The first thing I thought of when I saw this was that 13 is a composite number. It is (2+3i)(2-3i), and along with that, that 13 is the sum of two squares, 2^2+3^2. So I tried 2 and 3 in the equation and I get 1/2+1/3 = 5/6 in both cases, so 2 and 3 are roots. This is familiar because I know to get 5/6 cup for a recipe, I add a half cup and a third cup to get the 5/6 cup. Of course this does not give the answers involving sqrt(481) which require more work.
Very interesting problem, not very straight forward. When you just solve as usual (as you would want) you get a polynomial of high degree. I solved this using a substitution x=sqrt(13)*sin(a). Using this method I got the equation 1/(sin(a))+1/(cos(a))=5sqrt(13)/6. Then I cross multipied, squared both sides and used another substitution sin(a)*cos(a)=y, getting a quadratic equation with two solutions (two cases): 6/13 and -6/25. But then something weird happenned when I back substituted and I eventually got 8 solutions, but only 2 of them were valid: sin(a)=3/sqrt(13) and sin(a)=2/sqrt(13). So finally x=2 or x=3. Beautiful, easy solutions to this equation. Maybe I overpowered a little, but I didn's see a way to solve a polynomial of a high degree in the beginning.
A seemingly “innocent” equation with an “elegant” solution. I enjoyed your method very much. Good job! I solved it by other two methods: using a trigonometric substitution as well as converting the equation into a quartic equation. Got the same results but used a lot more paper! Your method saves the earth too! :)
When I plug the positive version of the root481 solution back into the equation, I'm getting the left side summing negative 5/6. So I only find x = {2, 3, (-root(481)-13)/10} as working answers.
Fun fact: this equation has 3 real roots (and only 3). However, if you apply the fixed point numerical method: x = [5/6 - 1/√(13-x^2)] ^ -1, you can try with many initial values between (-3.50, +2.99) but it only converges to x=2. I think that's a nice chance to discuss about the importance of Algebra and Calculus and don't trust blindly in Numerical Methods (they are too helpful but you need to be cautious).
I used a more classical approach and simplified the left side of the equation and got: (5x-6)SQRT(13-x^2) = 6X Squaring both sides I got the fourth grade equation: 25x^4-60x^3-253x^2+780x-468=0 Using the old Rational Zeros Method I got solutions x=2 and x=3. Dividing the 4th grade equation by (x-2)(x-3) I got the trinomial: 25x^2 + 65x - 78 = 0 from which I got two more real solutions: x = {-13 +/- SQRT(481) } / 10 The four solutions of: 25x^4-60x^3-253x^2+780x-468=0 are: x = 2 x = 3 x = 0.8931712199 x = -3.4931712199 I then plugged the values into the original equation and found that 3 answers are valid: Solution set = {2 ,3 ,-3.4931712199}. Nice problem. Have an excellent year 2022.
Notice two things: 6 is an oblong number which means it can be written as x ( x + 1 ). and 13 can be written as the sum of the squares of two consecutive numbers. SO as a result all we need to solve is sqrt ( 13 - x^2 ) = x + 1 13 - x^2 = ( x + 1 )^2 13 = x^2 + ( x + 1 )^2 13 = 2x^2 + 2x + 1 0 = 2x^2 + 2x - 12 0 = x^2 + x - 6 0 = ( x + 3 )( x - 2 ) x = 2 or x = - 3 since the square root gives a positive answer, the only valid solution is x = 2. similarly, we can solve sqrt ( 13 - x^2 ) = x - 1 and get x = 3.
tfw you express the the inverse root with its equivalent by isolation in accordance with the original equation, to then use it in the squared equation, to then end up multiplying for equal denominators, to then cross multiply with the right hand side, reorder it and equate it to zero, to then end up with a barely solvable quartic function, to then comfort yourself with Wolfram providing the same solutions for your quartic as shown, not having made a mistake in your futile effort
((5/6)-(1/x))^2=(1/√(13-x^2)) : : I got equation 25x^4-60x^3-253x^2+780x-468=0 at the original form, you will soon notice it is OK when x=2,3 so this equation can be devided by (x-2)(x-3) then you get factrization (x-2)(x-3)(25x^2+65x-78)=0
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as 13 - x^2 must e positive let's try some integers to remove the radical. first, x=2 and 1/root(13-x^2) = 1/3 = 2/6 then, 1/x = 1/2 so we got : 3/6 + 2/6 = 5/6. I do not find any other value for that could help remove the radical.
@@SyberMath Ok this one even more so, why WHY are you substituting a trig function when this doesnt look at all like any trig formula or function? I hope you can PLEASE respond especially this time.. thanks very much.I think Ramanujan too loke me would've been flummoxed by thus...
