find 4 solution for eguation

extraordinary said:
x^2+y=7
y^2+x=11 thanks beforehand
Click to expand...
No, most of us do not GIVE answers. As our name says, we provide HELP.

Big hint. The most general way to solve systems of simultaneous equations is the method of substitution. You use one equation to find one variable in terms of the other, and then you do what?
 
extraordinary said:
x^2+y=7
y^2+x=11 thanks beforehand
Click to expand...

Hint:

Plot y = 7-x2 and y = ±√(11-x) and look at the points of intersection.

Please read the post titled "Read before Posting".

We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.
 
JeffM said:
No, most of us do not GIVE answers. As our name says, we provide HELP.

Big hint. The most general way to solve systems of simultaneous equations is the method of substitution. You use one equation to find one variable in terms of the other, and then you do what?
Click to expand...
oh... what a pity(
"most of us do not GIVE answers -we provide HELP" if nt solutions, i can find such help by google. indeed i need the ways of solutions themselves.
You use one equation to find one variable in terms of the other, and then you do what?"- i found 1 solution by this way, but i need 3 more.
i hope everything will be ok. thank you again
 
really?
ah...ok, then can we solve such eguations as matrices? or can u suggest me 3 other way?
i could find just 1 solution, so i didnt stuck anywhere . i dont even now the other 3 ways to be stucked(
im so sorry, my english can be nt so good,so exuse my mistakes.
 
extraordinary said:
really?
ah...ok, then can we solve such eguations as matrices? or can u suggest me 3 other way?
i could find just 1 solution, so i didnt stuck anywhere . i dont even now the other 3 ways to be stucked(
im so sorry, my english can be nt so good,so exuse my mistakes.
Click to expand...

What is the equation that you got?

What is the one solution that you got?
 
Subhotosh Khan said:
What is the equation that you got?

What is the one solution that you got?
Click to expand...
i got it , but it is oka. it doesnt need any correction, it is ok.
i need 3 new solution, or though 3 ways of solutions.
i am sorry, im new here, so i dnt understand, why cant ppl write me solutions themselves? if there r such capable ppl, i ask u to help me. thx
 
I deleted my reply because I erred in guessing that substitution would lead to quadratic form.
 
Hi extraordinary

i am sorry, im new here, so i dnt understand, why cant ppl write me solutions themselves? if there r such capable ppl, i ask u to help me. thx
Click to expand...
HERE is a link to our posting guidelines. Cheers :cool:
Click to expand...


I am new here too. Reading the posting guidelines is a good idea.
There are very good teachers here, but they will not do the work.
We watch you do the work, and comment to let you know if you are going in the right direction.
If you hire a tutor (for $30 to $60/hr) they will come to your house and watch you work.
We can not see you.
That is why we need to ask you questions.
That is why we need you to show us your work .

If Language is a problem -- we will be patient -- you probably speak more languages than I do:)


This is an interesting problem!
Substitution was a great hint.
You know there are 4 solutions.
There will be many steps involving the quadratic formula.

We will keep helping as long as you show us what you are doing.
I like this problem!
 
ah...ok, then can we solve such eguations as matrices? or can u suggest me 3 other way?
i could find just 1 solution, so i didnt stuck anywhere .
Click to expand...

Matrices is a great thing to think about.
However, matrices require that we have linear equations in the unknowns {x,y}.
I don't see a way to re-write these equations to get rid of x2 and y2 terms.
Also, there would not be 4 solutions. I guess matrices are not useful here.

If you still have your work from the one solution, please share it...
Even if you think it is wrong.

The goal here is NOT to complete your homework for you.
We do want to...
1) Help avoid being lost as you do your homework.
2) Help you not waste time on approaches that will not work.
3) Help you get a good grade when you take your exams by yourself.


 
I assume you were able to substitute and obtain a 4th degree polynomial (in say x) and set it equal to 0. The rational roots theorem gives one root to the resulting polynomial. The remainder theorem then tells you how to obtain a cubic polynomial. Fortunately cubics aren't too terrible to solve by radicals (i.e. use the "cubic formula").

Another option is to use one of many approximation methods.
 
Graphing the quartic resulting from substitution shows that it has an integer root at x = 2 plus three other real roots. I presume that 2 is the answer that the OP got.

As daon says, from that point you can get an exact answer using the cubic formula or approximations using Newton's method and the results of graphing for initial values. The OP did not tell us whether he was to give an exact or approximate answer.
 
Bob Brown MSEE said:



I am new here too. Reading the posting guidelines is a good idea.
There are very good teachers here, but they will not do the work.
We watch you do the work, and comment to let you know if you are going in the right direction.
If you hire a tutor (for $30 to $60/hr) they will come to your house and watch you work.
We can not see you.
That is why we need to ask you questions.
That is why we need you to show us your work .

If Language is a problem -- we will be patient -- you probably speak more languages than I do:)-


This is an interesting problem!
Substitution was a great hint.
You know there are 4 solutions.
There will be many steps involving the quadratic formula.

We will keep helping as long as you show us what you are doing.
I like this problem!
Click to expand...
thank you "If Language is a problem -- we will be patient -- you probably speak more languages than I do"-yes, i do)
the common way is: x^4-14x^2+x+38=0 what 'd i do next? if in' x4 + Ax3 + Bx2 + Ex + D = 0"but in my eguation there is no any A?
 
extraordinary said:
thank you "If Language is a problem -- we will be patient -- you probably speak more languages than I do"-yes, i do)
the common way is: x^4-14x^2+x+38=0 what 'd i do next? if in' x4 + Ax3 + Bx2 + Ex + D = 0"but in my eguation there is no any A?
Click to expand...

