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Only 1% Can Solve This Math Problem

MindYourDecisions

63,804 views

There's a nice technique to solve this question.

Problem 1
https://www.facebook.com/photo.php?fb...
Reddit: does_anyone_know_how_to_solve_this_polynomial

Problem 2 (2019 NDA paper 2 mathematics q18)
https://www.vedantu.com/exams/nda-2-m...
https://askfilo.com/user-question-ans...

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20 Comments

@denvercheddie 7 months ago Pinned

6:31 You could have gotten the answer of 2 much faster. Recognize that a+a^2+a^3 = a*0 = 0. Same with beta. Then that leaves us with 1+1.

@Johnny-w5q3k 7 months ago

x^(n+1) - 1 = (x - 1)(x^n + ... x^2 + x + 1)

is useful trick to remember

@andresgradilla146 7 months ago

I went one step further in exercise 2.

When you have 1 + alpha + alpha² + alpha³ + 1 + beta + beta² + beta³ just take out alpha and beta as common term and you get

1+alpha(1+alpha+alpha²) +1+beta(1+beta+beta²) and you can simplify both sums in parentheses to 0 which multiplied by either alpha or beta will still be 0, so you end up with 1+1, which will of course be 2.

@ai314159 7 months ago

Missed opportunity to discuss cyclotomic polynomials and roots of unity.

@frendlyleaf6187 7 months ago (edited)

All you need is x^5 = 1 AND x ≠ 1 using these two the question is trivial. And obviously 12.

@renesperb 6 months ago

Write the equation as (x^5-1)/(x-1)=0 .Hence if x≠ 1 we have x^5 = 1 . The rest is as discussed in the video.

@old_grey_cat 7 months ago

I think the piece would have been improved by pointing out that the starting equation means we are dealing with complex numbers, so we won't try to solve for X itself.

@ADITYA2009-j2u 6 months ago

Learning-multiply with a specific term to eqn and subtract it with previous eqn to get the required value

Condition eqn should not contain coefficients so that terms should be cancelling.

@YennetiDhiraj 7 months ago

It was fun and an easy one, seemed hard at first glance but became easy as soon as I saw the given equation properly. Thanks for bringing questions for us to solve.

@EvilSandwich 7 months ago (edited)

Okay. Great video, but I was screaming that you could have simplified that Fourth Degree Polynomial in a single step by realizing that it was equal to (x^5 - 1)/(x-1) and multiplied both sides by (x-1). Since that quartic is one of the factors in the Difference of Two 5th powers formula.

Same with the quadratic at the end. Multiply both sides by (x-1).

@tommcintyre8391 7 months ago

I could see it, but not by being smart - only by being quite familiar with the trick of multiplying any geometric series of powers of x by (x-1) to collapse it to a binomial.
Fun problems!

@ckrylova 7 months ago (edited)

The first equation - there are 4 fifth roots of 1, complex ones. The fifth root is 1.
After that x^5=1, and x^(5a)=1. Eventually, the second expression is 12

@TimSiegrist-to5fv 7 months ago (edited)

Love all your stuff by the way...in case my other comment seems a little too harsh.

@deerh2o 7 months ago

Thanks for introducing me to the generalization that if the sum of the zero through n powers of x equals zero, then x^(n+1) = 1.

@TupperWallace 6 months ago

If you recognize the first equation as the fifth roots of unity forming a five pointed star on the complex plane, then raising x to a power is like counting off points on the star and cycling repeatedly. The powers 0, 11, 22, 33, 44, and 55 will take you to each of the points of the star, but only one is on the real number line. But the symmetry of the star makes the simplifying algebra possible projecting down to the real line. Incidentally, this is similar to how a youthful Gauss determined how to draw a 17 sided polygon using a construction of essentially x to the 16th power. His success led him to choose mathematics as a career, and he asked to have a 17-gon carved on his tombstone.

@S8EdgyVA 7 months ago

I can’t believe I tried to factor out the quadratic t^2+5t+6 (I was struggling with the problem and tried to factor out the bottom expression) as (t-2)(t-3) instead of (t+2)(t+3)