Can you solve this exam question from India?

[@disgruntledtoons]

If you're good enough at math the brute force method is actually quicker, especially once you recognize that the two roots of the quadratic are reciprocals of each other. That allows you to substitute one of the roots in place of the 1/x term in the second equation, giving you a lot of terms that cancel out.

[@aarushrath8827]

expanded with binomial, then there was a power 3 term, so expanded that, and got the answer maybe 2 mins or more but definitely not 6 so brute force works ig

[@f5673-t1h]

There are simpler ways of seeing these facts. First, the constant term is the prduct of the (negatives) of the roots, which is 1 here, so (after the two minus signs cancel each other out) the roots are inverses. (and of course, you need the leading coefficient to be zero) Second, you have a square root of 5 (and a denominator of 2) in the solutions to the quadratic. This should ring some bells for you, because the golden ratio has it. If you notice a bit further, they're 1 above the golden ratio(s), and we know that the golden ratio satisfies y^2 = y+1, so that means the two solutions of the original equation are the squares of the golden ratio (if you did not know by heart). The golden ratio is commonly used in such problems because it's easy to work with. So really, what the question is asking for is y^10 + 1/y^10, where y is the golden ratio. If you're familiar with the golden ratio, you know that raising it to the power of n will give f(n)y+f(n-1), where f(n) denotes the nth Fibonacci number (same holds with you replace y with 1/y). So just go 1, 1, 2, 3, 5, 8, 13, 21, 34, 55. So the answer is (55y+34)+(55/y+34), which simplifies to 55+2*34 (as y + 1/y = 1), which is 123.

[@NinjaCat-lb3vv]

I thought it was soooo simple. The method i did was solve the quadratic, but the substitutions in the second equation were crazy difficult.i gave up nearly immediately, but i can confirm that your method was far more elegant than brute force. I must admit though, that your method seemed a bit ambiguous and strange to begin wonder how you came up with it. Oh, and your neat little tricks are nice. Great!

[@OneShotAwayy]

Same!!! Second I saw x+1/x and the right side x^5+1/x^5 I initially thought of some sort of u substitution. But the way he extrapolated up the ladder was definitely not a method that crossed my mind. So simple too

[@danmerget]

I did the same thing, but I didn't find the second equation to be "crazy difficult". IMHO, it was about the same level of difficulty as Presh's answer. After solving the quadratic, we get x = (3 +- sqrt(5)) / 2. If we multiply the two possible values of x together, we get (9-5)/4 = 1. That means that we can assume one solution is x and the other is 1/x, and it doesn't matter which is which since x and 1/x are interchangeable in the second equation. WOLOG, assume x = (3 + sqrt(5)) / 2 and 1/x = (3 - sqrt(5)) / 2. Now solve for x^5. First, we square x to get x^2 = (7 + 3*sqrt(5)) / 2. Then we square x^2 to get x^4 = (47 + 21*sqrt(5)) / 2. Then multiply x^4 times x to get x^5 = (123 + 55*sqrt(5)) / 2. Finally, note that if we'd simply changed the sign of the sqrt(5) term at each step of the preceding paragraph, we would've been calculating (1/x)^5. So x^5 + (1/x)^5 = (123 + 55*sqrt(5)) / 2 + (123 - 55*sqrt(5) / 2 = 123.

[@courtneykachur9487]

I agree, in many if the advanced math problems i often think that folks just select absolutely random methods to solve. Its like mathmaticians have so many tools at their disposal that knowing which one to use when is just witchcraft

[@OneShotAwayy]

@courtneykachur9487 agreed, idk if your Calc 2 class was structured the same as mine, but ohhh god do I remember chapter 7. It was all the different types of tricks you could use for complicated integrals(methods of integration). It had by parts, u substitution, trig substitution, partial fractions (ohhhhh god pArTiAl FrAcTiOnS!!!!), integrals involving roots, involving quadratics etc. That chapter was a nightmare, opened my eyes to a completely separate difficulty with math, it wasn't the calculations, but instead it was exactly what you explained. Some times with math the hard part is knowing which trick to use when.

[@MarieAnne.]

It's not really that complicated, since product of roots = 1 (i.e. roots are reciprocals). So whatever value we use for x, 1/x is simply the other root. Since x = (3±√5)/2 then x^5 + 1/x^5 = ((3+√5)/2)^5 + ((3−√5)/2)^5 x^5 + 1/x^5 = 1/32 [(3+√5)^5 + (3−√5)^5] Notice that binomial expansions of (3+√5)^5 and (3−√5)^5 are almost identical, except that even terms cancel out since one is positive and the other is negative. Using binomial coefficients for raising to power of 5 (1, 5, 10, 10, 5, 1) we get x^5 + 1/x^5 = 1/32 [2 (1(3)^5 + 0 + 10(3)^3(√5)^2 + 0 + 5(3)(√5)^4 + 0)] x^5 + 1/x^5 = 1/16 (243 + 1350 + 375) x^5 + 1/x^5 = 1968/16 = 123

[@nasrullahhusnan2289]

x²-3x+1=0 --> x=½[3±sqrt(5)] =1+ß =ß² where ß=½[1±sqrt(5)], golden ratio in which ß²=ß+1 To find x⁵+(1/x)⁵ we can use the relationship of golden ratio to Fibonaci series.

[@kailashanand5086]

Wow nice one

[@UdayanaSaxena-o9t]

Can you explain it little bit

[@nasrullahhusnan2289]

@@UdayanaSaxena-o9t: x²-3x+1=½[3±sqrt{(-3)²-4(1)(1)}] =½[3±sqrt(9-4)] =½[3±sqrt(5)] =½[2+{1±sqrt(5)}] =1+½[1±sqrt(5)] +ve value of the 2nd term ½[1+sqrt(5)] is called golden ratio (due to inability to write the usual symbol here, I use ß) -ve value of rhe 2nd term, ½[1-sqrt(5), is -1/ß The property of ß is ß²=ß+1

[@bookert2373]

Math students: the useful takeaway here is that polynomials with symmetric coefficients always have roots that come in reciprocal pairs (x, 1/x, z, 1/z, etc). In this problem, that immediately means x+1/x = 3 (sum of roots always equals negative of second coefficient, etc). Those are the things to add to your toolkit - the part about x⁵ + 1/x⁵ was just put in to make it a harder test question - play with that if you want, but don’t let it divert you from learning and doing real math - focus on the tools, not the tricks.

[@marcosreal11]

So it comes down to learning the properties of polynomials.

[@PoppySuzumi1223]

There's another way to solve this by raising the power of (x + 1/x) to 5 at the beginning, i.e. (x + 1/x)⁵ As the Pascal Triangle Coefficient of Bionomial Theorem is: 1,5,10,10,5,1 (x⁵ + 5x³ + 10x + 10/x + 5/x³ + 1/x⁵) = 3⁵ = 243 Then recombine it, and you will see the pattern: (x⁵ + 1/x⁵) + (5x³ + 5/x³) + (10x + 10/x) = 243 x⁵ + 1/x⁵ = 243 - 5•18 - 10•3 = 123 Equation solved.

[@Orillians]

was thinking of this and I just thought to myself "Do they want me to use binomial?"

[@tracykinchen6919]

Where in the heck does 18 come from?

[@casualmarkit9969]

​@@tracykinchen6919 4:21

[@RaiymbekZhasulanuly]

I solved it exactly like that! Though I got 93

[@years-gq6dy]

This is just a more brute force calculation. More tedious to present as proof, honestly. You would still need to calculate the cube which you should've also shown.

[@ferdinandkuhn6975]

This elegant solution is so neat... I love it!

[@MichaelRothwell1]

Nice problem and solution. In fact, given that x²-3x+1=0, the expressions xⁿ+1/xⁿ are equal to the even numbered terms of the Lucas sequence (explained below), the first few terms being : 1, *3* , 4, *7* , 11, *18* , 29, *47* , 76, *123* where the 5th even numbered term tells us that x⁵+1/x⁵=123, as per the video. The Lucas sequence is a close relative of the more familiar Fibonacci sequence. The Fibonacci sequence is defined recursively by F₀=0, F₁=1 and Fₙ₊₁=Fₙ+Fₙ₋₁. The first few terms starting at F₁ are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 Note that terms with negative index can be defined so that the recursive relation still holds, using Fₙ₋₁=Fₙ₊₁-Fₙ, the first few terms being 1, -1, 2, -3, 5, -8 The Lucas sequence satisfies the same recursive relation Lₙ₊₁=Lₙ+Lₙ₋₁ as the Fibonacci sequence but has starting values L₀=2, L₁=1. It is straightforward to show that Lₙ=Fₙ₊₁+Fₙ₋₁ by induction, by checking the first two terms, and then using the recursive relation. From this, we get the first few terms of the Lucas sequence from those of the Fibonacci as above. Now let's why xⁿ+1/xⁿ=L₂ₙ. The golden ratio φ is the positive root of the quadratic equation x²-x-1=0 (the other root being -1/φ, as the product of the roots is -1), so φ²=φ+1 (φ²)²=(φ+1)² =φ²+2φ+1 =φ²+2(φ+1)-1 (φ²)²=3φ²-1 So φ² satisfies x²-3x+1=0, the other root being 1/φ² (as the product of the roots is 1) So if x²-3x+1=0, then x=φ² or x=1/φ² and xⁿ+1/xⁿ=φ²ⁿ+1/φ²ⁿ Now φⁿ=Fₙφ+Fₙ₋₁ (this is straightforward to prove by induction) and so (replacing n by -n) φ⁻ⁿ=F₋ₙφ+F₋ₙ₋₁ As F₋ₙ=(-1)ⁿ⁺¹Fₙ (also easily proved by induction), φ⁻ⁿ=(-1)ⁿ⁺¹Fₙφ+(-1)ⁿFₙ₊₁ =(-1)ⁿ(Fₙ₊₁-Fₙφ) So φⁿ+(-1/φ)ⁿ=φⁿ+(-1)ⁿφ⁻ⁿ =Fₙφ+Fₙ₋₁+(Fₙ₊₁-Fₙφ) =Fₙ₋₁+Fₙ₊₁=Lₙ So if x²-3x+1=0, then xⁿ+1/xⁿ=φ²ⁿ+1/φ²ⁿ=φ²ⁿ+(-1/φ)²ⁿ=L₂ₙ as claimed

[@BenDRobinson]

Yes, I spotted that the roots of the polynomial were the square and inverse square of the golden ratio, so I only needed the +/- tenth power... and indeed only one term of it. I happened to have the fourth and sixth powers in front of me from working on another problem... a quick bit of surd arithmetic (knowing I only needed the term that wasn't cancelling out) and I had the answer.

[@MichaelRothwell1]

@BenDRobinson you may have noticed that I managed to solve the problem without using a single surd :)

[@BenDRobinson]

@@MichaelRothwell1 yes, very nice, and you solved the much more general problem. I wasn't familiar with Lucas numbers per se, but have often enough used F(2n)=F(n)[F(n-1)+F(n+1)] so now I have an extra shorthand for part of that relation!

[@MichaelRothwell1]

@@BenDRobinson in that case, I recommend you watch the video "are these the only square Fibonacci numbers??" by Michael Penn, which shows how Lucas numbers can be fundamental to solving problems about Fibonacci numbers, in this case the result that the only Fibonacci numbers that are perfect squares are 0, 1, 144.

[@epicbird08]

Alternatively, use x + 1/x = (x^2 + 1)/x = (3x)/x = 3 and x^2 = 3x - 1, 1/x^2 = 3/x - 1. Then x^n + 1/x^n = 3(x^(n-1) + 1/x(n-1)) - (x^(n-2) + 1/x^(n-2)). Letting a_n = x^n + 1/x^n, we get a_n = 3a_(n-1) - a_(n-2), with a_0 = 2 and a_1 = 3. Now just list out the terms of the recurrence: 2,3,7,18,47,123. The answer is 123.

[@MichaelRothwell1]

Very neat!

[@MichaelRothwell1]

It turns out that the values of xⁿ+1/xⁿ are the even numbered terms in the Lucas sequence. Please see my comment for details.

[@mathmannix]

That was cool, and now I know how to solve a new class of problems easily. Thanks!

[@deerh2o]

Loved the methodology and how it can generalize into other problems, whether given in India or other countries :)

[@C.O._Jones]

Took me a few minutes, thank goodness so many terms cancel or the exponentiation would get ugly really quickly.

[@proboiz_50]

In my school, i learned these type of equations but I didn't learned these techniques yet My teacher said, in these equations we don't have to find the value of (x) , we just have to use formulas, multiply both sides, divide both sides to find the required value

[@timsmith8489]

Your initial idea of using the quadratic formula and then substituting in x^5 + 1/x^5 actually is pretty easy. Just note the the reciprocal of something of the form u +/- sqrt(v) = (u _/+ sqrt(v)) / (u^2 + v^2). From the quadratic formula we have x = 1/2 (3 +/- sqrt(v)). The reciprocal is 1/2(3 _/+ sqrt(5)). So we are looking for ((3 +/- sqrt(5))^5 + (3 _/+ sqrt(5))^5/2^5. Now note that if you were to expand those 5th power terms using the binomial theorem the terms with even powers of 3 have the same magnitude but opposite signs in the expansion of x^5 and 1/x^5 so cancel. The terms with odd powers of 3 are positive and the same magnitude and in those terms sqrt(5) is to an even power getting rid of the annoying square root. For each those odd power of 3 terms are C(5,0) 3^5 + C(5,2) 3^3 sqrt(5)^2 + C(5,4) 3^1 sqrt(5)^4 = 3^5 + 10 3^3 5 + 5 3^1 5^2 = 243 + 1350 + 375 = 1968. Add those odd power terms from both binomial expansions to get twice that, and divide by that 2^5 we have in the denominator and we get 123. NOTE: I'm using _/+ rather than what you'd expect because KZbin comment markup apparently takes the latter as starting strikeout mode.

[@dafyddthomas6897]

I spotted the Golden Section: x = (3 +- root5)/2 = 1 + (1 +- root5)/2 = 1 + Phi or 1- phi Phi nears 1.618; phi nears 0.618; x = Phi^2 or phi^2 I remebered that (Phi^n +phi^n) divided by some formula = nth Fibonacci And then I fell down the rabbit hole

[@arechb]

Good morning sir. Very nice explanation. And a very excellent way of thinking. Thanks for running this type of channel.

[@nordicexile7378]

Here's a way using (what I'm calling) "repeated substitution". given that x^2 = 3x - 1, repeatedly replace every instance of x^2 with (3x-1): x^5 = x(3x-1)(3x-1) = 9x^3 - 6x^2 + x = 9x(3x-1) - 6(3x-1) = 27x^2 - 26x + 6 = 27(3x-1) - 26x + 6 = 55x - 21 so x^5 = 55x - 21 now add x^5 + x^-5: (55x - 21) + 1/(55x - 21) = (55x - 21)^2/(55x - 21) +1/(55x - 21) = (3025x^2 - 2310x + 442)/(55x - 21) = (3025(3x-1) - 2310x + 442)/(55x - 21) = (6765x - 2583)/(55x - 21) = 123*(55x - 21)/(55x - 21) = 123 all the x terms cancelled out! Pretty cool, huh? 😊

[@mohitrawat5225]

This video made me remind of my childhood when I first got to this method.

[@AzureKyle]

I actually found something interesting about the equation while trying it. First, I did the quadratic formula to get x=(3+/-sqrt(5))/2, then unable to simplify, I instead used a calculator to calculate the numerical value of x for the 3+ answer first, raised it to the power of 5, divided 1 by it, and added both together to approximately get 123, since the calculator couldn't get all the digits. Then I did the same thing for the 3- answer, and realized the 3- answer raised to the power of 5 was very similar to the 1/3+ answer raised to the power of 5, and vice-versa. With the 3- answer giving me approximately 123 as the answer as well, but from the 122.9~ side, while the 3+ answer gave me the 123.000~1 side. So it doesn't matter what answer you use for x, you basically get the same result of ~123.

[@yurenchu]

The reason why this is so, is because the two solutions are each other's inverse. Or in other words: (3-sqrt(5))/2 = 1 / [ (3+sqrt(5))/2 ] This is easily verified: [ (3 - sqrt(5))/2 ] * [ (3 + sqrt(5))/2 ] = ... Note: (a/b) * (c/d) = (a*c)/(b*d) ... = (3 - sqrt(5)) * (3 + sqrt(5)) / 4 ... Note: (x - y) * (x + y) = x^2 - y^2 ... = ( 3^2 - (sqrt(5))^2 ) / 4 = ( 9 - 5 ) / 4 = (4)/4 = 1 which means [ (3-sqrt(5))/2 ] and [ (3+sqrt(5))/2 ] are each other's inverse.

[@erikkonstas]

I observed that c = 1, hence the two roots are reciprocals of each other, so that asks us to raise each root to the 5th and add those together, but the raising part is hard to do directly...

[@christopherlawyer4214]

I was today years old when I learned that if both A anC are 1 the solutions will be reciprocals of each other.

[@daniellerosalie2155]

Well, I minored in math about ten years ago. I was able to get through 70 percent of the equation, but I did get stuck. Your explanation was thorough.

[@mihir_666]

x + 1/x = 3 X² + 1/x² = 7 X^4 + 1/x^4 = 47 x3 + 1/x3 = (x+1/x)(x2 + 1/x2) = 18 Now (x + 1/x)(x4 + 1/x4) = 141 x5 + 1/x3 + x3 + 1/x5 = 141 x5 + 1/x5 = 141 - 18 = 123

[@nityajangra]

This was a basic question that is expected to come in jee mains exam But yeah students preparing for jee know these methods beforehand (i too!) Keep posting the question 💪🏻

[@yurenchu]

Another method: x⁵ + 1/x⁵ = ... note: (a⁵ + b⁵) = (a+b)(a⁴ - a³b + a²b² - ab³ + b⁴) ... = (x + 1/x) * ( x⁴ - x³(1/x) + x²(1/x)² - x(1/x)³ + (1/x)⁴ ) = (x + 1/x) * ( x⁴ - x³/x + x²/x² - x/x³ + 1/x⁴ ) = (x + 1/x) * ( x⁴ - x² + 1 - 1/x² + 1/x⁴ ) = (x + 1/x) * ( x⁴ + 2 + 1/x⁴ - x² - 1/x² - 1 ) ... note: (a² + 2 + 1/a²) = (a + 1/a)² ... = (x + 1/x) * ( (x² + 1/x²)² - (x² + 1/x² + 1) ) ... note: (a² + 2 + 1/a²) = (a + 1/a)² ==> (a² + 1/a²) = (a + 1/a)² - 2 ... = (x + 1/x) * ( ((x + 1/x)² - 2)² - ((x + 1/x)² - 2 + 1) ) ... substitute u = (x + 1/x) ... = u * ( (u² - 2)² - (u² - 1) ) From the given quadratic equation: x² - 3x + 1 = 0 x² + 1 = 3x x + 1/x = 3 ==> u = 3 Inserting this into the previous result: x⁵ + 1/x⁵ = = u * ( (u² - 2)² - (u² - 1) ) = 3 * ( (3² - 2)² - (3² - 1) ) = 3 * ( (9 - 2)² - (9 - 1) ) = 3 * ( (7)² - (8) ) = 3 * ( 49 - 8 ) = 3 * ( 41 ) = 123

[@randomscience8741]

It's so easy you can just do it like this . Firstly from the quadratic find the value of (x+1/×.) Which is equal to 3.Than brake ( x^5+1/x^5) into( x^2+1/x^2)(x^3+1/x^3)-(x+1/x).Than apply corollary of (a^2+b^2)and(a^3+b^3).After doing so just place the value of (x+1/x).

[@gwk3951]

x와 1/x을 두 근으로 하는 이차방정식을 만든 후 뉴턴항등식을 이용하는 방법도 있긴 합니다

[@robertlunderwood]

I don't even bother with that. My usual methods is to abuse x² = 3x - 1. Multiply by x to get x³, substitute 3x² with 3(3x-1). Then multiply x² and x³ to get x⁵. Another substitution and then you're good to solve.

[@yurenchu]

Eh, that explanation is not complete. How do you find the value of (x^5 + 1/x^5) then? How do you reduce the 1/x^5 term?

[@robertlunderwood]

@yurenchu It wasn't supposed to be complete. x⁵ will be a linear term with a constant. The sum is simple algebra.

[@yurenchu]

@@robertlunderwood Yes, x^5 will be a linear term with a constant; but what about 1/x^5 ? How do you find (x^5 + 1/x^5) if you can't tell me how to reduce the 1/x^5 term ?

[@thomasrl194]

1/x^5 can be reduced to a linear term likewise. Substitute 1=3x-x^2 into 1/x^5 = (3x-x^2)/x^5=... After a few substitutions you get x^5=55x-21 and 1/x^5=144-55x. This is also a very nice and straightforward method.

[@yurenchu]

@@thomasrl194 Of course it can be reduced, in many different ways. My point is that he didn't do the legwork and bother to check if the x will indeed vanish in the end. (Why was he adamant that he wouldn't eventually need to know the explicit value of x in order to arrive at the answer of 123 ?)

[@YineMiBen]

x2 = 3x -1, reduce x5 to x.(3x-1)2=> x5=55x-21

[@smitpatel5107]

Done, in less than a minute perhaps....Love from India❤❤

[@encounteringjack5699]

Nice! I did the quadratic way and plugged it into calculator for both to get the answer.

[@sluggaming9832]

very similar to Mathcounts 2021 state sprint problem 26: "If t is a real number such that t +1/t = 3, what is the value of t ^5+1/t^5? It just removes the step of dividing by x in the first step and rearranging.

[@michaelblankenau6598]

If I was Ramanujan the answer would have just appeared to me . But I’m not , so it didn’t .

[@daddykhalil909]

6:23 excellent

[@tambuwalmathsclass]

Beautifully solved

[@jeffweber8244]

Neat problem and solution. Liked this one.

[@dhpbear2]

This raises the question, is there an equation for the solution to "x^^n + 1/(x^^n)" , for the given exponent, 'n'? I'm guessing it would be *recursive* !

[@pihedron]

It is recursive. In the form x^n + x^-n = k, you define f(n) = kf(n - 1) - f(n - 2) and f(0) = 0, f(1) = 1. Make a table. The answer is f(n + 1) - f(n - 1). 144 - 21 = 123.

[@yurenchu]

Let p[n] = xⁿ + 1/xⁿ . Here are _three_ valid formulas that each give the solution: Recursive formula: p[n+2] = 3*p[n+1] - p[n] , with p[0] = 2 and p[1] = 3 Closed-form formula: p[n] = ((3+√5)/2)ⁿ + ((3-√5)/2)ⁿ Another exact formula: p[n] = round( ((3+√5)/2)ⁿ ) , for any positive integer n . Explanation: Let p[k] = xᵏ + 1/xᵏ . Then p[1] * p[k] = = (x + 1/x) * (xᵏ + 1/xᵏ) = xᵏ⁺¹ + 1/xᵏ⁺¹ + xᵏ⁻¹ + 1/xᵏ⁻¹ = p[k+1] + p[k-1] ==> p[k+1] = p[1] * p[k] - p[k-1] ... given p[1] = 3 ... p[k+1] = 3*p[k] - p[k-1] This is the recursive formula. Starting with p[1] = 3 and p[0] = x⁰ + 1/x⁰ = 1 + 1/1 = 2 , we get the sequence p[0] = 2 p[1] = 3 p[2] = 3*3 - 2 = 7 p[3] = 3*7 - 3 = 18 p[4] = 3*18 - 7 = 47 p[5] = 3*47 - 18 = 123 p[6] = 3*123 - 47 = 322 p[7] = 3*322 - 123 = 843 p[8] = 3*843 - 322 = 2207 etcetera. Characteristic formula in recursion (define q = forward shift operator : qp[n] = p[n+1] ): q² = 3q - 1 q² - 3q = -1 q² - 3q + 9/4 = -1 + 9/4 (q - 3/2)² = 5/4 q - 3/2 = ±√(5/4) q = 3/2 ± (1/2)√5 q = [ 3 ± √5 ]/2 ==> two roots: q₁ = [ 3 + √5 ]/2 = 2.6180339887... q₂ = [ 3 - √5 ]/2 = 0.3819660113... Note: q₁ * q₂ = ([3+√5]/2) * ([3-√5]/2) = (3+√5)*(3-√5)/4 = (9 - 5)/4 = 4/4 = 1 ==> q₁ = 1/q₂ p[k] = A*(q₁)ᵏ + B*(q₂)ᵏ , for some constants A and B . Determine A and B from p[0] and p[1] : p[0] = 2 ==> A*(q₁)⁰ + B*(q₂)⁰ = 2 ==> A*1 + B*1 = 2 ==> A+B = 2 ==> B = (2-A) . p[1] = 3 ==> A*(q₁)¹ + B*(q₂)¹ = 3 ==> A*([3+√5]/2) + B*([3-√5]/2) = 3 ==> A*(3+√5) + B*(3-√5) = 3*2 ==> 3(A + B) + (A - B)√5 = 6 ==> 3*(2) + (A - B)√5 = 6 ==> (A - B)√5 = 0 ==> A=B ==> A = 2-A ==> 2A = 2 ==> A = 1 , B = 1 . ==> p[k] = (q₁)ᵏ + (q₂)ᵏ where q₁ = (3 + √5)/2 = (2.6180339887...) q₂ = (3 - √5)/2 = (0.3819660113...) This is the closed-form formula. Notice that for integer k ≥ 1 , |(q₂)ᵏ| = |(0.3819660113...)ᵏ| < 0.5 , hence | p[k] - (q₁)ᵏ | < 0.5 and since p[k] must be an integer (as follows from the recursive relation, which has only integer coefficients and integer start values), we can conclude p[k] = round( (q₁)ᵏ ) where q₁ = (3 + √5)/2 = (2.6180339887...) round(X) rounds X to the nearest integer. This is also an exact formula (even though it uses a rounding function). [end of proofs]

