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

[anemone]

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]

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

$\chi$ $\sigma$

 

[anemone]

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, 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.

 

[anemone]

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:

View attachment 1585

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! :)

 

[CellCoree]

Hi there,

This is definitely a challenging problem! It appears to involve a cubic equation with a variable coefficient, which can be difficult to solve. One approach you could try is to use the Rational Root Theorem to find possible rational solutions for y, and then use synthetic division to test these solutions and see if any of them satisfy the equation.

Another approach is to use a graphing calculator or software to graph the equation and see if there are any obvious real solutions within the given range for y. This could potentially give you a starting point for finding the exact solutions.

If you are unable to find a solution using these methods, it is possible that the problem may not have a simple or exact solution. In this case, you could try using numerical methods such as Newton's method or the bisection method to approximate the solutions.

I hope this helps and good luck with your problem-solving!