@@leif1075 You're underestimating Ramanujan! 😁 This is a technique often used in Calculus as well as algebra. If you have sqrt(a^2-x^2) in the expression, a usually being an integer, then replacing x with asin(alpha) turns the problem into a more manageable one in most cases.
@@SyberMath I never saw it in calculus..but if you didnt knownit beforehand would you or could you have deduced it? I wouldnt so i doubt Ramanujan wouldve either..
Interesting problem. With a technique you have used in other videos, the problem shortens. This solution avoids denesting a radical in going from x² to x. 1/x + 1/√(13-x²) = 5/6 y = √(13-x²) // y > 0 x²+y² = 13 1/x + 1/y = 5/6 (x+y)/(xy) = 5/6 x+y = 5t xy = 6t (x+y)² - 2xy = x²+y² = 13 25t²-12t-13 = 0 // quadratic formula t = (6±19)/25 t = 1,-13/25 t = 1, x+y = 5, xy = 6 (x,y) = (2,3) (3,2) t = -13/25 find roots x & y = u // quadratic formula u² -(x+y)u + xy = 0 u² - (5t)u + 6t = 0 u = (5t±√(25t²-24t))/2 // y = + root x = -(13+√481)/10
I solved in a slightly different manner, first substitute y = sqrt(13-x^2) to get system: 1/x +1/y = 5/6 and x^2 + y^2 = 13. Then another substitution: a = x+y, b = xy to get the system 6a = 5b and a^2 - 2b = 13, this is quadratic so easily solvable (two solutions: a = 5 b = 6 and a = -2.6 b = -3.12) and from there you can get x directly (again quadratic so finally 4 solutions: 2, 3, -1.3+-sqrt(4.81)). Of course you are mistaken speaking that radical is always non-negative, radical may be negative since -1^2 = 1. It is just by agreement we pretend radical is always positive but it is incorrect. So the solution x = -1.3 + sqrt(4.81) is also valid, you just take negative radical of sqrt(13-x^2) :)
@@SyberMath thank you. I want to make content like this, but with my language. I'm math teacher from Indonesia. Your content is so interesting. I get many idea from this
Очень сложно! Обозначим sqrt(13-x^2)=y>0 x^2+y^2=13 1/x+1/y=5/6 (x+y)/xy=5/6. В квадрат 6(x^2+y^2)+72xy=25 (xy)^2 25(xy)^2-72 xy-468=0 Дальше очевидно.
There are three solution . The "other" solution is solution to (1/x)+1/sqrt(13-x^2)=-5/6. Because lim_{x->0^-}( x+1/sqrt(13-x^2)=-infinity. And lim_{x->-(√13)^+}(x+1/sqrt(13-x^2)=+infinity. And lim
Elkin Campos Actually, the “other” solution x=(sqrt(481)-13)/10 is a solution to (1/x)-1/sqrt(13-x^2)=5/6. If you plug in this number in the left side of the original equation you’ll get 5*sqrt(481)/78 ~ 1.4 which is clearly not -5/6. Check your results with WolframAlpha or something before posting a video or a comment.
Yes! The only solutions that verify the original equation are: x=2; x=3 (both trivial); x= - √[(13/50)*(25+√481] ~ -3.49317. The conjugates are fake solutions.
It's fascinating how there is no apparent way to predict the number of real roots in radical equations. Of 8 possible roots, only 3 are true and the original eqn have only a sqrt
The first thing I thought of when I saw this was that 13 is a composite number. It is (2+3i)(2-3i), and along with that, that 13 is the sum of two squares, 2^2+3^2. So I tried 2 and 3 in the equation and I get 1/2+1/3 = 5/6 in both cases, so 2 and 3 are roots. This is familiar because I know to get 5/6 cup for a recipe, I add a half cup and a third cup to get the 5/6 cup. Of course this does not give the answers involving sqrt(481) which require more work.
Very good!
Very interesting problem, not very straight forward. When you just solve as usual (as you would want) you get a polynomial of high degree. I solved this using a substitution x=sqrt(13)*sin(a). Using this method I got the equation 1/(sin(a))+1/(cos(a))=5sqrt(13)/6. Then I cross multipied, squared both sides and used another substitution sin(a)*cos(a)=y, getting a quadratic equation with two solutions (two cases): 6/13 and -6/25. But then something weird happenned when I back substituted and I eventually got 8 solutions, but only 2 of them were valid: sin(a)=3/sqrt(13) and sin(a)=2/sqrt(13). So finally x=2 or x=3. Beautiful, easy solutions to this equation. Maybe I overpowered a little, but I didn's see a way to solve a polynomial of a high degree in the beginning.
A nice way of solving it!
A seemingly “innocent” equation with an “elegant” solution. I enjoyed your method very much. Good job!
I solved it by other two methods: using a trigonometric substitution as well as converting the equation into a quartic equation. Got the same results but used a lot more paper!
