Solve the given simultaneous equation

[chwala]

Homework Statement

Solve for $x$ and $y$ given;
$\sqrt x+ y=7$
$x+\sqrt y=11$

Relevant Equations

Simultaneous equations

I saw this problem on you tube ... The working to solution is much clear to me. Find the link here;

I am now trying to look at an alternative approach. My thinking is as follows;
$\sqrt x=7-y$
$x=(7-y)^2$

$(7-y)^2+\sqrt y=11$
$49-14y+y^2+\sqrt y=11$
i then let $\sqrt y=m$ giving me,
$m^4-14m^2+m+38=0$ at this point i am now exploring ways of solving quartic equations...looking at it...then will respond on my finding...i hope i am on the right path...

 

[malawi_glenn]

What is the point of this thread?

 

[chwala]

What is the point of this thread?

Am I on the right path? My post is clear. I am exploring on alternative approach.

 

[malawi_glenn]

My post is clear

There was no question.

I can give you a hint regarding the quartic equation. Can you find a simple positive m-value which will solve it?
Also, you can with derivative method figure out how many positive roots that quartic equation will have.
You can also use that $m < 7$ to pinpoint down the only solution you care about in that quartic equation.

 

[chwala]

Yes, I am looking into this ...looking at the numerical methods...I'll get back once done...

 

[malawi_glenn]

I just solved it following your work with no aid of any calculator.
Was kinda fun :)

 

[Steve4Physics]

I am now trying to look at an alternative approach.

EDITED

Is an algebraic solution essential? Can I point out (if not already evident) that the problem can be solved using some 'guesswork'.

Click on the spoiler below for hints...

From inspection:
- x and y look likely to be not only integers but also perfect squares; we can try this to see where it leads
- values of 0 or 1 for x and y are easily excluded;
- y must be less than 6 giving a solution for y.

 

[chwala]

I noticed a mistake in the video...on the factorisation part; ought to be
...
$(\sqrt x-\sqrt y)(1-\sqrt x+\sqrt y)=-4$ ...

And it looks like trial and error approach as the method on the video assumes solutions then works backwards to determine the value of variables $x$ and $y$. I would rather a step-step working to solution. I will look at this today...

 

[chwala]

There are various techniques in solving quartic equations; phew a bit technical...Ferrari method, Durand- Kerner method ...need to read on them.

 

[malawi_glenn]

You don't need those

 

[Delta2]

This polynomial we end up with $$m^4-14m^2+m+38$$ is very interesting, it has 4 roots ALL real...

 

[chwala]

This polynomial we end up with $$m^4-14m^2+m+38$$ is very interesting, it has 4 roots ALL real...

I have seen all the $4$ roots with help of calculator...
ie $-3.2831859, -1.8481265, 2$ and $3.1313125$

 

[chwala]

I just solved it following your work with no aid of any calculator.
Was kinda fun :)

Arrrrrrgh! I can now see that we can get the roots by using Factor theorem! Eeeeeeeeish!

 

[malawi_glenn]

Arrrrrrgh! I can now see that we can get the roots by using Factor theorem! Eeeeeeeeish!

I wrote that in post #4, don't you read the replies?

You first try with some solutions, you must have that 0 < m² < 7, so try with some solutions for m.
You will find pretty quick that m = 2 works.
Then do polynomial division to get a cubic polynomial.
Show that the resulting polynomial does not equal zero for any positive m less than sqrt(7), I used some basic curve sketch analysis with derivative method.

 

[malawi_glenn]

This polynomial we end up with $$m^4-14m^2+m+38$$ is very interesting, it has 4 roots ALL real...

What is so interesting with a quartic polynomial with four real roots?

 

[Delta2]

What is so interesting with a quartic polynomial with four real roots?

Well I don't know to be honest, for some reason my intuition told me that most polynomials of high degree have at least one complex root.

 

[chwala]

I got it! I will post my step-step solution later in the day...I used the quartic approach- converted to cubic terms...with some algorithm involved. Bingo talk later...

 

[chwala]

I wrote that in post #4, don't you read the replies?

You first try with some solutions, you must have that 0 < m² < 7, so try with some solutions for m.
You will find pretty quick that m = 2 works.
Then do polynomial division to get a cubic polynomial.
Show that the resulting polynomial does not equal zero for any positive m less than sqrt(7), I used some basic curve sketch analysis with derivative method.

This approach is straightforward man...i will post the approach that is used generally for quartic equations ..cheers

 

[malawi_glenn]

most polynomials of high degree have at least one complex root.

How do you compare the amounts of polynomals to quantify "most"?

This approach is straightforward man.

Yes it is, I give my 16-17y old students similar problems which are solved using this approach.

 

[Delta2]

How do you compare the amounts of polynomals to quantify "most"?

Pure intuition, there are infinite polynomials but my intuition told me at at least half of them have a complex root.