Be polite - we have many multilingual helpers here - for example I know of at least one who can read, write and speak fluently in four languages. You know more languages than that? I can curse people in seven/eight languages, thanks to the diversity of graduate school.

Now about your question - you do have 'A'. In your case, A = 0. Following daon's suggestion, you can see that one of the solutions is (2, 3).

The other three solutions are irrational and those are approximately (found through numerical analysis - Newton_Raphson method):

(-1.8481265269644,.............3.58442834033049)
(-3.28318599128617,..........-3.7793102533777)
(3.13131251825057,...........-2.80511808695275)
 
There are at least five methods to solve this problem if you take the route of substitution. There may be more that do not involve substitution.

As you have found out, substitution leads to the quartic equation \(\displaystyle x^4 + (0 * x^3) - 14x^2 +x + 38 = x^4 - 14x^2 + x + 38 = 0.\)

Good job.

Through some method that you have not disclosed, you appear to have found out on your own that x = 2 is a solution to that quartic, which entails that x = 2 and y = 3 is ONE EXACT solution to your original problem.

One method to find other EXACT answers is to use either the formula for the quartic or the formula for the depressed quartic. Those, however, are horribly inefficient ways to solve the problem exactly and are highly error prone.

Starting from the quartic equation, a preliminary step is to graph it.

If you do so, you will see that there are four real roots at approximately - 3, - 2, and 3 and what appears to be a fourth root exactly at 2. You can test that 2 is indeed a root.

\(\displaystyle 2^4 - (14 * 2^2) + 2 + 38 = 16 + 2 + 38 - (14 * 4) = 56 - 56 = 0.\)

If you have a good graphing calculator, you can zoom in on the other roots to get approximate answers for the other three roots.

If you do not have such a calculator, you can use numerical methods to get an approximation to any desired degree of accuracy. As Subhotosh Khan said, the obvious one to use is the Newton-Raphson method because, from your graph, you already have good initial approximations.

Those two methods do not give exact answers. If you are required to give exact answers, you can use the other method of solution suggested by daon.

\(\displaystyle f(x) = x^4 - 14x^2 + x + 38\ and\ f(2) = 0 \implies \exists\ g(x) = ax^3 + bx^2 + cx + d\ such\ that\ (x - 2) * g(x) = f(x) \implies\)

\(\displaystyle g(x) = \dfrac{f(x)}{x - 2} = \dfrac{x^4 - 14x^2 + x + 38}{x - 2} = x^3 + 2x^2 - 10x - 19.\)

You can then find exact solutions (involving radicals) for x by using the cubic formula. Click on the following url for an explanation of the cubic formula:http://www.math.vanderbilt.edu/~schectex/courses/cubic/
 
JeffM said:
There are at least five methods to solve this problem if you take the route of substitution. There may be more that do not involve substitution.

As you have found out, substitution leads to the quartic equation \(\displaystyle x^4 + (0 * x^3) - 14x^2 +x + 38 = x^4 - 14x^2 + x + 38 = 0.\)

Good job.

Through some method that you have not disclosed, you appear to have found out on your own that x = 2 is a solution to that quartic, which entails that x = 2 and y = 3 is ONE EXACT solution to your original problem.

One method to find other EXACT answers is to use either the formula for the quartic or the formula for the depressed quartic. Those, however, are horribly inefficient ways to solve the problem exactly and are highly error prone.

Starting from the quartic equation, a preliminary step is to graph it.

If you do so, you will see that there are four real roots at approximately - 3, - 2, and 3 and what appears to be a fourth root exactly at 2. You can test that 2 is indeed a root.

\(\displaystyle 2^4 - (14 * 2^2) + 2 + 38 = 16 + 2 + 38 - (14 * 4) = 56 - 56 = 0.\)

If you have a good graphing calculator, you can zoom in on the other roots to get approximate answers for the other three roots.

If you do not have such a calculator, you can use numerical methods to get an approximation to any desired degree of accuracy. As Subhotosh Khan said, the obvious one to use is the Newton-Raphson method because, from your graph, you already have good initial approximations.

Those two methods do not give exact answers. If you are required to give exact answers, you can use the other method of solution suggested by daon.

\(\displaystyle f(x) = x^4 - 14x^2 + x + 38\ and\ f(2) = 0 \implies \exists\ g(x) = ax^3 + bx^2 + cx + d\ such\ that\ (x - 2) * g(x) = f(x) \implies\)

\(\displaystyle g(x) = \dfrac{f(x)}{x - 2} = \dfrac{x^4 - 14x^2 + x + 38}{x - 2} = x^3 + 2x^2 - 10x - 19.\)

You can then find exact solutions (involving radicals) for x by using the cubic formula. Click on the following url for an explanation of the cubic formula:http://www.math.vanderbilt.edu/~schectex/courses/cubic/
Click to expand...
but i didnt get how we can divide it by (x-2)? i forgot it(.
thank you very much!
 
Subhotosh Khan said:
Be polite - we have many multilingual helpers here - for example I know of at least one who can read, write and speak fluently in four languages. You know more languages than that? I can curse people in seven/eight languages, thanks to the diversity of graduate school.
/QUOTE]
i will take these all into account,but i am so sorry, these do nothing to do with my theme. and thank you for your help in math.
Click to expand...
 
Last edited:
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thanks to all people that helped me. Khan i am grateful to you very much. I dont say i am clever, no, of course u are more clever than me. thanksto all of you
 
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