Your method saves the earth too! :)
Glad it helped! 😁
When I plug the positive version of the root481 solution back into the equation, I'm getting the left side summing negative 5/6. So I only find x = {2, 3, (-root(481)-13)/10} as working answers.
That's right! There should be three solutions.
That comes from one of them being extraneous due to squaring the equation.
Fun fact: this equation has 3 real roots (and only 3). However, if you apply the fixed point numerical method: x = [5/6 - 1/√(13-x^2)] ^ -1, you can try with many initial values between (-3.50, +2.99) but it only converges to x=2. I think that's a nice chance to discuss about the importance of Algebra and Calculus and don't trust blindly in Numerical Methods (they are too helpful but you need to be cautious).
How does the substitution with x = sqrt(13) * sin alpha work in the end
to derive the three solutions ?
I used a more classical approach and simplified the left side of the equation and got:
(5x-6)SQRT(13-x^2) = 6X
Squaring both sides I got the fourth grade equation:
25x^4-60x^3-253x^2+780x-468=0
Using the old Rational Zeros Method I got solutions x=2 and x=3. Dividing the 4th grade equation by (x-2)(x-3) I got the trinomial:
25x^2 + 65x - 78 = 0
from which I got two more real solutions:
x = {-13 +/- SQRT(481) } / 10
The four solutions of: 25x^4-60x^3-253x^2+780x-468=0
are:
x = 2
x = 3
x = 0.8931712199
x = -3.4931712199
I then plugged the values into the original equation and found that 3 answers are valid:
Solution set = {2 ,3 ,-3.4931712199}.
Nice problem. Have an excellent year 2022.
Notice two things: 6 is an oblong number which means it can be written as x ( x + 1 ). and 13 can be written as the sum of the squares of two consecutive numbers.
SO as a result all we need to solve is sqrt ( 13 - x^2 ) = x + 1
13 - x^2 = ( x + 1 )^2
13 = x^2 + ( x + 1 )^2
13 = 2x^2 + 2x + 1
0 = 2x^2 + 2x - 12
0 = x^2 + x - 6
0 = ( x + 3 )( x - 2 )
x = 2 or x = - 3
since the square root gives a positive answer, the only valid solution is x = 2.
similarly, we can solve sqrt ( 13 - x^2 ) = x - 1 and get x = 3.
"why? Because, you see 😂" you always say that and I always love
😁
@@SyberMath 🤣
tfw you express the the inverse root with its equivalent by isolation in accordance with the original equation, to then use it in the squared equation, to then end up multiplying for equal denominators, to then cross multiply with the right hand side, reorder it and equate it to zero, to then end up with a barely solvable quartic function, to then comfort yourself with Wolfram providing the same solutions for your quartic as shown, not having made a mistake in your futile effort
😁
Absolutely magnificent📐
This can be written as
1/x + 1/√(13-x*x) =1/2 +1 /3
Here x = 2, 3 are the solutions
((5/6)-(1/x))^2=(1/√(13-x^2))
:
:
I got equation
25x^4-60x^3-253x^2+780x-468=0
at the original form, you will soon
notice it is OK when x=2,3
so this equation can be devided
by (x-2)(x-3) then you get
factrization
(x-2)(x-3)(25x^2+65x-78)=0
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It says this product is not available for purchasing when i open the link
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15 seconds got some solutions but now to do it systematically to look for more obscure ones
Nice!
as 13 - x^2 must e positive let's try some integers to remove the radical.
first, x=2 and 1/root(13-x^2) = 1/3 = 2/6 then, 1/x = 1/2 so we got :
3/6 + 2/6 = 5/6. I do not find any other value for that could help remove the radical.
oup! I missed the last value that let remove the radical: x=3, obvious
Nice method. So, in my trigonometric solution I just excluded one of the solutions foo fast and that's why I had only 2 of them instead of 3.
Nice work!
@@SyberMath Ok this one even more so, why WHY are you substituting a trig function when this doesnt look at all like any trig formula or function? I hope you can PLEASE respond especially this time.. thanks very much.I think Ramanujan too loke me would've been flummoxed by thus...
@@leif1075 You're underestimating Ramanujan! 😁
This is a technique often used in Calculus as well as algebra.
If you have sqrt(a^2-x^2) in the expression, a usually being an integer, then replacing x with asin(alpha) turns the problem into a more manageable one in most cases.
@@SyberMath I never saw it in calculus..but if you didnt knownit beforehand would you or could you have deduced it? I wouldnt so i doubt Ramanujan wouldve either..
@@SyberMath Hope you can respond again when you can to my follow up here
If you look at rhs..
Its 5/6. Now 6 is lcm of 2 and 3. I got the solution as 2 & 3 within 10 sec.
Interesting problem. With a technique you have used in other videos, the problem shortens. This solution avoids denesting a radical in going from x² to x.
