5

I am running into a problem of solving the following system of 2 equations and 2 unknows $x,y$ over the real

$x \left(4 x^2+1\right)+(y-3) \sqrt{5-2 y}=0$ and $4 x^2+2 \sqrt{3-4 x}+y^2-7=0$

This problem belong to the Vietnamese University Entrance exam of Block A (Math, Physic and Chemistry) in 2010

My questions are:

1/ How to solve this challenging system

a/ Exactly in a systematic way

b/ Using numerical method

2/ Is there a general way to solve system with radical like this ?

Note that the solution is $(x,y)=(\frac{1}{2} , 2)$

The link to the full exam paper is

https://toanmath.com/2015/07/de-thi-va-dap-an-mon-toan-khoi-a-nam-2010.html

Thank you for your enthusiasm

Edit: To solve this system exactly, one can follow the method dictate from this post:

General Principles of Solving Radical Equations

Tuong Nguyen Minh
  • 363
  • I’m commuting so cannot solve it yet. But IMO, the answer for is no, this kind of problem doesn not have systematic way, cannot be solved by numerical method and there is no geberal way – NN2 Oct 29 '22 at 14:05
  • 1
    One trick would be that you set $5-2y=3-4x$ and check if that leads to a solution. – garondal Oct 29 '22 at 14:41
  • @garondal it seems to work but what's next ? – Tuong Nguyen Minh Oct 29 '22 at 15:05
  • 2
    The first equation is equivalent to $5-2y=4x^2$, with $x\ge0$, but it's not really obvious how to get there... – Théophile Oct 29 '22 at 15:07
  • 1
    Using my ansatz you would get $y-1=2x$ and then you can replace every $y$ with $x$ and then get rid of the square root term. But you will still end up with $0=-4x^3+4x^2+11x-6$. So I think this is not the nicest solution. – garondal Oct 29 '22 at 15:53
  • 1
    Algebraic geometry is not an appropriate tag for this question. From the tag description: "This tag should not be used for elementary problems which involve both algebra and geometry." Please read the tag descriptions before using them in the future. – KReiser Oct 29 '22 at 18:34
  • @garondal From there, in the hope that the answer is rational, you could use the rational root theorem to find it. There are $24$ cases to check ($\pm\frac11, \pm\frac12, \ldots, \pm\frac64$), but we can eliminate almost all of them by using the constraint $0 \le x \le \frac34$. Only three cases remain: $\frac12$, $\frac14$, and $\frac34$. All of this, however, represents a lot of work based on lucky guesses... – Théophile Oct 29 '22 at 18:45
  • 1
    Was this the actual question in the exam, or was this something that came out when you worked on a problem in, say, analytic geometry? – Jyrki Lahtonen Oct 29 '22 at 19:19
  • 1
    A "general" method here is to remove the square roots and get a system of two polynomial equations (knowing that squaring may introduce extra solutions). Then you can eliminate one of the variables. Gröbner basis techniques allow us to eliminate $x$ and we end up with a degree $24$ polynomial equation on $y$. That has only two real roots, $y=2$ and $y=-2$, giving points $(1/2,2)$, $(-3/2,-2)$, but the latter is a false solution (wrong sign for square root). The other 22 solutions are complex. I doubt that even in Vietnam students are taught this :-). That's why I asked the previous question. – Jyrki Lahtonen Oct 29 '22 at 19:32
  • 2
    In theory kids could be trained to spot the solution in @Théophile's comment (once you spot it, it is not hard to show that it is the only real solution for $y$ in the first equation)., but that is a bit outlandish. That's why I came up with the theory that you were working on a problem, and one approach lead to this system of equation, when another approach would give a smoother ride. – Jyrki Lahtonen Oct 29 '22 at 19:36
  • 1
    Isolating the radical in the first equation, you get a cubic in $y$ which has a factor linear in $y$ and a factor which has no real solutions. The first factor is the one Theophile found, and I agree that spotting this is a bit rough, so I wonder if Jyrki's hypothesis may be true. – KReiser Oct 29 '22 at 19:42
  • If one eliminates $x$ using $3-4x=5-2y$, and then the square roots, one gets in the end $(y^2-4)(y-5)=0$. – Intelligenti pauca Oct 29 '22 at 20:34
  • Thanks for adding the link! So I was wrong. Sorry about that. – Jyrki Lahtonen Oct 30 '22 at 05:09
  • What's wrong with the solution proposed on the page you linked? – Intelligenti pauca Oct 30 '22 at 14:04
  • @Intelligentipauca the solution from the ministry of education follows the line of thinking "solving by knowing the answer" . This is not really fair. – Tuong Nguyen Minh Oct 30 '22 at 18:59
  • To me that solution doesn't look worse than the solutions proposed here. – Intelligenti pauca Oct 30 '22 at 22:09

