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

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The discussion revolves around finding the number of real ordered pairs (y, a) that satisfy a specific cubic equation within the range 0 ≤ y ≤ 2.5. A participant suggests treating y as the independent variable to express a as a function of y, leading to a derived formula for a. After further analysis and graphing the function, it is concluded that there are infinitely many real ordered pairs (y, a) for the specified range, except for the case when a equals 1.3453. The problem appears to be resolved successfully.
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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 $$100y^3-3\left( 150+\frac{a}{8} \right)y^2+3\left( 125+\frac{5a}{8} \right)y-3\left( \frac{59a}{96}-\frac{75}{2} \right)=0$$ 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.
 
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anemone said:
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 $$100y^3-3\left( 150+\frac{a}{8} \right)y^2+3\left( 125+\frac{5a}{8} \right)y-3\left( \frac{59a}{96}-\frac{75}{2} \right)=0$$ 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

$\chi$ $\sigma$
 
chisigma said:
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:o, 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:

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

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.
 
Hmm...if I express $a$ as the subject of the formula, I get:

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

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

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

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

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! :)
 

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