- #1
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
[tex] sqrt(x-1) + sqrt(y-2) = sqrt5[/tex]
thx
Last edited:
Solve x & y in sqrt(x-1) + sqrt(y-2) = sqrt5
- Thread starter mohlam12
- Start date
In summary, the conversation discusses a problem where the goal is to solve for x and y in the equation sqrt(x-1) + sqrt(y-2) = sqrt5. The solution is found to be (1,7) and (6,2) when x and y are limited to positive integers. However, if x and y can be any real numbers, there are an infinite number of solutions. The conversation also mentions the importance of fully stating the problem and using precise terminology.
- #2
HallsofIvy
Science Advisor
Homework Helper
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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.
- #3
mohlam12
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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
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
- #4
HallsofIvy
Science Advisor
Homework Helper
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mohlam12 said:
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.
- #5
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Okay,so which is the "inequation"...?(v.title)
Daniel.
Daniel.
- #6
mohlam12
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im sorry, well I am an exchange student so i dun know how to call that :) thanks anyways!
- #7
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Call it what it is:AN EQUATION...
Daniel.
Daniel.
1. How do you solve for x and y in the equation sqrt(x-1) + sqrt(y-2) = sqrt5?
To solve for x and y in this equation, we can use a technique called isolating the variables. First, we need to subtract sqrt(y-2) from both sides of the equation, leaving us with sqrt(x-1) = sqrt5 - sqrt(y-2). Then, we can square both sides of the equation to eliminate the square root, giving us x-1 = 5 - 2sqrt5sqrt(y-2) + y-2. Simplifying this, we get x+y = 6 + 2sqrt5sqrt(y-2). Finally, we can isolate y by subtracting x from both sides and then dividing by 2sqrt5, leaving us with y = (6-x)/(2sqrt5) + (x-1)/(2sqrt5). This gives us the solutions for both x and y in terms of each other.
2. Can this equation be solved algebraically?
Yes, this equation can be solved algebraically by using the steps outlined in the previous question. However, there are also other methods, such as graphing or using a calculator, that can be used to find the solutions.
3. What are the solutions to this equation?
The solutions to this equation are dependent on the value of x. As we found in the first question, the solution for y is y = (6-x)/(2sqrt5) + (x-1)/(2sqrt5). Therefore, the solutions for x and y will vary depending on the specific value of x.
4. Can this equation have imaginary solutions?
Yes, this equation can have imaginary solutions. When solving for x and y, we may encounter a negative value under the square root, which would result in an imaginary solution. This is not uncommon in equations involving square roots.
5. Are there any restrictions on the values of x and y in this equation?
Yes, there are restrictions on the values of x and y in this equation. Since we cannot take the square root of a negative number, the expressions inside the square root must be greater than or equal to 0. This means that x-1 must be greater than or equal to 0, and y-2 must be greater than or equal to 0. These restrictions can help us narrow down the possible solutions for x and y.
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