[@JeffGoris]

If you're allowed a calculator then plug in either root and simply calculate. Ie, ((3−√5)÷2)^5 + (2÷(3−√5))^5 gives you 123. It surprised me that it is the same answer with either root so ((3+√5)÷2)^5 + (2÷(3+√5))^5 = 123 also. It then got me thinking and I realised the 2 roots (P and q) when multiplied together must equal 1 since (x-p)(x-q) = x^2-3x +1 will mean that pq=1. It then becomes clear that x^5+x^-5 = ((3+√5)÷2)^5+((3−√5)/2)^5 and many of the terms will cancel when expanded.

[@garywagner2616]

Wow! How simple - that was as easy as 1, 2, 3!

[@bibe74]

Since the quantity requested is (x^10+1)/x^5, I tried the polynomial division of the numerator and denominator by (x^2-3*x+1), and dividing the remainders of both divisions (since the divisor is equal to zero). I got that (x^10+1)/(x^2-3*x+1) gives a remainder of 6765*x-2583, which is 123(55*x-21), and x^5/(x^2-3*x+1) gives a remainder of exactly 55*x-21. Hence the required quantity is 123.

[@Qermaq]

It's elegant and simple and I hate these still. They will never be easy. I must learn to love their difficulty.

[@Communalसनातनी]

123 It is easy just it required 2 identities (A+b)^2 (A+B)^3 And Polynomial class 10

[@Ragnarok411]

We can also do it by using Newton's formula easily

[@3nelad]

3,7,18,47,123,.... This sequence is a(n+2)=3a(n+1) - a(n) with a(1)=3, a(2)=7.

[@kailashanand5086]

How did you find this?

[@yurenchu]

​@@kailashanand5086 That's easy: Let a[n] = xⁿ + 1/xⁿ . Then a[1] * a[n] = = (x + 1/x) * (xⁿ + 1/xⁿ) = xⁿ⁺¹ + 1/xⁿ⁺¹ + xⁿ⁻¹ + 1/xⁿ⁻¹ = a[n+1] + a[n-1] ==> a[n+1] = a[1] * a[n] - a[n-1] ... Since given (x² - 3x + 1) = 0 ==> (x² + 1) = 3x ==> (x + 1/x) = 3 , we know a[1] = 3 ... a[n+1] = 3*a[n] - a[n-1] This is the recursive formula. Starting with a[1] = 3 and a[0] = x⁰ + 1/x⁰ = 1 + 1/1 = 2 , we get the sequence a[0] = 2 a[1] = 3 a[2] = 3*3 - 2 = 7 a[3] = 3*7 - 3 = 18 a[4] = 3*18 - 7 = 47 a[5] = 3*47 - 18 = 123 a[6] = 3*123 - 47 = 322 a[7] = 3*322 - 123 = 843 a[8] = 3*843 - 322 = 2207 etcetera.

[@vcvartak7111]

No recursive formula generated for ( x^n + (1/x)^n) for all natural nos.

[@mudit2960]

Couldn’t we use the binomial expansion here after finding x^3+1/x^3=18? (x+1/x)^5 = X + 5(x^3+1/x^3) + 10(x+1/x) 243 = X + 90+ 30 X=123 where X = x^5+1/x^5

[@FaerieDragonZook]

x^5 + x^-5 = (x+ 1/x)^5 - 5x^3 - 10 x -10/x - 5/x^3 = (x+ 1/x)^5 - 5(x + 1/x)^3 + 5(x + 1/x) = 3^5 - 5* 3^3 + 5*3 = 243 - 135 + 15 = 123.

[@raynewport9395]

I'm having so much trouble trying to solve it here, so I'm fairly sure I won't be able to solve it from India.

[@Dharun-ge2fo]

Raise both sides to the 5th power and apply binomial theorem could be another approach

[@devondevon4366]

123 A different approach. x^2 -3x + 1 =0 x^2 + 1 = 3x (add 3x to both sides of equation) x + 1/x = 3 (divide both sides by x (x + 1/x)^3 = 3^3 (cube both sides) x^3 + 1/x^3 + 3 x* 1/x( x + 1/x) = 27 3 (3) = 27 x^3 + 1/x^3 = 27 -9 = 18 (x + 1/x)^5 = 3^5 (raised both sides to the fifth power since trying to find x^5) x^5 + 1/x^5 + 5x* 1/x ( x^3 + 2x^2*1/x + 2x* 1/x^2 +1/x^3) = 243 ( 3^5=243) x^5 + 1/x^5 + 5 ( x^3 + 2x + 2/x + 1/x^3) = 243 x^5 + 1/x^5 + 5 ( 18 + 2(3) =243 ( since 1/x+ x =3 , then 2/x + 2x = 2(3) or 6) (and x^3 + 1/x^3 = 18) x^5 + 1/x^5 + 5 ( 24)=243 as 18 + 6 = 24 x^5 + 1/x^5 = 243 - 120 as 5 (24) = 120 x^5 + 1/x^5 = 123 Answer as 243 minus 120 = 123 Answer

[@Nebarus]

42!

[@ferdinandkuhn6975]

One day, my math teacher will give me a point for that...

[@Acousticguitar-s8j]

Factorial of 42?

