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Find the number of real ordered pairs (y,a)

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,712
Hi MHB,

This is another interesting problem that I tried to work on it with no luck so far to break the code...

Problem:

Find the number of real order pairs of $(y, a)$ that satisfy \(\displaystyle 100y^3-3\left( 150+\frac{a}{8} \right)y^2+3\left( 125+\frac{5a}{8} \right)y-3\left( \frac{59a}{96}-\frac{75}{2} \right)=0\) such that $0 \le y \le 2.5$.

At first glance, my mind draws a blank and this is not good because whenever this happens to me, chances are I will not be able to solve the problem. And here I am now, asking for a modicum of help or ideas to crack this problem.


Thanks.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hi MHB,

This is another interesting problem that I tried to work on it with no luck so far to break the code...

Problem:

Find the number of real order pairs of $(y, a)$ that satisfy \(\displaystyle 100y^3-3\left( 150+\frac{a}{8} \right)y^2+3\left( 125+\frac{5a}{8} \right)y-3\left( \frac{59a}{96}-\frac{75}{2} \right)=0\) such that $0 \le y \le 2.5$.

At first glance, my mind draws a blank and this is not good because whenever this happens to me, chances are I will not be able to solve the problem. And here I am now, asking for a modicum of help or ideas to crack this problem.


Thanks.
If y is the independent variable You can to arrive to write a single valued function $\displaystyle a = \varphi(y)$ because Your equation is linear in a, so that is...

$\displaystyle f(y)\ a + g(y) = 0 \implies \varphi(y) = - \frac{g(y)}{f(y)}\ (1)$

Kind regards

$\chi$ $\sigma$
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,712
If y is the independent variable You can to arrive to write a single valued function $\displaystyle a = \varphi(y)$ because Your equation is linear in a, so that is...

$\displaystyle f(y)\ a + g(y) = 0 \implies \varphi(y) = - \frac{g(y)}{f(y)}\ (1)$

Kind regards

$\chi$ $\sigma$
Thanks for your reply, chisigma.

To be honest with you:eek:, I have actually tried that (I am sorry for not showing my attempt in the first place) and found that there really is nothing much to do afterwards but I might be wrong.

According to your suggestion, I get:

\(\displaystyle 100y^3-3(150)y^2+3(125)y+3(\frac{75}{2})=a\left( \frac{3y^2}{8}-\frac{15y}{8}+\frac{59}{32} \right)\)

And I even sketched the functions of $f(y)=100y^3-3(150)y^2+3(125)y+3(\frac{75}{2})$ and $g(y)=\dfrac{3y^2}{8}-\dfrac{15y}{8}+\dfrac{59}{32}$ but I don't see the relation between $a$ and the given restriction where $0 \le y \le 2.5$ and the newly formed equation, and hence how to find the number of real order pairs of $(a, y)$ based on that.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,712
Hmm...if I express $a$ as the subject of the formula, I get:

\(\displaystyle a=\frac{100y^3-3(150)y^2+3(125)y+3(\frac{75}{2})}{\large \left( \frac{3y^2}{8}-\frac{15y}{8}+\frac{59}{32} \right)}\)

\(\displaystyle \;\;\;=\frac{3200y^3-14400y^2+12000y+3600}{12y^2-60y+59}\)

\(\displaystyle \;\;\;=\frac{800y}{3}+\frac{400}{3}+\frac{12800}{3} \left(\frac{y-1}{12y^2-60y+59} \right)\)

\(\displaystyle \;\;\;=\frac{800y}{3}+\frac{400}{3}+\frac{12800}{3} \left(\frac{y-1}{12(y-3.6547)(y-1.3453)} \right)\)

And if I plot the sketch of $a$ versus $y$, I get:

Find (a,y).JPG

From the graph, we can tell there are infinitely many real order pairs of $(y, a)$, with $a\ne 1.3453$ for $0 \le y \le 2.5$.

I think I have solved this problem! :)