Solve x & y in sqrt(x-1) + sqrt(y-2) = sqrt5

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Discussion Overview

The discussion revolves around solving the equation sqrt(x-1) + sqrt(y-2) = sqrt5 for the variables x and y. Participants explore the conditions under which solutions exist, considering both integer and real number solutions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the equation and seeks help in finding values for x and y.
  • Another participant points out that one equation cannot generally solve for two unknowns and suggests squaring the equation to derive a quadratic form.
  • A participant shares their approach, providing ranges for x and y based on the equation, and lists potential pairs of (x, y) values.
  • There is a clarification that if x and y must be positive integers, the only solutions found are (1, 7) and (6, 2).
  • Another participant questions the terminology used, indicating a lack of understanding of the mathematical terms.
  • A participant humorously corrects the terminology, emphasizing that it is an equation.

Areas of Agreement / Disagreement

Participants generally agree on the integer solutions (1, 7) and (6, 2) if x and y are constrained to positive integers. However, there is disagreement regarding the interpretation of the problem, particularly whether x and y can be any real numbers, which would lead to an infinite number of solutions.

Contextual Notes

There is uncertainty regarding the constraints on x and y, as the initial problem did not specify whether they must be integers or can be real numbers. This affects the interpretation of the solutions presented.

mohlam12
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hey eveybody, i have a problem for findin the answer of this, we have to solve for x and y.
[tex]sqrt(x-1) + sqrt(y-2) = sqrt5[/tex]

thx
 
Last edited:
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Okay, what is your problem? What have you tried? You understand, of course, that you CAN'T generally solve one equation for two unknowns? You can get a quadratic equation for x and y by squaring twice.
 
ok, here is what i did but I am not sure if it is right.
we have 6>=X>=1 and 7>=y>=2 (because sqrt(x-1)=sqrt5 - sqrt(y-2) and sqrt(y-2)=sqrt5 - sqrt(x-1) )
now we have that x could be {1,2,3,4,5,6} and y could be {2,3,4,5,6,7}
therefore,
if x=1 --> y=7
if x= 2 --> y=8-2sqrt5 (not with the intervalle i gave)
if x=3 --> y/= the intervalle
if x=4 --> y/= the intervalle
if x=5 --> y/= the intervalle
if x=6 --> y=2

finally, the solution would be : (1,7) and (6,2)

is it right? i dun know
 
mohlam12 said:
ok, here is what i did but I am not sure if it is right.
we have 6>=X>=1 and 7>=y>=2 (because sqrt(x-1)=sqrt5 - sqrt(y-2) and sqrt(y-2)=sqrt5 - sqrt(x-1) )
now we have that x could be {1,2,3,4,5,6} and y could be {2,3,4,5,6,7}
therefore,
if x=1 --> y=7
if x= 2 --> y=8-2sqrt5 (not with the intervalle i gave)
if x=3 --> y/= the intervalle
if x=4 --> y/= the intervalle
if x=5 --> y/= the intervalle
if x=6 --> y=2

finally, the solution would be : (1,7) and (6,2)

is it right? i dun know

If you ask us for help, don't hide part of the problem! You did not say before that x and y had to be positive integers! If that is, in fact, the case, then yes, the simplest way to do this problem is to try each value and find that the only solutions are (1, 7) and (6,2). If x and y can be any real numbers, then there are an infinite number of solutions.
 
Okay,so which is the "inequation"...?(v.title):confused:

Daniel.
 
im sorry, well I am an exchange student so i dun know how to call that :) thanks anyways!
 
Call it what it is:AN EQUATION...:smile:

Daniel.
 

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