[@ferdinandkuhn6975]

@@Acousticguitar-s8j It's in reference to a movie called "The Hitchhiker's Guide to the Galaxy", where, I one scene, an advanced civilization tries to find the answer behind life, the universe, everything, and their super computer spits out the answer "42". That movie is really damn funny, you should definitely check it out! You'll not regret it, worth every penny 👍

[@C.O._Jones]

You don’t even need to check x = 0, since you have that 1/x^5 term.

[@BleuSquid]

I disagree... because the answer could always be "there is no solution"

[@years-gq6dy]

It is the easiest way to make the argument without needing much words.

[@yurenchu]

Nope, we do need to check x = 0 . Because if x = 0 is the solution to the given equation (x^2 - 3x + 1 = 0), then it would mean that x^5 + 1/x^5 is not a (finite) number.

[@C.O._Jones]

@ No, x cannot be = 0. You cannot divide by zero.

[@C.O._Jones]

@ Okay, go ahead and divide by zero. Waste your time.

[@preethasarts3379]

Quite impressive🎉

[@Z-eng0]

Brilliant method, though the math might be a bit faster, I think the brute force way would be simpler for those who don't have the time to think about it

[@1a1u0g9t4s2u]

Nice, another tool ( math trick) to add to my tool box.

[@j.r.1210]

I used a *third* method: Rearrange the quadratic to get x^2 = 3x - 1. Square both sides to get x^4 = 9x^2 - 6x + 1. Substitute 3x - 1 for x^2 again to get x^4 = 21x - 8. Multiply both sides by x to get x^5 = 21x^2 - 8x. Use the substitution trick again to get x^5 = 55x - 21. Now it's just a matter of adding (55x - 21) and 1/(55x - 21). Work it out, using the substitution trick yet again along the way, and you end up with (6765x - 2583)/(55x - 21). Remarkably, this reduces to 123.

[@JohnJones-pu4gi]

I did that. But you can divide the quadratic by x (since it isn't zero) to get x^-1 = 3-x and then proceed as before to get x^-5. This removes the unpleasant 1/(55x-21) term.

[@michaelpurtell4741]

Presh great problem: in order to do it, I spent the last two days playing with and learning about the relationship between the Golden ratio and Fibonacci numbers it was a lot of fun X= phi squared ..phi^n=Fn*phi+Fn-1…1/phi^n=Fn+1-phi*n…phi^n+1/phi^n= Fn-1+Fn+1 or 2*Fn-1+Fn so Fn-1=34 and Fn=55 so answer 123

[@GodbornNoven]

Since we're in india that means no calculator... Tricky. Well if its necessary to give a simple simplified answer then its really difficult if not then just solve the first eq to get x=(3±√5)/2 Then x⁵+1/x⁵ is thus equal to [((3±√5)/2)¹⁰ + 1)]/([3±√5]/2)⁵ =123 This looks really ugly tho

[@GodbornNoven]

oh i figured it out. The product of the roots = 1. Their sum is 3 p=1 implies the roots are reciprocals. S=3 means (x+1/x)² = 3² x²+2+(1/x)²=9 x²+1/x²=7 Square it again x⁴+2+1/x⁴=49 x⁴+1/x⁴=47 Now finally multiply both sides by x/1x (x⁴+1/x⁴)(x + 1/x)= x⁵+x³+1/x³+1/x⁵ = 141 (x²+1/x²)(x+1/x) = 7×3=21 x³+x+1/x+1/x³=21 x+1/x=3 x³+1/x³=21-3=18 x⁵+1/x⁵=141-18=123

[@robheale9860]

Wouldn't there be two solutions given that the quadratic equation has two solutions.

[@ArchibaldCoke]

It's weird, but they both work out to 123.

[@MarieAnne.]

No, there's only one solution, and here's why. If you use x = (3+√5)/2, then 1/x = 2/(3+√5) = (3−√5)/2. so we get x^5 + 1/x^5 = ((3+√5)/2)^5 = ((3−√5)/2)^5 If you use x = (3−√5)/2, then 1/x = 2/(3−√5) = (3+√5)/2. so we get x^5 + 1/x^5 = ((3−√5)/2)^5 = ((3+√5)/2)^5 So we get same result whatever value of x we use. In general, when you have quadratic equation ax² + bx + c = 0, then product of roots = c/a So for quadratic equation in the form x² + bx + 1 = 0, with roots x = p and x = q, then pq = 1/1 = 1 q = 1/p and p = 1/q Therefore: p^n + 1/p^n = p^n + (1/p)^n = p^n + q^n q^n + 1/q^n = q^n + (1/q)^n = q^n + p^n

[@undercoveragent9889]

There _are_ two solutions for 'x': (3+√5)/2), and (3−√5)/2) and they are the reciprocal of each other. Pretty cool.

[@PratibhaBhadoriya-k2s]

Thanks

[@MiklosKoncsek]

It's not the point to determine what X is... although it is perfect ratio (phi) squared !

[@onradioactivewaves]

@@MiklosKoncsek that's exactly how those convenient properties work here 🙂

[@DJCray8472]

I was using squaring the term (x^2+1/x^2)=7 -> (x^4+1/x^4)=47 and then multiply with (x+1/x) ... then I found out, that you still need to know the term (x^3+1/x^3) ... anyhow (x^5 +1/x^5) would be 3*47-18=123

[@qwang3118]

x + 1/x = 3. = => x^3 + 1/x^3 = 27 - 9 = 18 = => x^5 + 1/x^5 = 243 - 30 - 90 = 123.

[@angrytedtalks]

I can't imagine any practical application for this. Surely 1/x^5 is just x^-5 no? x=(the golden ratio)+1=2.618034 Fibonacci would have something to say about this question.

[@abirhossainshanto4900]

This is also a common problem in 10th grade in Bangladesh🇧🇩

[@Complexitor37]

challenge: how do you solve for x when the xth root of 2 is equal to x.

[@hikari1690]

You don't! Just give up after finding x = 1 doesn't work 😂

[@C7ZI]

You're gonna have to use some complex functions and graphs to solve it so not recommended.

[@Complexitor37]

There is a possible solution with just a handheld calculator.

[@C7ZI]

@@Complexitor37 What is it?

[@hikari1690]

@@C7ZI you can't. There are no real solutions. The other guy took my joke too far or tried to google the question and gemini spit out a rubbish answer. The solution is as you said complex plain so the typical lambert w function stuff. You definitely can't do that on a handheld calculator other than the ln(2) bits which is just part of the solution.

