Find real solutions to a system

  • Context: MHB 
  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    System
Click For Summary

Discussion Overview

The discussion revolves around finding all real solutions to the equation $\dfrac{2xy}{x+y}+\sqrt{\dfrac{x^2+y^2}{2}}=\sqrt{xy}+\dfrac{x+y}{2}$, where $x$ and $y$ are positive real numbers. The focus is on the mathematical reasoning and steps involved in solving the equation.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant proposes substituting $a = y/x$ to simplify the equation and notes potential issues with $x = 0$ and $a = 0$.
  • The participant derives a new equation in terms of $a$ and discusses isolating square roots and simplifying expressions.
  • Another participant mentions that the rational root theorem indicates $b = \pm 1$ as the only rational solutions, leading to the conclusion that $a = 1$ results in $y = x$.
  • The participant emphasizes that $x = 0$ is not a valid solution due to violating the original equation.
  • There is a repeated comment from the same participant about their recent success in solving similar problems.
  • A participant humorously questions the use of spoilers in the conversation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the overall approach to the problem, and while one participant presents a solution, there is no agreement on the validity of the steps taken or the implications of the findings.

Contextual Notes

The discussion includes assumptions about the positivity of $x$ and $y$, and the implications of setting $x = 0$ or $y = 0$ are noted but not resolved.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Suppose that $x$ and $y$ are positive real numbers. Find all real solutions of the equation $\dfrac{2xy}{x+y}+\sqrt{\dfrac{x^2+y^2}{2}}=\sqrt{xy}+\dfrac{x+y}{2}$.
 
Mathematics news on Phys.org
[sp]
As usual with me I can't find an efficient (or in this case even elegant) way to do this. But I had an idea...

[math]\dfrac{2xy}{x + y} + \sqrt{ \dfrac{x^2 + y^2}{2} } = \sqrt{xy} + \dfrac{x + y}{2}[/math]

Let a = y/x.
Two points:
1) Watch out for x = 0 later.
2) I tacitly multiplied the first fraction by a/a. So watch out of a = 0, as well.

Then, after a bit
[math]\dfrac{2a}{a + 1} + \dfrac{1}{ \sqrt{2} } ~ \sqrt{a^2 + 1} = \sqrt{a} + \dfrac{1}{2} (a + 1)[/math]

Isolate the [math]\sqrt{a + 1}[/math] and square. After some more simplifying and clearing the fractions
[math]2 (a^2 + 1)(a + 1)^2 = 16 a^2 + 4 a(a + 1)^2 + (a + 1)^4 - 8 a (a + 1)^2 + ( -16 a(a + 1) + 4 a (a + 1)^3 ) \sqrt{a}[/math]

Expanding and simplifying
[math]2a^4 + 4a^3 + 4a^2 + 4a + 2 = (a^4 + 14a^2 + 1) + (4a^3 - 4a^2 - 4a + 4) \sqrt{a}[/math]

[math]a^4+ 4a^3 - 10a^2 + 4a + 1 = (4a^3 - 4a^2 - 4a + 4) \sqrt{a}[/math]

So far my approach for finding a is pretty standard. Isolate one square root, simplify, then the next step would be to isolate the other square root and simplify. But there is an issue with this procedure: If a = 1 (as it will be when we solve it) then we have both sides of this equal to 0! I have chosen to change the variable to [math]b^2 = a[/math]. This means we don't have to square out that LHS, which is ugly enough. But again, we have to be careful about values of b now. (Don't worry, it'll all come together.)

So with the substitution and (yet more) simplifying:
[math]b^8 - 4 b^7 + 4 b^6 + 4 b^5 - 10 b^4 + 4 b^3 + 4 b^2 - 4 b + 1 = 0[/math]

The rational root theorem says that [math]b = \pm 1[/math] are the only rational solutions. Instead of some fancy theorems to prove these are the only real solutions, I simply graphed it. It was faster.

So, [math]b = \pm 1[/math] are the only real solutions. Time to work backwards. [math]a = b^2 = 1[/math]. Thus [math]\dfrac{y}{x} = 1[/math] and we get that the solutions to the original equation are all y = x. What about the x = 0 thing? y = x = 0 violates the original equation.

So the final solution is all real [math]y = x, ~ x \neq 0[/math].
[/sp]

-Dan
 
Last edited by a moderator:
Well done, topsquark! You know, I sense your obsession lately with my POTWs and challenge problems, hehehe...I hope so far you found nothing but fun in tackling all those problems!
 
anemone said:
Well done, topsquark! You know, I sense your obsession lately with my POTWs and challenge problems, hehehe...I hope so far you found nothing but fun in tackling all those problems!
[sp]
It's not so much that I've been obsessed, it's more that I have actually been able to solve some recently. I usually try to work them out.
[/sp]

-Dan
 
topsquark said:
[sp]
It's not so much that I've been obsessed, it's more that I have actually been able to solve some recently. I usually try to work them out.
[/sp]

-Dan

giphy.gif
 
[sp](Whispers) Why are we having a conversation using spoilers?[/sp]

-Dan
 
Because it is fun! (Bow)(Emo)(Drink)
 

Similar threads

MHB Solve the Equation involving $4^{\sqrt{log_2 x}}$
  • · Replies 1 ·
Replies
1
Views
2K
MHB Inequality involving positive real numbers
  • · Replies 1 ·
Replies
1
Views
1K
High School Given an equation, find the value of this expression
  • · Replies 17 ·
Replies
17
Views
2K
Undergrad ##x\lt y \lt z## positive integer numbers and if ##\dfrac 1x+\dfrac 1y+\dfrac 1z=\dfrac 13##
  • · Replies 6 ·
Replies
6
Views
2K
MHB Absolute value of real numbers
  • · Replies 1 ·
Replies
1
Views
1K
Rationalize the denominator: ##\sqrt{\dfrac{1}{2x^3y^5}}##
  • · Replies 6 ·
Replies
6
Views
3K
MHB 7.8.99 find PS, VS Period, graph
  • · Replies 7 ·
Replies
7
Views
2K
MHB 7.t.27 write eq for a sinusiol graph
  • · Replies 2 ·
Replies
2
Views
1K
MHB Finding Min Value of $\dfrac{|b|+|c|}{a}$ from Roots of Cubic Equations
  • · Replies 4 ·
Replies
4
Views
2K
MHB Prove Inequality for $x,y,z$ Positive Real Numbers
  • · Replies 1 ·
Replies
1
Views
2K