A very nice olympiad question | How to solve (4 + \sqrt{5})^x + (4 - \sqrt{5})^× | Algebra |
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- เผยแพร่เมื่อ 27 เม.ย. 2024
- See the way I breakdown the solution of this question. There is a lot you can learn from this video.
How to solve (4 + \sqrt{5})^x + (4 - \sqrt{5})^×
. ENJOY
If this is your first time to my channel, here, I shared simple step by step method of solving Algebra with a simple trick.
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Bruh. Why all these channels solve really easy questions and say them olympiad question. Please solve "real" olympiad questions.
Awesome!
Wonderful explanation.
I solve it in less than 5 minutes, found the same trick used. Thank you.
One can guess the answer almost immediately. Assuming of course, the solver knows something about reciprocals and radicals.
Nice 🎉
I look forward for more videos like this. Great job on the solution and explanation
Thanks, more to come!
Very complicated but extraordinary explanation. Thanks brother 🎉
I'm glad you enjoyed it.
I've one TH-cam channel, brother
youtube.com/@venkatesanmathsacademy8904
Legal, muito bem explicado!
Thanks. I am glad you enjoyed it.
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???
well done sir you make this problem so very easy
I'm glad to hear that.
Brilliant!!
Thanks.
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.
Wonderful ❤
Thank you 😄
Nice
You specifically are too great to praise. Really sir
Thank you very much. I am grateful.
Amazing
I'm glad you enjoyed it.
Excelente. Bonito ejercicio....
Thanks bro. I'm glad you enjoyed it.
I like the way you teach 🎉🎉
I am glad to hear that.
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.
amaizing
Thank you. I'm glad you enjoyed it.
Great 👍 job
I am grateful, man. Glad you enjoyed it.
Super cool 👌👍
Thank you! Cheers!
Glad you enjoyed it.
Thank you
You're welcome, sir.
Very clever!
Thank you! Cheers!
You can use 62^2-2^2
=(60)(64) in the square root. It might be easier.
❤❤❤❤❤
I am glad you enjoyed it.
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.
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.
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
62square-4, could have been written as (62+2)(62-2)=8square*4*15
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.
(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.
I like how didnt search any math but i learnt this today from myteacher and now it gets recommended to me
Yeah, man. TH-cam just kinda knows what you want and show it to you.
Yea lol
Now i understand the -2 solution
I am glad you did.😊
Masemasic (pronuncia inglesa) no português. Recaptulation (pronuncia: recapituleichan )em portugues. Moral da história: Ensinando pai nosso a vigario.
I did it in 2 minutes by estimating x= 2. Because of the √1̅5̅, x has to be even to create an integral quantity. It has to be low because even 4⁴ > 62.
Nice one
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.
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
Just do rationalisation of 4-root15
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
Instead of solving of solving so long I would have tried assuming the x as 2 or 3 or 4 and try to get 62
And how would you have found the negative value of x?
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!
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.
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
Excellent. You nailed it. 👍
-2;2
x = 2 ; x = -2
X=2
3:33 why don't you raise it to the power of 1 with x?
I think it's not necessary. It will be equal 1.
Sehr einfach 4 - 15^ = 1/ 4 + 15^
Por favor, podrias repetirla, pero un poco mas despacio?
You can reduce the playback speed.
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.
@@user-hz5ne2rl5e I see your point. Thanks for the explanation.
Is this olympiad problem? Oh, you are right if only for 7-8 graders.
The Olympiad title is really just clickbait.
He will never mispronounce Sistine Chapel
Assuming 2 lowest positive insurd answer comes quickly how long
Solved by just hit & trial i.e.
62= 32+30=> 16+16+ 30 =>
16+ 16+15+15
16 & 15 will come from squaring
binomial theorem would have made it easier and shorter.
That's wonder. What would be your approach?
x=2; x=-1/2.
Pretty stereotype question
A very nice Olympiad question: (4 + √15)ˣ + (4 - √15)ˣ = 62; x = ?
62 > (4 + √15)ˣ > (4 - √15)ˣ > 0
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
I'm slow, but this is way too slow for me.
U forgot to raise 1 to the power x
1 to any power is 1. Therefore, 1 to power x is still 1.
x=3
No
А где доказательство, что других решений не существует? Задача не решена.
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?
Isn't that a 10th grade question?
This is typically a problem made complicated in appearance but convenient for solving when you know the trick. Not much real value.
I agree with you.
Assume the power of x to be 2 and then open up the squares you will get the answer in less than 10 seconds😂😂😂
When you say the “x” it sounds to me like “ess” instead of “eks.”
did it in my mind lol, it's easy, 4 - ✓15 and 4 + ✓15 are just reciprocals of each other and solve quadratic equation
1
2
無
Then you look at the original equation and think what of we put x=2 . Magic you solve the problem in 1 min ahahahahah
아유 골치 아프네
Take logarithm base 4+sqrt(15) of 4-sqrt(15) on ur computer and you will probably solve it easily
But anyway,if you don’t notice this then just watch the video
in Việt Nam , this question is too easy for a student in 16-17 age 😂😂😂😂
Absolutely. This isnt a problem for 16-17. Its for 10-12. For 16-17 check ioqm,jee advance papers for the correct level.
@@MazziniFan I said that but it's really too easy. I don't understand why they left the text " olympiad question " =))) Is their math always so easy?
4x+4x ➖ =8x^2 15x+15x ➖ 30x^2 {8x^2+30x^2}=38x^4 2^19 x^2^2 1^1 x1^2 (x ➖ 2x+1) (4)^2=16 (15)^2=225 {2 25 ➖ 16}=209 10^20 3^2.2^5 5^4 2^1 1^2^2 3^2 1^1 1^1^1 3^2" (x ➖ 3x+2)
0) n=2 , 3 , 4 , 5 ….
1) [n+sqrt(n^2-1)]*[n-sqrt(n^2-1)]==1 ;
2)[n+sqrt(n^2-1)]^2+[n-sqrt(n^2-1)]^2==N==2*[n^2+n^2-1]=2*[2*n^2-1]
3) n= 2| 3 |4 |5 | …..
N=14|34|62|98| …
4)[n+sqrt(n^2-1)]^x+[n-sqrt(n^2-1)]^x=N
x1=2 ;x2=-2
5) [n+sqr(n^2-1)]^x=p>0 ; ONLY !! : x=lg(p)/lg(n+sqrt(n^2-1) ) . p^2-N*p+1=0 . ONLY !! : p=N/2+-sqrt([N/2]^2-1) !!
With respect , Lidiy
That's awesome. You nailed it.
This is looks so awesome. But I do have some questions: what is the meaning of your line(1), where did you get that?
@@boredomgotmehere раскройте скобки и убедитесь !
С уважением Лидий
X=2
x=2
You are very correct, brother. But there is still one more solution to it.
X=2