1/x + 1/√(13-x²) = 5/6
y = √(13-x²)
// y > 0
x²+y² = 13
1/x + 1/y = 5/6
(x+y)/(xy) = 5/6
x+y = 5t
xy = 6t
(x+y)² - 2xy = x²+y² = 13
25t²-12t-13 = 0
// quadratic formula
t = (6±19)/25
t = 1,-13/25
t = 1, x+y = 5, xy = 6
(x,y) = (2,3) (3,2)
t = -13/25
find roots x & y = u
// quadratic formula
u² -(x+y)u + xy = 0
u² - (5t)u + 6t = 0
u = (5t±√(25t²-24t))/2
// y = + root
x = -(13+√481)/10
Nice!
I solved in a slightly different manner, first substitute y = sqrt(13-x^2) to get system: 1/x +1/y = 5/6 and x^2 + y^2 = 13. Then another substitution: a = x+y, b = xy to get the system 6a = 5b and a^2 - 2b = 13, this is quadratic so easily solvable (two solutions: a = 5 b = 6 and a = -2.6 b = -3.12) and from there you can get x directly (again quadratic so finally 4 solutions: 2, 3, -1.3+-sqrt(4.81)). Of course you are mistaken speaking that radical is always non-negative, radical may be negative since -1^2 = 1. It is just by agreement we pretend radical is always positive but it is incorrect. So the solution x = -1.3 + sqrt(4.81) is also valid, you just take negative radical of sqrt(13-x^2) :)
Nice!
Brilliant!!!!
I hope I can make amazing math content like this
Sure you can!
@@SyberMath thank you.
I want to make content like this, but with my language.
I'm math teacher from Indonesia.
Your content is so interesting.
I get many idea from this
Super sir
I like you
You r a great math teacher
Thank you! 🥰
Wonderful.
Many thanks!
Nice problem!!
Thank you! Cheers!
-13squire root
three solutions from the computer:
*Solve[1/x + 1/Sqrt[13 - x^2] == 5/6]*
Good Job, computer! 😁
I solved it in a different way the fraction 1/x can be written as √1/√x^2 and then it's up to u guys 😊😊😊
It's not the end of us story... 😂
1/ x+{1/√(13- x^ 2)}=5/6
=(1/2) +(1/3)
1/ x=1/2
1 / x=1/3
x=(2,3) satisfies the above equation.
Is that the only solution?
@@SyberMath u suppose addition of two irrational number equal to rational number?
@@-basicmaths862 it could
Please avoid bad visibility
X=3
also, x=2
Is that it?
Syber maths I solved this using pure algebra
(-sqrt(481)+13) / 10 is not x
5/6=1/2+1/3
That's right!
Очень сложно!
Обозначим sqrt(13-x^2)=y>0
x^2+y^2=13
1/x+1/y=5/6
(x+y)/xy=5/6. В квадрат
6(x^2+y^2)+72xy=25 (xy)^2
25(xy)^2-72 xy-468=0
Дальше очевидно.
Nice!
Normal people use the second method!
X=2,3
Are those all the solutions?
did I just do this in my head in 10 seconds 😷
Nvm didn’t get last solution
Np
Sybma
I think so
What do you mean?
I mean it is easy
@@Biblapghosh Thanks!
Of course x=2
Is that the only solution?
@@SyberMath and, of course, not the only solution.
But it is easy and interesting problem
Yes! I like radicals.
There are three solution . The "other" solution is solution to (1/x)+1/sqrt(13-x^2)=-5/6. Because lim_{x->0^-}( x+1/sqrt(13-x^2)=-infinity. And lim_{x->-(√13)^+}(x+1/sqrt(13-x^2)=+infinity.
And lim
Elkin Campos Actually, the “other” solution x=(sqrt(481)-13)/10 is a solution to (1/x)-1/sqrt(13-x^2)=5/6. If you plug in this number in the left side of the original equation you’ll get 5*sqrt(481)/78 ~ 1.4 which is clearly not -5/6.
Check your results with WolframAlpha or something before posting a video or a comment.
@@V1tal1t1 uh, but is -√481+-13 not √481+-13 "9:18" .
@@V1tal1t1
Define f(x)= 1/x+1/(sqrt(13-x^2) then f'(x)=-1/x^2-(1/2)*(√13-x^2)^3*(-2*x)= -1/x^2+x/(√13-x^2)^3
Elkin Campos That is correct, and this solution is -(13+sqrt(481))/10
Take it easy!
It
?
י
I solved quartic , one of the solutions was false
So there was an extraneous solution?
Yes! The only solutions that verify the original equation are: x=2; x=3 (both trivial); x= - √[(13/50)*(25+√481] ~ -3.49317. The conjugates are fake solutions.
It's fascinating how there is no apparent way to predict the number of real roots in radical equations. Of 8 possible roots, only 3 are true and the original eqn have only a sqrt