(4+Sqrt15)^x+(4-Sqrt15)^x=62 Due to (4-Sqrt15)*(4+Sqrt15)=1 or 4-Sqrt15=1/(4+Sqrt15), (4+Sqrt15)^x+[1/(4+Sqrt15)]^x=62 (4+Sqrt15)^x+[(4+Sqrt15)^(-1)]^x=62 (4+Sqrt15)^x+(4+Sqrt15)^(-x)=62 After using y=(4+Sqrt15)^x, so y+1/y=62 (y^2+1)/y=62 y^2+1=62y y^2-62y+1=0 Hence, y1=31-8Sqrt15 and y2=31+8Sqrt15 are solutions. 1) For y=31-8Sqrt15, (4+Sqrt15)^x=31-8Sqrt15, (4+Sqrt15)^x=(4-Sqrt15)^2 (4+Sqrt15)^x=[(4+Sqrt15)^(-1)]^2 (4+Sqrt15)^x=(4+Sqrt15)^(-2) x1=-2 is solution. 2) For y=31+8Sqrt15, (4+Sqrt15)^x=31+8Sqrt15 (4+Sqrt15)^x=(4+Sqrt15)^2 x2=2 is solution.
Let a=4-√15 as a constant Then a^x+1/a^x=62 Consider the following transformation (a^(x/2)+1/a^(x/2))^2=a^x+1/a^x+2=62+2=64 so a^(x/2)+1/a^(x/2)=√64=8 Let u=a^(x/2) u+1/u=8 u²+1=8u u=(8±√(8²-4))/2=(8±√60)/2=4±√15 (4+√15)^(x/2)=4±√15 x/2=±1 The answer: x=±2
I solved in 2 mints. I put those terms as roots of a quadratic equation. Then using the quadratic formula, formed the eqn, and luckily it worked on square.
That started pretty complicated. You can do it quickly by approximation. root 15 = bit less than root 16, = bit less than 4. so you got (bit less than 8) squared + (bit more than 0 squared) = 62. so x can only be 2, anything else is way too big, or way too small. i.e. 8^2 = 64 and 7^2 = 49. plus you're squaring so it can be +2 or -2.
This is not the point of the problem, the point is to learn how to reach the answer rigorously and mathematically. Also, how can you prove those are the only solutions? You can’t, which is why, in any real setting, you can’t approximate your way out of there
@@17-harshitbhatt57 You’re pretty right. I was also criticising the long answers before, and the reason was that I didn’t know how to get to the answer systematically and proving it. Better I try to better myself and talk later. You are right here.
It's easier to work the determinant of the quadratic equation in p as sqrt[(-62)²-4]=sqrt[(62)²-2²] =sqrt(64×60) =8×2sqrt(15) instead of squaring 62 then substracting with 4, having a sqrt of a large number. =(62+2)(62-2) =64(60) =8²2²15
One can quickly see that all the cross terms (terms containing V15) will cancel, so one can see only the squared terms count. These squared terms are the same for each of the squared expressions, so one can take either one and half the 62 to get 31. 16 (4 squared) plus 15 (V15 squared) is 31, so x is 2. QED
Inspection the number types of the equation reveals (a) 4 and 62 are rational numbers, (b) root(15) is irrational, (c) If x is non-integer, root(15)^x is irrational, (d) if root(15) is raised to an odd integer, the result is irrational, Showing x is an even integer. Inspecting the numerical values of the equation (a-b)^2 = a^2 -2ab + b^2, ignoring b^2 as a first approximation, Root(15) is (4 - delta)^2 giving approximately 16 - 8 delta = 15, making delta = 1/8 with an estimated error of 1/512. ((1/8)^2 /8) Substituting the approximation (4 - 1/8) for root(15) gives for the 1st bracket (approximately 8 - 1/8) and for the second bracket (approximately 1/8). Investigating the numbers 1. Assume the 1st bracket is dominant, approximately 8. Applying even integer powers of x to 8 gives 8^2 = 64, 8^4 = 64^4. 2. Assume the 2nd bracket is dominant, approximately 1/8. Applying even integer powers of x to 1/8 gives (1/8)^-2 = 64, (1/8)^-4 = 64^4 (Dividing the log(left hand side)/log(right hand side), and take the nearest even integer as x for the general case.) In this instance x = 2 or -2. Substituting x =2 into the equation gives (a+b)^2 + (a-b)^2 = 2(a^2+b^2) substitute a = 4, b = root(15). 2(16 + 15) = 62.
@@SpencersAcademy At the Olympiad level, it is beatable, 1. No proof (irrational number)^(irrational number) is irrational. 2. No proof of why the approximation for root(15) gives the same numerical result as root(15) when delta^2 is ignored For example 2((4)^2 + (4 - 1/8)^2 )= 2(16 + (16 -1 +1/64), which gives 62 - 1/64, ignoring delta^2, is 62. 3. No back substitution of x = -2.
let p = 4+sqrt（15) ,q = 4- sqrt（15） then notice that p+q = 8，p-q=2sqrt（15），pq=1 then notice that p^2+q^2={(p+q)^2+(p-q)^2}/2=62 so x=2 because pq =1 so x=-2 also satisfies the equation at the same time p=1/q f(x)=p^x+q^x=p^x+1/(p^x) when x>0 df(x)/dx >0 when x
the solution for p has the sqrt of (62-2)*(62+2) =60*64=15*2^2*8^2 since there is a difference of squares under the sqrt....so there is no need for your leap of faith in placing a 15 under the sqrt, it comes by itself
1) Expand both binomial in Newton bonimal 2) Note that most of them cancel each other OBS: Consider x natural and test for x = {0,1, 2, 3, 4, ...} At least i tried 🙂
That was more than wonderful, and the explanation and interpretation are very excellent. Allah ❤bless your beautiful thinking. What a beauty in algebra and mathematics. Thank you, dear professor.
f(x)=(4 + sqrt(15))^x + (4 - sqrt(15))^x is symmetrical around x=0. This is easy enough to show that f(x)=f(-x). Also, f(x) is continuous for all x. Therefore, graphs of f(x) and y=62 interest twice. An obvious solution is x=2 f(2)=62, then the second solution for the symmetrical function f(x) around x=0 is x=-2. Solutions: x=-2, x=2.
Your reply is awesome but I do have some questions: How did you know the graph is symmetrical around x=0? Also, how did you figure that the obvious solution of x = 2, is 62? Hope you don’t mind replying. Thanks in advance.
@@boredomgotmehere This is a known result to me that a function of the form f(x)= (a+sqrt(b))^x +(a-sqrt(b))^x is even. It can be verified that f(x)=f(-x) in two lines. So f(x) is symmetrical. x=2 being a solution is also obvious to me by looking at the equation. Also, in my past experience, such school exercises as this one, often have one simple solution. If x=2, and x=-2 for a symmetrical function, then the axis of symmetry is x=0.
I multiplied everything by (4+sqrt(15)) and lucked out on getting 1^x then getting your quadratic equation and instead of doing your guessing process I did (4+sqrt15)^x=31- sqrt(960) I treated it as a logarithmic equation
Root 15 is closed तो root16 ie 4(लिटिल less than4) So let put value 4 इन place of 4 (4+4)^ x +0 = 62 For x= 2, lhs become 64 which is close to 62. Now let us check the given eqn with x = 2, (4 + _ /15)^2 + ( 4 - _/15)^2 = 2( a^2 + b^2) = 2( 16+ 15) = 62 =rhs So x=2 is the ans
f(x) = (4+sqrt(15))^x + (4 -sqrt(15))^x As 4-sqrt(15) = 1/(4+sqrt(15) , we have f(x) = a^x + a^-x, with a = 4 +sqrt(15), so f(x) = f (-x) for any real x So, if x is solution of f(x) = 62, then -x is also solution. The function f strictly increases on R+, so if the equation f(x) = 62 has a solution on R+, then it is unique. Now, as x =2 is evident solution, it is the unique solution on R+, and thanks to the initial remark: -2 is the unique solution on R-. Conclusion: x = 2 or x = -2 are the only solutions.
@@noname-ed2un An elementary algebraic identity is: a^2 - b^2 = (a - b) (a + b) Substitute: a = 62 , b = 2 and you'll get: 62^2 - 2^2 = (62 - 2) (62 + 2) = (60) (64) Have a nice day !
I started from 62=2×(4^2+15) which means x=2 is a solution. Then I reformed the problem like p^x+p^(-x)=62, so I found -2 is also another solution. Considering the shape of cosh, these are the only solutions in real number. I have no idea in complex number :|
p+1/p=62 is obvious after the first step since (4+sqrt(15))(4-sqrt(15))=1 or (4+sqrt(15)) = 1/ (4+sqrt(15)) so the quadratic is immediate in 2 steps Far too much beating about the bush, way too many redundant steps. The other thing that is obvious is that 1< max(x)
(4+Sqrt15)^x+(4-Sqrt15)^x=62
Due to (4-Sqrt15)*(4+Sqrt15)=1 or 4-Sqrt15=1/(4+Sqrt15),
(4+Sqrt15)^x+[1/(4+Sqrt15)]^x=62
(4+Sqrt15)^x+[(4+Sqrt15)^(-1)]^x=62
(4+Sqrt15)^x+(4+Sqrt15)^(-x)=62
After using y=(4+Sqrt15)^x, so
y+1/y=62
(y^2+1)/y=62
y^2+1=62y
y^2-62y+1=0
Hence, y1=31-8Sqrt15 and y2=31+8Sqrt15 are solutions.
1) For y=31-8Sqrt15,
(4+Sqrt15)^x=31-8Sqrt15,
(4+Sqrt15)^x=(4-Sqrt15)^2
(4+Sqrt15)^x=[(4+Sqrt15)^(-1)]^2
(4+Sqrt15)^x=(4+Sqrt15)^(-2)
x1=-2 is solution.
2) For y=31+8Sqrt15,
(4+Sqrt15)^x=31+8Sqrt15
(4+Sqrt15)^x=(4+Sqrt15)^2
x2=2 is solution.
Verrrry verrrty good tank you🙏🙏🙏
Nicely you explained the problem.❤❤
Very smart teacher...
@@haroldosantiago819 Pattern is
a + b = c
b=1/a
a+1/a=c
a^2+1=ac...
Nothing special.
Let a=4-√15 as a constant
Then
a^x+1/a^x=62
Consider the following transformation
(a^(x/2)+1/a^(x/2))^2=a^x+1/a^x+2=62+2=64
so
a^(x/2)+1/a^(x/2)=√64=8
Let u=a^(x/2)
u+1/u=8
u²+1=8u
u=(8±√(8²-4))/2=(8±√60)/2=4±√15
(4+√15)^(x/2)=4±√15
x/2=±1
The answer:
x=±2
That's a fantastic solution you have here.
I like the way you teach 🎉🎉
I am glad to hear that.
Easier? Define Z(n) = (x^n)+ 1/(x^n), Z(0)=2, then Z(n+1)= Z(n)Z(1) - Z(n-1). Here Z(1) = 8. Z(2) = Z(1)*Z(1)-Z(0) = 62. Hence n=2.
Not really. For we have two answers: n = 2; n = -2
You specifically are too great to praise. Really sir
Thank you very much. I am grateful.
Excellent. Thanks for this. Keep them coming.
I'm glad you enjoyed it.
I solved in 2 mints.
I put those terms as roots of a quadratic equation. Then using the quadratic formula, formed the eqn, and luckily it worked on square.
Really nice trick, indeed; thanks!
I'm glad you enjoyed it 😊
I admit to lucking my way into a solution. Add 2 to both sides and take the square root and it becomes clear that x/2=1 works.
Thanks. You explained very nicely.
I'm glad it was helpful!
well done sir you make this problem so very easy
I'm glad to hear that.
Good work my fellow Mathematician ❤ keep up with the content I have subscribed
Thank you! 😊. You're welcome aboard.
@SpencersAcademy happy for you
Very complicated but extraordinary explanation. Thanks brother 🎉
I'm glad you enjoyed it.
I've one UA-cam channel, brother
youtube.com/@venkatesanmathsacademy8904
Legal, muito bem explicado!
Thanks. I am glad you enjoyed it.
I look forward for more videos like this. Great job on the solution and explanation
Thanks, more to come!
Awesome!
By hit and trial put x=2 it will be (a+b)^2 + (a-b)^2 which will be 2(a^2+b^2) which is 2*(4^2+√15^ 2)= 2*(16+15)= 2*31= 62 which statisfies so x=2
You missed the possibility that x=-2.
That was Smooth
Thanks.
Nicely explained
Thank you so much 🙂
It will become super easy if we use "logarithmic with base 2" both side but before that we need to add 2 both side
That started pretty complicated. You can do it quickly by approximation. root 15 = bit less than root 16, = bit less than 4. so you got (bit less than 8) squared + (bit more than 0 squared) = 62. so x can only be 2, anything else is way too big, or way too small. i.e. 8^2 = 64 and 7^2 = 49. plus you're squaring so it can be +2 or -2.
Its not about the answer its about the process and out of the box thinking by this way you're killing the question itself.
This is not the point of the problem, the point is to learn how to reach the answer rigorously and mathematically. Also, how can you prove those are the only solutions? You can’t, which is why, in any real setting, you can’t approximate your way out of there
@@17-harshitbhatt57
You’re pretty right. I was also criticising the long answers before, and the reason was that I didn’t know how to get to the answer systematically and proving it. Better I try to better myself and talk later. You are right here.
😊
@@Maran108???
amazing!
Thank you! Cheers!
Excellent ❤
Thanks 😊
1 step: because 4+sqrt{15} near 8 and 4-sqrt{15}
Thanks
You're welcome 😊
Nice work.
Thank you! Cheers!
It's easier to work the determinant of the quadratic equation in p as
sqrt[(-62)²-4]=sqrt[(62)²-2²]
=sqrt(64×60)
=8×2sqrt(15)
instead of squaring 62 then substracting with 4, having a sqrt of a large number.
=(62+2)(62-2)
=64(60)
=8²2²15
I solve it in less than 5 minutes, found the same trick used. Thank you.
Great job...
Thanks 😊
Great 👍 job
I am grateful, man. Glad you enjoyed it.
Solve the equation
(4+√15)^x + (4-√15)^x = 62
Note that (4+√15) and (4-√15) are both positive, hence for real values of x , (4+√15)^x and (4-√15)^x exist and are also positive.
Note also that (4+√15)(4-√15) = 4² - (√15)² = 16 - 15 = 1 , hence (4+√15) = 1/(4-√15) .
Multiply both sides of the equation by (4-√15)^x :
[(4+√15)^x] * (4-√15)^x + [(4-√15)^x] * (4-√15)^x = 62 * (4-√15)^x
[(4+√15)(4-√15)]^x + [(4-√15)^x]² = 62 * (4-√15)^x
[1]^x + [(4-√15)^x]² = 62 * (4-√15)^x
... substitute u = (4-√15)^x ...
... note: 1^x = 1 for any real x ...
1 + u² = 62*u
1 - 62u + u² = 0
1 - 2*31u + u² = 0
961 - 2*31u + u² = 960
31² - 2*31u + u² = 64*15
(31-u)² = 64*15
31-u = ±8√15
u = 31 ± 8√15
... remember u = (4-√15)^x ...
(4-√15)^x = 31 ± 8√15
x = ln(31 ± 8√15) / ln(4 - √15)
Well, actually we can do better. Note that
31 ± 8√15 =
= 16 + 15 ± 2*4√15
= (√16)² + (√15)² ± 2*(√16)*(√15)
= (√16 ± √15)²
= (4 ± √15)²
Hence,
(4 - √15)^x = 31 - 8√15
(4 - √15)^x = (4 - √15)²
x = 2
and
(4 - √15)^x = 31 + 8√15
(4 - √15)^x = (4 + √15)²
... remember (4+√15) = 1/(4-√15) ...
(4 - √15)^x = 1/(4-√15)²
(4 - √15)^x = (4-√15)⁻²
x = -2
So there are two solutions: x = 2 and x = -2 .
Fantastic
A=(4+15^1/2 )^x , B =(4- 15^1/2 )^x. Then AB =1, A+B=62. Therefore A, B are two solutions of x^2 -62x +1=0.
.. And so what? We are looking for solutions for x?!
@@ft7339 x= 31 + - sqrt(31^2 - 1)
Respected Sir, Good evening
Same to you
Wonderful explanation.
Or, for integer x, every 2nd term of binomial expansion is cancelled out. Now try x=1 LHS=8. Now try x=2 LHS=16+15+16+15=62.
One can quickly see that all the cross terms (terms containing V15) will cancel, so one can see only the squared terms count. These squared terms are the same for each of the squared expressions, so one can take either one and half the 62 to get 31. 16 (4 squared) plus 15 (V15 squared) is 31, so x is 2. QED
Nice one 👍
2nd term is reciprocal of 1st term. Let A=1st term. So A times eqn is AA+1=62A. So, we can use the quadratic formula to find the solutions.
Excelente. Bonito ejercicio....
Thanks bro. I'm glad you enjoyed it.
Very intelligent
Wonderful ❤
Thank you 😄
Great job
Keep it up
I would like to see trigonometry and calculus olympiad
Thanks.
AWESOME
Thank you! Cheers!
Entertaining!
Thanks 😊
Both answers worked; and, it seems to me that (a-b)^x = (a+b)^(-x); but I really need to explore that conjecture more, to verify it 👍🏻.
That's true only when a²-b² = 1 (which in this particular case, it is), so that (a-b)(a+b) = 1 .
x=2,-2 did it in mind
Amazing
I'm glad you enjoyed it.
Inspection the number types of the equation reveals
(a) 4 and 62 are rational numbers, (b) root(15) is irrational,
(c) If x is non-integer, root(15)^x is irrational,
(d) if root(15) is raised to an odd integer, the result is irrational,
Showing x is an even integer.
Inspecting the numerical values of the equation
(a-b)^2 = a^2 -2ab + b^2, ignoring b^2 as a first approximation,
Root(15) is (4 - delta)^2 giving approximately 16 - 8 delta = 15,
making delta = 1/8 with an estimated error of 1/512. ((1/8)^2 /8)
Substituting the approximation (4 - 1/8) for root(15) gives
for the 1st bracket (approximately 8 - 1/8) and for the second bracket (approximately 1/8).
Investigating the numbers
1. Assume the 1st bracket is dominant, approximately 8.
Applying even integer powers of x to 8 gives 8^2 = 64, 8^4 = 64^4.
2. Assume the 2nd bracket is dominant, approximately 1/8.
Applying even integer powers of x to 1/8 gives (1/8)^-2 = 64, (1/8)^-4 = 64^4
(Dividing the log(left hand side)/log(right hand side), and take the nearest even integer as x for the general case.)
In this instance x = 2 or -2.
Substituting x =2 into the equation gives
(a+b)^2 + (a-b)^2 = 2(a^2+b^2)
substitute a = 4, b = root(15).
2(16 + 15) = 62.
This is very impressive, I'm not gonna lie. You nailed it. 👏
@@SpencersAcademy
At the Olympiad level, it is beatable,
1. No proof (irrational number)^(irrational number) is irrational.
2. No proof of why the approximation for root(15) gives the same numerical result as root(15) when delta^2 is ignored
For example
2((4)^2 + (4 - 1/8)^2 )= 2(16 + (16 -1 +1/64), which gives 62 - 1/64, ignoring delta^2, is 62.
