I came up with this question in my spare time for pleasure but have hit a brick wall, ironically enough. It is a simple-looking simultaneous equation with two variables x and y, where we solve for all real solution pairs for x and y. Here are the simultaneous equations in question:



*Sorry I just joined Stackexchange so I still have no idea how to format the question like how I did it on paper.

I have tried substituting the first equation into the second, vis-a-vis having $x^{1/2}=a$:


Following through with the Quadratic Formula and attaining a real solution for y, I substituted it back into the second equation to get a fairly long new equation. After extensive number crunching and other processes, including long division and Newton's Method (the equation eventually got to the 10th exponent), I ended up with something along the lines of $y=~0.00734$, which makes $x=~0.25005$.

*There is no decimal place/significant digit requirement for this question, as I chose to write each answer to the $5$th decimal place for convenience

*I have also run the entire process one more time, this time substituting with x, but it also didn't seem to go anywhere

However, upon checking with Wolfram Alpha and graphing with Desmos, I found out that the simultaneous equations DO have $2$ sets of real solutions, but the one I've attained was NOT one of them.

I am aware that I could simply punch these equations into a computer and have them run the numbers for me and spit out a solution, but I'd still like to solve it by hand and, most importantly, get to know the methods and processes involved in reaching the solution. Therefore, I would appreciate anyone's help in order to tame this little monster I created.

  • $\begingroup$ Hello :-). I have formatted your question a little. You can learn how to format from math.meta.stackexchange.com/questions/5020/… and then complete rest of formatting $\endgroup$ Commented Aug 29, 2021 at 4:44
  • 1
    $\begingroup$ Yes, you get a polynomial equation of degree $10$ with no "nice" roots, and, yes, you probably made an error somewhere in your calculations. It's hard to tell more since you don't show those calculations. $\endgroup$
    – dxiv
    Commented Aug 29, 2021 at 4:49

1 Answer 1


Take the first equation : this is a quadratic in $\sqrt x$; its two roots are $$x_1=\frac{1}{8} \left(4 y^2+1-\sqrt{8 y^2+1}\right)\qquad x_2=\frac{1}{8} \left(4 y^2+1+\sqrt{8 y^2+1}\right)$$ Plug in the second equation to get long and nasty expressions in $y$ (after squaring, they could be transformed into polynomials of degree $11$ !!). Plot them.

With $x_1$, you will see one single real root around $y=-5.3$.

With $x_2$, you will see two real roots around $y=-2.5$ and $y=0.3$.

Zoom more and more around the roots or use Newton-Raphson method.

The problem is that squaring introduces extra solutions. To better see your problem consider the norm $$\Phi=\Big[2 x-y^2-\sqrt{x} \Big]^2+\Big[\frac{y}{x}+y^3-1+x^2 \Big]^2$$ and make a countour plot of $\Phi$. You should notice that the only roots correspond more or less to $(0.33,0.28)$ and $(4.2,-2.5)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .