If $x,\,y$ are positive integers such that $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}+\dfrac{1}{11!11!}+\dfrac{1}{13!9!}+\dfrac{1}{15!7!}+\dfrac{1}{17!5!}+\dfrac{1}{19!3!}+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$.
Find $x,\,y$.
Showing posts with label simplify. Show all posts
Showing posts with label simplify. Show all posts
IMO Practice Problem: Find the exact real root for the equation $10x^3-12x^2-6x-1=0$
Find the exact real root for the equation $10x^3-12x^2-6x-1=0$.
The given cubic cannot be factored easily and beautifully, so the method of factoring the given polynomial is out of the question.
I hear you, the next best approach might be to try out the substitution method, with the hope that after the substitution, we have less variable terms in our newly set equation. But there seems no suitable substitution is available so to make simpler the given equation.
The given cubic cannot be factored easily and beautifully, so the method of factoring the given polynomial is out of the question.
I hear you, the next best approach might be to try out the substitution method, with the hope that after the substitution, we have less variable terms in our newly set equation. But there seems no suitable substitution is available so to make simpler the given equation.
How to obtain the cubic polynomial from our previous problem?
In the previous problem's post ((2) Evaluate The Sum Of 1/xy+z-1+1/yz+x-1+1/xz+y-1 ), we said if we let $k=\dfrac{1}{\dfrac{4}{x}+x-1}$, we will then have $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$.
In this post, we will reveal the logic and reason behind it and guide you step-by-step in achieving the other cubic polynomial where $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$.
In this post, we will reveal the logic and reason behind it and guide you step-by-step in achieving the other cubic polynomial where $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$.
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