Math Olympiad | How to solve an Octic Equation ? | Only for math genius !

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#exponentialproblems #matholympiad #maths #octic

[@prime423]

Use the rational root theorem!!Much easier than the method employed!!Try simplest method first.

[@vijaymaths5483]

Thanks for the tip!

[@angelmendez-rivera351]

The rational root theorem only finds 2 of the 8 roots.

[@Rocio62154]

SENSATIONAL PROFESSOR!!

[@waygonedon]

Why did you choose to divide through by x^4?

[@sumit-mn6ys]

Amazing tutorial 🙌

[@vijaymaths5483]

Thank you 🙌

[@crazyindianvines1472]

Nice explanation

[@vijaymaths5483]

Keep watching

[@superiorlyrics8326]

👏👏👏

[@vijaymaths5483]

👍

[@mihaipuiu6231]

VIJAI! Please remove "only for math genius!" from the title because you make me laugh.

[@vijaymaths5483]

No sir 😀
As per me you are a genius person ⚘️

[@vijaymaths5483]

Laughter is the best medicine brother 🌝

[@NadiehFan]

@@vijaymaths5483 It is easy to see by looking at the ratio between the coefficients read from left to right and read from right to left that this a palindromic equation in disguise. Specifically, if we substitute
x = √2·y
we get a genuine palindromic equation of degree 8, the solution of which can be reduced to solving an equation of degree 4 in a variable z = y + 1/y.

[@RealQinnMalloryu4]

2/x^8 =4 9/x^7 =2 20/x^6.3 33/x^5 =6.3 46^/x^4 11.2 66/x^3=22 80/x^2 =40 72/x=36 32/x=:16 2^2 2^1 3^2.3^1 3^2.3^1 11^1 .2^1 2^11 2^20 6^6 4^4 1^1 1^1 1^1.1^1 1^1.1^11^1. 1^1 1^11^1 1^5^4 3^2^3^2 2^2^2^2 1^2^2 1^1^3^1 1^1^1^1 3^1 1^12 32 (x ➖ 3x+2)

[@franciscook5819]

2*x^8-9*x^7+20*x^6-33*x^5+46*x^4-66*x^3+80*x^2-72*x+32=0
Looking at the equation (coefficients) it seemed possible that low integers were solutions so I tried both x=1 and 2 and both gave zero. (You can look at the rational roots theorem => rational roots p/q where p | 32 (constant term) and q | 2 (highest power multiplier) for values to guess).
Dividing by (x^2-3*x+2) gives 2*x^6-3*x^5+7*x^4-6*x^3+14*x^2-12*x+16 which is positive for x≤0,
0≤x≤0.5
2*x^6-3*x^5=x^5*(2*x-3) > 1/32*(-3) > -1
7*x^4-6*x^3=x^3*(7*x-6) > 1/8*(-6) > -1
14*x^2-12*x=2*x*(7*x-6) > 1*(-6) > -6 so 16 -1 -1 -6 > 0
0.5≤x≤1
2*x^6-3*x^5=x^5*(2*x-3) > 1*(-2) = -2
7*x^4-6*x^3=x^3*(7*x-6) > 1*(-2.5) = -2.5
14*x^2-12*x=2*x*(7*x-6) > 2*(-2.5) > -5 so 16 -2 -2.5 -5 > 0
for x≥1 it is positive (In fact the minimum is about 13 near 0.5 or 0.6) so no further real roots.
In fact, the equation factorises thus:
2*x^8-9*x^7+20*x^6-33*x^5+46*x^4-66*x^3+80*x^2-72*x+32=(x^2-3*x+2)*(x^2-2*x+2)*(x^2-x+2)*(2*x^2+3*x+4)=0;

[@angelmendez-rivera351]

This is all well, but your analysis has no explanation for how to find the roots or the factorization. All you proved was that all the remaining roots are complex.

[@franciscook5819]

@@angelmendez-rivera351 The question was to find real roots. I did so via an alternative method There was no need to find all the roots - just prove that, after factoring out the known solutions, the factored curve doesn't cross the x-axis, so no real solutions.