 

[chwala]

Homework Statement:: Solve for $x$ and $y$ given;
$\sqrt x+ y=7$
$x+\sqrt y=11$
Relevant Equations:: Simultaneous equations

I saw this problem on you tube ... The working to solution is much clear to me. Find the link here;

I am now trying to look at an alternative approach. My thinking is as follows;
$\sqrt x=7-y$
$x=(7-y)^2$

$(7-y)^2+\sqrt y=11$
$49-14y+y^2+\sqrt y=11$
i then let $\sqrt y=m$ giving me,
$m^4-14m^2+m+38=0$ at this point i am now exploring ways of solving quartic equations...looking at it...then will respond on my finding...i hope i am on the right path...

We have;
$m^4-14m^2+m+38=0$

On Using the approach given on this link https://www.1728.org/quartic2.htm

We shall have;
$f=-14$
$g=1$
$h=38$

Our cubic equation is;
$y^3-7y^2+\dfrac{44}{16}y-\dfrac{1}{64}=0$

On solving, we have

$y_1=6.583$
$y_2=0.412$
$y_3=0.0058$

giving us;

$p=2.566$
$q=0.6419$
$r=-0.0758$

substituting into the equations, we end up with;

$x_1=p+q+r-s=3.132$
$x_2=-p+q-r-s=-1.8483$
$x_3=-p-q+r-s=-3.2837$
$x_4=p-q-r-s=1.999999≅2$ which is the envisaged value that we are looking for...from here the working to solution will follow...

 

[malawi_glenn]

$x_4=p-q-r-s=1.999999≅2$

how come you don't get exactly 2?

 

[chwala]

how come you don't get exactly 2?

The approach is an approximate method (Numerical method) and not exact.

 

[WWGD]

Pure intuition, there are infinite polynomials but my intuition told me at at least half of them have a complex root.

There are uncountably-many polynomials over the Reals, uncountably many with Complex roots( degree 2 or higher) ; those that are multiples of $x^2+a ; a>0$ alone are enough*, and uncountably-many with Real roots: $( x-r_1)(x-r_2)(x-r_3)(x-r_4); r_i \in \mathbb R$
*Similar argument if you want all roots to be Real.

 

[WWGD]

I'd suggest in general to get a feel for it, as someone had suggested. And, maybe too, a more geometric approach using parabolas ( after squaring; set ## \sqrt u:= u, etc), though you'd have to go back-forth in the change of variables.

 

[Delta2]

There are uncountably-many polynomials over the Reals, uncountably many with Complex roots( degree 2 or higher) ; those that are multiples of $x^2+a ; a>0$ alone are enough*, and uncountably-many with Real roots: $( x-r_1)(x-r_2)(x-r_3)(x-r_4); r_i \in \mathbb R$
*Similar argument if you want all roots to be Real.

Hmm, can we say that the probability a random polynomial to have only real roots is 50%? Hard though to imagine how we define this probability as if we try to define it as a fraction, both the numerator and the denominator are uncountable many...

 

[anuttarasammyak]

I made wolfram draw graphs of y=11-x^2 and x=7-y^2 where $\sqrt{x},\sqrt{y}$ in the problem statement is transferred to x, y.

I could confirm that there is only one solution for x,y>0, obvious x=3, y=2.

Equation for x is
$$(11-x^2)^2=7-x$$
$$x^4-22x^2+x+114=0$$
$$(x-3)(x^3+3x^2-13x-38)=0$$

 

[MidgetDwarf]

How do you compare the amounts of polynomals to quantify "most"?Yes it is, I give my 16-17y old students similar problems which are solved using this approach.

Descarte's Rule Of Signs.

 

[malawi_glenn]

Descarte's Rule Of Signs.

Yeah but we do this in context with derivative of polynomial functions.
But thanks for the reminder! That is very useful trick.

 

[Orodruin]

Hmm, can we say that the probability a random polynomial to have only real roots is 50%?

Not without specifying a probability distribution on the polynomials. The result will depend on the distribution you impose.

Since both sets are aleph one, it is possible to construct a bijection between them.

 

[malawi_glenn]

Since both sets are aleph one, it is possible to construct a bijection between them.

I was thinking about that too. I recalled a problem in an real analysis book of mine which I did not read that long time ago which asked how many algebraic vs. trancendetal numbers (over the rationals) there are. If I remember correctly, there should be countable many ($\aleph_0$) algebraic numbers and uncountable ($\aleph_1$) trancendental numbers... I should revisit my notes at some points

 

[WWGD]

Hmm, can we say that the probability a random polynomial to have only real roots is 50%? Hard though to imagine how we define this probability as if we try to define it as a fraction, both the numerator and the denominator are uncountable many...

I guess we'd have to choose a distribution to determine probabilities.

 

[neilparker62]

Homework Statement:: Solve for $x$ and $y$ given;
$\sqrt x+ y=7$
$x+\sqrt y=11$
Relevant Equations:: Simultaneous equations

Presuming this is a Diophantine (integer only) equation , we don't have too many choices:

1 + 6
2 + 5
3 + 4

One number has to be a perfect square. Exclude 2 + 5. If x=1, y=6 and that doesn't work either.