2 Answers2

2

First, what can we say about the number of possible solutions? Well, when $y > 0$, the first curve has positive slope and the second negative. So there is at most $1$ solution with $y > 0$. Could there be a solution with $y \le 0?$ From the first equation, that would mean $x(4x^2+1) \ge 3\sqrt{5} > 5$, i.e., $x > 1$. But that would make the radical in the second equation complex, so we can't have $y \le 0$. Thus the problem has at most 1 solution, and it has positive $y$.

Let's get more bounds on the possible solutions. Going back to the first equation, we note that $y \le 5/2$ is required for the radical to be real, and as such $(y-3)\sqrt{5-2y}$ is never positive. Therefore $x(4x^2+1)$, and hence $x$ itself, is nonnegative. And of course, from the radical in the second equation, we have $x \le 3/4$. The solution is now restricted to the interval $(x, y)\in [0, 3/4]\times(0, 5/2]$.

Now since this is an exam problem, the radicals probably can't be too "bad". They're likely to be in $\mathbb Z$, very likely to be in $\mathbb Q$, and, if it comes to it, almost certainly in $\mathbb Q[\sqrt{n}]$. Starting with $\mathbb Z$, the only options consistent with the bounds are $\sqrt{4-3x} \in \{0, 1\}$ and $\sqrt{5-2y}\in \{0, 1, 2\}$. Trying all 6 possible combinations finds a solution: $(x, y) \in (1/2, 2)$.

eyeballfrog
  • 23,253
2

1/ After replacing $x\rightarrow\frac{x}{2}$ and $y\rightarrow2y$(for which we need to understand before that $\left(\frac{1}{2},2\right)$ it's one of solutions) we obtain the following system: $$\begin{cases}\frac{x}{2}(x^2+1)+(2y-3)\sqrt{5-4y}=0 \\ x^2+2\sqrt{3-2x}+4y^2-7=0 \end{cases}. $$ The first equation gives $$x^3+x+(4y-6)\sqrt{5-4y}=0.$$ Let $\sqrt{5-4y}=t$.

Thus, $$x^3+x+(-t^2-1)t=0,$$ which gives $x=t$ because the expression $t^3+t$ increases.

Id est, $$x=\sqrt{5-4y}$$ or $$y=\frac{5-x^2}{4}$$ and from second equation we obtain $$x^2+2\sqrt{3-2x}+\frac{(5-x^2)^2}{4}-7=0,$$ where $0\leq x\leq\frac{3}{2},$ or $$x^4-6x^2-3+8\sqrt{3-2x}=0$$ or $$x^4-6x^2+5+8\left(\sqrt{3-2x}-1\right)=0$$ or $$(x-1)\left((x+1)(5-x^2)+\frac{16}{1+\sqrt{3-2x}}\right)=0,$$ which gives $x=1$, $y=1$ and we solved the starting system: $\left\{\left(\frac{1}{2},2\right)\right\}$

2/ I think does not exist a general way

Michael Rozenberg
  • 203,855
  • 1
    This pretty much is a valid solution. That's neat! – Hitesh Mar 06 '25 at 10:13
  • 1
    Yes this is pretty much follow the same spirit of the solution from the Ministry of Education. Solving by knowing the answer and then legitimize them. This is how most exam in my country goes. So I hope that there could be a general method to tackle this. Thank you, I have already upvoted your answer ! – Tuong Nguyen Minh Mar 06 '25 at 11:59
  • @Tuong Nguyen Minh It's known and standard trick. For example, the equation $x^{x^{2025}}=2025$ we can solve by the same trick. – Michael Rozenberg Mar 06 '25 at 12:23
  • What would be an effective way to guest the solution in the first place ? – Tuong Nguyen Minh Mar 09 '25 at 06:35