3. No back substitution of x = -2.
@@SpencersAcademy
I would love to learn what the answers were to this questions in the Olympiad.
With thanks for your appreciation.
Thinks
You're welcome 😊
Glad you enjoyed it
let p = 4+sqrt（15) ,q = 4- sqrt（15）
then notice that p+q = 8，p-q=2sqrt（15），pq=1
then notice that p^2+q^2={(p+q)^2+(p-q)^2}/2=62
so x=2
because pq =1 so x=-2 also satisfies the equation
at the same time p=1/q
f(x)=p^x+q^x=p^x+1/(p^x)
when x>0 df(x)/dx >0
when x
Excellent. You nailed it. 👍
the solution for p has the sqrt of (62-2)*(62+2) =60*64=15*2^2*8^2 since there is a difference of squares under the sqrt....so there is no need for your leap of faith in placing a 15 under the sqrt, it comes by itself
1) Expand both binomial in Newton bonimal
2) Note that most of them cancel each other
OBS: Consider x natural and test for x = {0,1, 2, 3, 4, ...} At least i tried 🙂
Yeah, you did.
That was more than wonderful, and the explanation and interpretation are very excellent. Allah ❤bless your beautiful thinking. What a beauty in algebra and mathematics. Thank you, dear professor.
I am really grateful for this. I'll continue to do my best, Sir.
👏👏👏
New commer sir
You're welcome here. 🥳
Let
a = (4 + √15) ^ (x/2)
b = (4 - √15) ^ (x/2)
Tnen
a² + b² = 62
Calc
(a + b)² = a² + b² + 2ab = 62 + 2 = 64 => (a+b) = 8 (because a, b > 0)
Has the system
a² + b² = 62
a + b = 8
Find (a-b)
(a-b)² = a² + b² - 2ab = 62 - 2 = 60
Therefore
(1) a + b = 8
(2) a - b = ±2√15
From (1) and (2)
2a = 8 ±2√15
or
a = 4 ± √15
or
(4 + √15) ^ (x/2) = 4 ± √15
case 1.
(4 + √15) ^ (x/2) = 4 + √15 => (x/2) = 1 => x = 2 is solution
case 2
(4 + √15) ^ (x/2) = 4 - √15
or
[ (4+ √15)(4-√15)/(4-√15)]^(x/2) = 4 - √15
or
[1/(4-√15)]^(x/2) = 4 - √15
or
(4-√15)^(-x/2) = 4-√15 => (-x/2) = 1 => x = -2 is solution
All solutions are
x = ±2
Excellent work!!!
Saw the same question in Nust entry test past papers, couldn't figure it out. Now I know it.
Awesome. I'm glad it helped.
Bruh. Why all these channels solve really easy questions and say them olympiad question. Please solve "real" olympiad questions.
Do you know any channel which actually solve olympiad level question
Dude do you want a slap? Shut up and keep stepping..my man the professor has this channel under control.
Polish AGH olympiad is really hard
Same method applies, as illustrated
I agree. Too easy, why not using a title such as solving easy exponential equations instead of words such as Olympiad
Now i understand the -2 solution
I am glad you did.😊
Brilliant!!
Thanks.
Multiplying both sides by (4 - sqrt(15))^x:
(4 + sqrt(15))^x * (4 - sqrt(15))^x + (4 - sqrt(15))^x * (4 - sqrt(15))^x = 62 * (4 - sqrt(15))^x.
((4 + sqrt(15)) * (4 - sqrt(15)))^x + ((4 - sqrt(15))^x)^2 = 62 * (4 - sqrt(15))^x.
1^x + ((4 - sqrt(15))^x)^2 = 62 * (4 - sqrt(15))^x.
1 + ((4 - sqrt(15))^x)^2 = 62 * (4 - sqrt(15))^x.
Let u = (4 - sqrt(15))^x.
1 + u^2 = 62 * u.
u = 31 + 8 sqrt(15) or u = 31 - 8 sqrt(15).
x = log_[4 - sqrt(15)](31 + 8 sqrt(15)) or x = log_[4 - sqrt(15)](31 - 8 sqrt(15)).
