X^2y+y^2x >6 inequality

  • Thread starter evagelos
  • Start date
  • #1
evagelos
315
0
For what values of x and , y is the following inequality satisfied:


[tex]x^2y+y^2x >6[/tex]

I tried to give a proof and i went as far:

xy(x+y)>6 then what?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
43,021
970


For what values of x and , y is the following inequality satisfied:


[tex]x^2y+y^2x >6[/tex]

I tried to give a proof and i went as far:

xy(x+y)>6 then what?
The best way to handle complicated inequalities is to look first at the associated equality. Here, we start by looking at [tex]x^2y+ y^2x= 6[/itex]. You can think of that as a quadratic equation in y and solve using the quadratic formula:
[tex]y= \frac{-x^2\pm\sqrt{x^4+ 24x}}{2x}[/tex]

Graph that on, say, a graphing calculator and it shows the boundary between "> 6" and "< 6". Putting in one (x, y) point for each region will tell you which regions are "> 6".
 
  • #3
evagelos
315
0


The best way to handle complicated inequalities is to look first at the associated equality. Here, we start by looking at [tex]x^2y+ y^2x= 6[/itex]. You can think of that as a quadratic equation in y and solve using the quadratic formula:
[tex]y= \frac{-x^2\pm\sqrt{x^4+ 24x}}{2x}[/tex]

Graph that on, say, a graphing calculator and it shows the boundary between "> 6" and "< 6". Putting in one (x, y) point for each region will tell you which regions are "> 6".

You mean then, that there is no solid proof for the inequality but only graphic procedures?

How about the inequality[tex]x^3y+yx^3>10[/tex] can we use graphic procedures??
 
  • #4
uart
Science Advisor
2,797
21


You mean then, that there is no solid proof for the inequality but only graphic procedures?

No that's not what Halls said. He found the boundary which precisely identifies the region of the x,y plane you are looking for. Graphing it is optional (though very helpful in my opinion).

For x>0 it reduces to the union of :

[tex]
y > \frac{-x^2 + \sqrt{x^4+ 24x}}{2x}
[/tex]

and

[tex]
y < \frac{-x^2 - \sqrt{x^4+ 24x}}{2x}
[/tex]
 
Last edited:

Suggested for: X^2y+y^2x >6 inequality

  • Last Post
Replies
0
Views
365
  • Last Post
Replies
6
Views
470
  • Last Post
Replies
8
Views
493
Replies
4
Views
275
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
900
Replies
4
Views
985
  • Last Post
Replies
13
Views
822
Replies
6
Views
1K
  • Last Post
Replies
2
Views
1K
Top