[@lupus.andron.exhaustus]

Whatever happened to the music at the end of your videos? 🤔

[@lasalleman6792]

Just use the quadratic equation. I get x = 2.618 and x = .3819 Both roots give an answer of 123. Seems there should be a positive and a negative, but here there are two positive solutions. Strange, but maybe I'm missing something.

[@onradioactivewaves]

@@lasalleman6792 because the solutions are reciprocals of one another and we are evaluating one solution plus the reciprocals, either solution yields the same result. Also the properties of the Golden ratio here 😉

[@jasimmathsandphysics]

Let 1/x+x=a then 1/x^5+x^5=a^5-5a^3+5a. From the first equation a=3 so you get 123.

[@markusbanach-stb5892]

Sadly, if you just use a pocket calculator, you can solve this in mere seconds and not learn the beatiful math behind it.

[@RhinoTheTerrible]

And you didn't even need to know what X was.

[@Metal_Master_YT]

I just completed the square and got x = 1.5 ± the sqare root of 1.25 and I just popped it into a calculator and then put both answers into the final equation, only for them both to arrive at the same answer, which meant there was only one answer of 123.

[@Neodynium.the_permanent_magnet]

What happened to the music at the end?

[@TheBugkillah]

Nope, wasn’t easier.

[@LeoLiss]

I'm not feeling smart anymore. That's good.

[@jc5c515]

I missed the music at the end

[@muhammadal-nahhal8174]

Same here

[@TheChamp1971]

The solution was as easy as 123!!

[@onradioactivewaves]

@@TheChamp1971 don't forget the abc

[@tatererer9747]

I don't think it was as easy as 1.038x10^103 was it?

[@TheChamp1971]

@@tatererer9747 I knew someone would read that as a double factorial!!

[@nasrullahhusnan2289]

x²--3x+1=0 --> x+(1/x)=3 Note that [x+(1/x)]²=x²+(1/x)²+2 9=x²+(1/x)²+2 --> x²+(1/x)²=7 (1) [x+(1/x)]³=x³+(1/x)³+3[x+(1/x)] 3³=x³+(1/x)³+9 --> x³+(1/x)³=18 (2) Multiply (1) and (2): [x²+(1/x)²][x³+(1/x)³]=7×18 7×18=x⁵+(1/x)⁵+(1/x)+x 126=x⁵+(1/x)⁵+3 --> x⁵+(1/x)⁵=123

[@jda4844]

Brilliant

[@markbothum4338]

And x = 2.618. Approximately.

[@ExtraMedium-]

Or 0.382, approximately

[@KeithAllen-pg8ep]

The Lucas sequence and the Fibonacci sequence are generated by powers of the Golden Ratio, φ. Writing s for the square root of 5, then φ = (1 + s)/2, φ^2 = (3 + s)/2, φ^3 = (4 + 2s)/2, φ^4 = (7 + 3s)/2, φ^5 = (11 + 5s)/2, etc. The Lucas sequence is given by the first terms of each bracket, and the Fibonacci sequence is given by the coefficients of s in each bracket. So, L = {1,3,4,7,11,18,29,47,76,123,199,322, ...}; F = {1,1,2,3,5,8,13,21,34,55, ..}. Notice the 'even-positioned' terms of L are 3,7,18,47,123, and the next is 322 - which is the value of x^6 + 1/(x^6). Also note that both L and F are fibonacci sequences, as is {6,7,13,20,33,53,86, ...} for example. All such sequences have the property that {(n+1)th term}/{nth term} rapidly approaches the value of φ.

[@vmadhavan435]

I did it this way too!

[@s-squaremaths6055]

( 18 × 7 ) - 3 = 123 Sir .

[@garvsenshivpuri9891]

Which country you are from ??

[@AmirgabYT2185]

123 🔥

[@satrajitghosh8162]

x + 1/x = 3 x^2 + 1/ x^2 = 3 ^2 - 2 x * ( 1/x) = 7 x^3 + 1/ x^3 = ( x + 1/x)^3 - 3 x * (1/x)( x + 1/x) = 3 ^3 - 3 * 3 = 18 ( x^2 + 1/x^2)( x^3+ 1/x^3) = ( x^5 + 1/x^5) + ( x + 1/x) Hereby ( x^5 + 1/x^5) = 7 * 18 - 3 = 123

[@chillwhale07]

Shouldn't there be two solutions?

[@MarieAnne.]

No, there's only one solution, and here's why. If you use x = (3+√5)/2, then 1/x = 2/(3+√5) = (3−√5)/2. so we get x^5 + 1/x^5 = ((3+√5)/2)^5 = ((3−√5)/2)^5 If you use x = (3−√5)/2, then 1/x = 2/(3−√5) = (3+√5)/2. so we get x^5 + 1/x^5 = ((3−√5)/2)^5 = ((3+√5)/2)^5 Therefore we get same result whatever value of x we use. In general, when you have quadratic equation ax² + bx + c = 0, then product of roots = c/a So for quadratic equation in the form x² + bx + 1 = 0, with roots x = p and x = q, then pq = 1/1 = 1 q = 1/p and p = 1/q Therefore: p^n + 1/p^n = p^n + (1/p)^n = p^n + q^n q^n + 1/q^n = q^n + (1/q)^n = q^n + p^n

[@chillwhale07]

@MarieAnne. Thank you 😊

[@ManojkantSamal]

Ans : 123.......May be ^=read as to the power *=read as square root As per question X^2 -3x+1=0 X^2+1=3x....... Eqn1 Let, X=a, (1/x)=b So, a+b=x+(1/x)=(x^2+1)/x=3x/x=3 a+b=3......eqn2 ab=x(1/x)=1......eqn3 We know, (a+b)^5=a^5+b^5+[5ab(a+b){(a+b)^2-ab}] Put eqn2 & eqn3 3^5=a^5+b^5+[(5×3){3^2-1}] =a^5+b^5+(15×8) =a^5+b^5+120 So, a^5+b^5+120=3^5 a^5+b^5=3^5-130 =243-120=123 So, (X^5)+(1/x)^5=123 (X^5)+(1/x^5)=123

[@yurenchu]