(4 - sqrt(15))^2 = 16 - 2 * 4 * sqrt(15) + (sqrt(15))^2 = 16 - 8 * sqrt(15) + 15 = 31 - 8 * sqrt(15).
So, in the latter case, x = log_[4 - sqrt(15)](31 - 8 sqrt(15)) = log_[4 - sqrt(15)]((4 - sqrt(15))^2) = 2.
In the former case, x = log_[4 - sqrt(15)](31 + 8 sqrt(15)) = log_[4 - sqrt(15)]((4 + sqrt(15))^2) = ln((4 + sqrt(15))^2) / ln(4 - sqrt(15)) = ln((4 + sqrt(15))^2) / ln((4 + sqrt(15))^(-1)) = 2 * ln(4 + sqrt(15)) / (-1 * ln(4 + sqrt(15))) = -2.
Excellent delivery
Thank you. @@SpencersAcademyWould it be ok for me to upload my own version?
Guessed in less than 1second 😎😎😎 x=2
You wont get marks for guessing. You have to solve it.
You missed the solution x=-2.
4+sqrt(15)^x
4+sqrt(16-1)^x
x=4
1/62(x^2-sqrt(2)sqrt(x^2-1)^2)^y=1
62=(4^2-(sqrt(2)sqrt(4^2-1))^2)^y
Close to 1.8
I like how didnt search any math but i learnt this today from myteacher and now it gets recommended to me
Yeah, man. UA-cam just kinda knows what you want and show it to you.
Yea lol
f(x)=(4 + sqrt(15))^x + (4 - sqrt(15))^x is symmetrical around x=0. This is easy enough to show that f(x)=f(-x). Also, f(x) is continuous for all x. Therefore, graphs of f(x) and y=62 interest twice. An obvious solution is x=2 f(2)=62, then the second solution for the symmetrical function f(x) around x=0 is x=-2. Solutions: x=-2, x=2.
You're absolutely correct. I like your approach.
Your reply is awesome but I do have some questions: How did you know the graph is symmetrical around x=0? Also, how did you figure that the obvious solution of x = 2, is 62? Hope you don’t mind replying. Thanks in advance.
@@boredomgotmehere This is a known result to me that a function of the form f(x)= (a+sqrt(b))^x +(a-sqrt(b))^x is even. It can be verified that f(x)=f(-x) in two lines. So f(x) is symmetrical. x=2 being a solution is also obvious to me by looking at the equation. Also, in my past experience, such school exercises as this one, often have one simple solution. If x=2, and x=-2 for a symmetrical function, then the axis of symmetry is x=0.
@@AlexMarkin-w6c I see your point. Thanks for the explanation.
I multiplied everything by (4+sqrt(15)) and lucked out on getting 1^x then getting your quadratic equation and instead of doing your guessing process I did (4+sqrt15)^x=31- sqrt(960) I treated it as a logarithmic equation
same and it was easier but i didn't get the -2
2(a²+b²)=(a+b)²+(a-b)²=62
So, X= 2
Easy
Root 15 is closed तो root16 ie 4(लिटिल less than4)
So let put value 4 इन place of 4
(4+4)^ x +0 = 62
For x= 2, lhs become 64 which is close to 62.
Now let us check the given eqn with x = 2,
(4 + _ /15)^2 + ( 4 - _/15)^2
= 2( a^2 + b^2)
= 2( 16+ 15)
= 62
=rhs
So x=2 is the ans
Fantastic. 👍
For luck, x is integer.
f(x) = (4+sqrt(15))^x + (4 -sqrt(15))^x
As 4-sqrt(15) = 1/(4+sqrt(15) , we have f(x) = a^x + a^-x,
with a = 4 +sqrt(15), so f(x) = f (-x) for any real x
So, if x is solution of f(x) = 62, then -x is also solution.
The function f strictly increases on R+, so if the equation f(x) = 62 has a solution on R+, then it is unique.
