sqrt((96)(97)(98)(99)+1))

In this video I used the palindromic property of the resulting polynomial to compose a perfect square which solves the task

[@shivanshnigam4015]

If you multiply x with x+3 and x+1 with x+2 then you can say t= x²+3x and it will be even more easier to calculate

[@marineintelligence]

shouldn't you add the 1 to get t = x²+3x + 1 ?

[@Amonboyev_Mehriddin]

yes yes yes very goooood. I agree with this opinion

[@hammadsirhindi1320]

It will be even more easy if we write it sqrt(100-4)(100-3)(100-2)(100-1)

[@shivanshnigam4015]

@@marineintelligence you can do that

[@chengshengway]

Yes, it should be done this way so that it showed that the product is 1 less than a perfect square

[@baselinesweb]

The excitement you embody when you say 'Do you think this is a perfect square' is exactly the excitement that anybody should feel when they get to that point. You humanize all that is good with math!

[@josephacri6855]

Really beautiful how numbers work,very essence of life,thanks for sharing this,I never saw this palindromic principle demonstrated like this before

[@Maiyut27tgb]

Very good 👍❤

[@fangliren]

At the point at which you’re looking at x^4 + 6x^3 + 11x^2 + 6x + 1, I would be tempted to “try” (x^2 + cx + 1)^2 and solve for c if it looks promising. We expect a square, by the way the question is posed. Also, all coefficients are positive, so a little thought tells us we can’t have any negative coefficients in the factor, so we don’t have to try -1 (or -x^2). This gives us x^4 + 2cx^3 + (c^2 + 2)x^2 + 2cx + 1, immediately giving 2c=6, c=3, which thankfully is compatible with c^2+2=11. So the guess is right, and we just plug 96 into x^2+3x+1

[@Yamcha1717]

I did it an unconventional way (and I don't know if it's "legal" in maths) : I've tried with x=1, and the result was 25, 5², or (1*4 +1)² Then I tried with x=2, and the result was 121, 11² or (2*5 +1)² So I theorized the result could be (x*(x+3) + 1)² and I wrote the calcul : x(x+1)(x+2)(x+3) + 1 = (x(x+3) + 1)^2 I developped both sides and got x^4 + 6x^3 + 11x^2 + 6x + 1 = x^4 + 6x^3 + 11x^2 + 6x + 1 The same calcul on both sides, so my assumption seems correct. Put under the square root, it gives x(x+3) + 1. Replace x with 96, I have 96*99 + 1, or 9504 + 1 or 9505. Edit : minor corrections

[@PrimeNewtons]

You'll need to show that x(x+d)(x+2d)(x+3d) +d⁴ is a perfect square

[@BH2K6]

I'm so happy I found this channel

[@saifhossain9462]

U are Awesome I like your unique problem solving methods ❤

[@nidoking042]

I gave this one a try and set x to be the second number of the four, because (x-1)(x+1) was a nice, easy product to deal with. The factoring was a bit guess-and-check, but since it started with x^4 and ended with 1, the square root was obviously x^2+ax(+/-)1, and the coefficients I had were all ones or twos, so I only had four meaningful options for a. The end result was x^2+x-1, which looks like a nice, neat answer.

[@holyshit922]

You can use long division method , just like for numbers sqrt(1+6x+11x^2+6x^3+x^4) = 1+3x+x^2 -1 6x+11x^2+6x^3+x^4|(2+3x)*(3x) -(6x+9x^2) 2x^2+6x^3+x^4|(2+6x+x^2)(x^2) -(2x^2 +6x^3+x^4) 0 (1+6x+11x^2+6x^3+x^4) = (1+3x+x^2)^2 There also is method for elminating repeated roots by calculating gcd(P(x),P'(x))

[@BetaLoversYT]

I wish I had you as teacher in University classes, for real.

[@vitotozzi1972]

Amazing!

[@paraskumar9850]

😳 woah!

[@dannieee333]

fav teacher

[@FlynnFromTaiga]

That's sooooo cool. Like wtf, algebra , how can you do this

[@user-wm7wi1ow9c]

My way for this state is: Suppose those numbers ( 99,96,97,98 )that are 100 So sq(100^4) is 100^2=10000 But we know those numbers are not really 100 ,they less &soon Then the answer is less&soon 10,000

[@icetruckthrilla]