_Answer_ : 123 _Calculation_ (from the thumbnail): x² - 3x + 1 = 0 x² + 1 = 3x ... Note: x=0 is not a solution to the original equation; therefore, we can safely divide both sides by x ... x + 1/x = 3 (x + 1/x)³ = 3³ x³ + 3x + 3/x + 1/x³ = 3³ x³ + 3(x + 1/x) + 1/x³ = 3³ x³ + 1/x³ = 3³ - 3(x + 1/x) x³ + 1/x³ = 3³ - 3*(3) x³ + 1/x³ = 27 - 9 = 18 (x + 1/x)⁵ = 3⁵ x⁵ + 5x³ + 10x + 10/x + 5/x³ + 1/x⁵ = 3⁵ x⁵ + (1/x)⁵ + 5x³ + 5/x³ + 10x + 10/x = 3⁵ x⁵ + (1/x)⁵ + 5(x³ + 1/x³) + 10(x + 1/x) = 3⁵ x⁵ + (1/x)⁵ = 3⁵ - 5(x³ + 1/x³) - 10(x + 1/x) x⁵ + (1/x)⁵ = 3⁵ - 5(18) - 10(3) x⁵ + (1/x)⁵ = 243 - 90 - 30 x⁵ + (1/x)⁵ = 123

[@RAFAELSILVA-by6dy]

Note that if x is a solution to the quadratic, then so is 1/x. The roots, therefore, are reciprocals. This doesn't necessarily lead to a simpler solution, but it's another way to see that x + 1/x = 3 (= -b in the quadratic). And to see that x^3 + 1/x^3 = 27 - 3(3) = 18 etc. PS in general if x is a solution to x^2 -bx + 1 = 0, then x^5 + 1/x^5 = b^5 -5b^3 +5b

[@onradioactivewaves]

I see what you are saying here but it's still not obvious. Your first and second statement simply says if x is a solution to the quadratic then so is 1/x, therefore the reciprocal is also a solution ( this is circular logic). There are quadratic with solutions in which the roots are not reciprocals. You seem to be implying we are interested in equations in which x and 1/x are solutions, but rather than staging it that way it is first assumed with circular reasoning and then later implied. Perhaps I'm still not getting it. May I suggest that you just explicitly state the condition, start out with " quadratic equations in which a=c will have roots which are reciprocal of one another" would make a lot more sense to me. When you say ls if x is a solution to x^2-bx+1=0" how can x be a solution when it is the variable. I find this statement to be confusing, but again, maybe that's just on me. I know technical writing can be difficult. I suggest you try to make your statements make sense on their own without needing any other context as much as possible, that way anyone can read it having context or not, it will still make sense. Respectfully, this is just my own constructive criticism for you based on my own prior experience. Thank you for your comment.

[@Patrik6920]

@@onradioactivewaves consider ax²+bx +c rewritten as x+1/x=3 in the video ax+c/x=b the solution to x⁵+1/x⁵ =? is b⁵/2 +b/2 wich is 3⁵/2 + 3/2 = 123 works for any exponent and real number solution or a general solution [ax²+bx +c, a=1] -> xⁿ+c/xⁿ = ( -bⁿ / 2c) + ( -b / 2c ) [ax²+bx +c, a≠1] -> xⁿ+c/xⁿ = ( (-b/a)ⁿ / ( 2(c/a) ) ) + ( -b / ( 2a(c/a) ) )

[@onradioactivewaves]

@Patrik6920 thank you, I kind of followed already. Its been a good 10 years that I went through these mathematics. I've been trying to learn more and brush up for years. I also have a kid who is learning algebra now and learning anything in math always comes back around full circle. When I got into my 300 level classes for EE they were basically all applied math classes and I used every bit of mathematics I ever learned, I can say that in all honesty. And the more I've advanced in mathematics, the more I understand and appreciate how complex algebra really is. I'm here reviewing and learning more on a continuous basis, thank you for the feedback, I appreciate it.

[@Patrik6920]

@@onradioactivewaves ya very much enyoy these too, brush up and learn some new stuff once in awhile .. ppl come up with quite clever solutions sometimes ..

[@years-gq6dy]

I dunno. Seems kinda unnecessary if you can just show it by manipulating the equation. If you presented that argument, it will be too unnecessarily wordy.

[@KekusMagnus]

after seeing this i'm not sure that the first method is much slower

[@sunrevolver]

So, repetitive simple aritmetic is perceived to be faster and not as tedious as solving a single more complex arithmetic.. Interesting...

[@Nothingx303]

Juicy really 😋 😊😊

[@hiffi5g588]

It becomes bit easy after using binomial theorem 🙃

[@RobbieHatley]

Cute, but I solved it in seconds from looking at the thumbnail (before even launching the video) by doing a little algebra and doing a little Perl programming. perl -E 'my @r; $r[0]=(3-sqrt(5))/2; $r[1]=((3+sqrt(5))/2); say($_**5+1/$_**5) for @r;' 123 123 So there are two roots for the given quadratic, but (x^5+1/x^5) is 123 for both roots. Very odd. The method you use in this video seems *_waaaaay_* too complicated to me, though. Just use the quadratic formula to get the two roots (3+√5)/2 and (3-√5)/2, then plug those into (x^5+1/x^5) and both give 123.

[@Tahngarthor]

I prefer the plug n' chug method (aka brute force) :p

[@OfficialEIonMusk]

Lol i am from India and had the same question in my coaching test in class 9

[@69Hauser]

I solved it by both methods. First option (orthodox way): solving the cuadratic and replacing in the second ecuation we have: ((3±√5)/2)^5 + 1/((3±√5)/2)^5 = (123±55√5)/2 + 2/(123±55√5) = =(15129±6765√5) / (123±55√5) = (15129±6765√5)(123∓55√5) / (123^2 -5·55^2) = = 492/4 = 123 Easy and more enjoyable than the "smart" method

[@robertveith6383]

That is *not* "easy." Thatt is very cumbersome.

[@69Hauser]

@@robertveith6383 I have to say that the 1st method only took me 4 or 5 min more than the 2nd. Is it cumbersome?, maybe, that's just the reason why solving that way is so satisfying. And by no means it's difficult (if you're used to).🙂

[@Effect_channel]

You actually still doing the complicated way? Put 123 and viola!

[@melwinalm]

Greatly explained

[@mustafaakbas3723]

👍

[@riteshkumarsingh8174]

A question from my country and I solved it in 5 mins 💀