Now, as x =2 is evident solution, it is the unique solution on R+, and thanks to the initial remark: -2 is the unique solution on R-.
Conclusion: x = 2 or x = -2 are the only solutions.
Great job!
Very clever!
Thank you! Cheers!
You can use 62^2-2^2
=(60)(64) in the square root. It might be easier.
Exactly. It might be much easier. How could he miss such an elementary point ?!
Can you explain this a little bit more
@@noname-ed2un An elementary algebraic identity is:
a^2 - b^2 = (a - b) (a + b)
Substitute: a = 62 , b = 2 and you'll get:
62^2 - 2^2 = (62 - 2) (62 + 2) = (60) (64)
Have a nice day !
x = 0 ===>> 1 + 1 = 62 (F)
x = 1 ===> 4 + 4 = 62 (F)
x = 2 ===> 30 + 32 = 62 (V)
S = {2}
You missed the possibility that x=-2.
Where did the plus sign inbetween brackects vanish to? We can use difference of square method ignoring a plus inbetween.
You're thinking too hard.
You don't get the solution
You are a wonderful prof.....keep it up to help many young aspiring mathematicians.....sholom
I'm slow, but this is way too slow for me.
amaizing
Thank you. I'm glad you enjoyed it.
Super cool 👌👍
Thank you! Cheers!
Glad you enjoyed it.
Simple answer 2
X=2
2{(4)^2+(root 15)^2}
=2(16+15)=62
From 31+8V15 =(4+V15)^x .Just using a^x = b = > x = (logb)/loga . Sub number yield x = 2. Save a lot of time.
Yeah, but you get 50% credit only since -2 is also an answer. That too is obvious once you know the answer!
And how do you get the other solution of x = - 2..?
Nice 🎉
❤❤❤❤❤
I am glad you enjoyed it.
62square-4, could have been written as (62+2)(62-2)=8square*4*15
Approximation method,x=2 insertion can be easy.But if the question x=more than 10,you need more difficult tackling.
A very nice Olympiad question: (4 + √15)ˣ + (4 - √15)ˣ = 62; x = ?
Let: a = 4 + √15, b = 4 - √15, ab = 16 - 15 = 1; a = 1/b, b = 1/a
a² + b² = (1/b)² + (1/a)² = a⁻² + b⁻² = (4 + √15)² + (4 - √15)²
= 2(16 + 15) = 62 = (4 + √15)ˣ + (4 - √15)ˣ; x = 2 or x = - 2
Answer check:
x = ± 2: (4 + √15)ˣ + (4 - √15)ˣ = 62; Confirmed as shown
Final answer:
x = 2 or x = - 2
By inspection, 62= 15 + 15+ 16 +16, so x=2.
(4 + roo15) ² + (4-root15)²=2(16+15)=62 so x=2
Thank you
You're welcome, sir.
You should explain 1^x always makes 1.
1 reased to any power is 1. Hence, 1^x = 1.
In. Korea. Middle. School.
x=2 16+16+30=62
x=-2 1/（31+8√15）+1/（31-8√15）
=31-8√15+31+8√15=62
x=±2
Excellent
You’ve missed out in or omitted what can be generalized in this set of techniques that can be applied to other problems
(There are no “tricks” in mathematics, only techniques!)
I started from 62=2×(4^2+15) which means x=2 is a solution. Then I reformed the problem like p^x+p^(-x)=62, so I found -2 is also another solution. Considering the shape of cosh, these are the only solutions in real number. I have no idea in complex number :|
That's a good one, man.
62/log8 is the answer
Your coleque show this before a few days here! Didnt you see it?!
Use hit and trial method first put x=1 and then 2
There has to a way to further simplify this even the working out.
2
Just do rationalisation of 4-root15
p+1/p=62 is obvious after the first step since (4+sqrt(15))(4-sqrt(15))=1
or (4+sqrt(15)) = 1/ (4+sqrt(15)) so the quadratic is immediate in 2 steps
Far too much beating about the bush, way too many redundant steps.
The other thing that is obvious is that 1< max(x)