364
1
$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
4 hours ago
1
$begingroup$
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 |
1
$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
4 hours ago
1
$begingroup$
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
1
1
$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
What's wrong with just using the solution from Mathematica?
$endgroup$
– David G. Stork
4 hours ago
1
1
$begingroup$
It is a problem from a Olympiad contest. You can't use solvers there.
$endgroup$
– shewlong
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
$begingroup$
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
$$
50 (-1 + sqrt 2 + sqrt2 sqrt 2 - 1) approx 88.320
$$
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 4 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
3 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 4 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
3 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
$$
50 (-1 + sqrt 2 + sqrt2 sqrt 2 - 1) approx 88.320
$$
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 4 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
3 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...
answered 4 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
3 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...
answered 4 hours ago
lhflhf
167k11172404
$endgroup$
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 4 hours ago
lhflhf
167k11172404
edited 3 hours ago
answered 4 hours ago
lhflhf
167k11172404
answered 4 hours ago
lhflhf
167k11172404
answered 4 hours ago
lhflhf
167k11172404
167k11172404
$begingroup$
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
3 hours ago
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
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
3 hours ago
$begingroup$
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
$begingroup$
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
$begingroup$
@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
3 hours ago
$begingroup$
OK... I guess we can assume $x in mathbbQ$.
$endgroup$
– David G. Stork
3 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 4 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
3 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 4 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
3 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 4 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 4 hours ago
Holding ArthurHolding Arthur
1,555417
answered 4 hours ago
Holding ArthurHolding Arthur
1,555417
answered 4 hours ago
Holding ArthurHolding Arthur
1,555417
answered 4 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
3 hours ago
$begingroup$
The right side is (x(100+x))^2.
$endgroup$
– shewlong
3 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
3 hours ago
$begingroup$
The right side is (x(100+x))^2.
$endgroup$
– shewlong
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
$begingroup$
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
$begingroup$
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
3 hours ago
$begingroup$
The right side is (x(100+x))^2.
$endgroup$
– shewlong
3 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
4 hours ago
1
$begingroup$
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