I liked your approach but I was tempted by how close the numbers were to 100 this is my attack on this problem sqrt(96*97*98*99 + 1) this is actually a bit simpler if you take h = 100 and substitute: sqrt((h-4)(h-3)(h-2)(h-1) + 1) then switch the pairs to make pairs with numbers that sum up to 5 sqrt( (h-4)(h-1) * (h-3)(h-2) + 1 ) then multiply the pairs out sqrt( (h^2 - 5h + 4)(h^2 - 5h + 6) + 1 ) then we notice the first two terms of each polynomial in the product are the same (h^2 - 5h) so substitute n = h^2 - 5h which substituting back 100 for h works out to n = h^2 - 5h = (100^2) - (5*100) = 10000 - 500 = 9500 so actually n = 9500 putting n in this expression works out to sqrt((n+4)(n+6) + 1) multiply the binomials out sqrt(n^2 + 10n + 24 + 1) adding the two constants sqrt(n^2 + 10n + 25) now we see that this is equivalent to sqrt(n^2 + 2 * 5 * n + 5^2) so we actually have a square of a binomial sqrt( (n + 5) ^ 2 ) the square and the square root cancel out n + 5 substitute back n = 9500 which we figured out when getting n from h^2 - 5h 9500 + 5 which makes the answer 9505

[@carlosfox8201]

Bless you

[@lawrencejelsma8118]

I was thinking of the product of sums for weighted choice of x at some middle coefficient to let x = 97 or x = 98 but that fails for even multiplied x sums. So this method turns out better. Plus when I divided out the x^2 to get an answer I forgot to remuktiply it back in. I was doing the solution in R^2 and then the multiplied 11 coefficient dividing out by x^2 I just ignored that in finding the final results I saw you get with R^2 = x^2 multiplied back in. I need this brainwork practicing because I do mathematics statements errors like that too often! 😬

[@unpredictible1104]

You are Cool

[@johnka5407]

Graphs of the function we start with and of the function we end up with are different.

[@surendrakverma555]

Very good

[@PrimeNewtons]

Thanks

[@kinshuksinghania4289]

So it's effectively 97² + 96 in the end

[@GreenMeansGOF]

Maybe x=97 is easier.

[@glorrin]

The formula works for all strictly positive integer. Does it works for 0 ? well, strangely enough yes, although, the proof doesn't work if x = 0 What about negative numbers ? for -3, -2, -1 and 0, R is 1, no need for a formula for any number below -3, we are multiplying 4 negative numbers, which is the same as multiplying 4 positive numbers, but |x| is now the biggest of the 4 so with y = -(x+3) R = y^2 + 3y +1 or R = (x+3)^2 - 3x -8

[@Risperdali]

But everybody is a sinner!

[@0lympy]

10:30 I swear I won't consent to sinners' proposals :D

[@cret859]

If you take the absolute value, you get a valid expression of R(x)= | x²+3x+1 | = √(x.(x+1).(x+2).(x+3)+1) for any positive, null or negative real x. The square root is always defined since the palindromic polynômial x⁴+6x³+11x²+6x+1 is never negative whatever the value of x. But the absolute value is mandatory in R(x) expression since polynômial x²+3x+1 is négative when x is in ] (-3-√5)/2 ; (-3+√5)/2 [. Note that R(x)=1 for x in { -3 -2 -1 0 } only. Furthermore R(x)=0 at x=(-3±√5)/2 respectively x ≈ -2.6180 and x ≈ -0.38197 next interesting problem is to found how positive and negative integers p>0 and n

[@WinsonWu-tb6rv]

Palindromic Property is a little complicated to me , so i try this one , lucky , it works. (x-d)xy(y+d)+d^4 where x+d=y =(x^2-xd)(y^2+yd)+d^4 =(xy)^2+x^2yd-xy^2d-xyd^2+d^4 =(xy)^2-xyd(y-x)-xyd^2+d^4 =(xy)^2-2xyd^2+d^4 =(xy-d^2)^2

[@Nishchaya.01stha]

If a,b,c and d are in arithmetic progression with common difference k, then abcd+k⁴ is a perfect square.

[@PrimeNewtons]

What theorem or lemma is this?

[@Nishchaya.01stha]

@@PrimeNewtons I am not sure about it. I had a discussion about this problem with my friend and we just discovered it.

[@PrimeNewtons]

I'm sure one can do a proof for it

[@Nishchaya.01stha]

​@@PrimeNewtonsYa I have already tried out the proof. It works!

[@Nishchaya.01stha]

@@PrimeNewtons Ya I have alreaday tried out the proof. It works!

[@WinsonWu-tb6rv]

96x97x98x99+1 =(97-1)x97x98x(98+1)+1 =(97^2-97)x(98^2+98)+1 =(97x98)^2+97^2x98-97x98^2-97x98+1 =(97x98)^2+97x98(97-98)-97x98+1 =(97x98)^2-2x97x98+1 =(97x98-1)^2