Conditions of the system of equations.

Question

Find m to the equation:$$\left\{ \begin{array}{l}2x^3-\left(y+2\right)x^2+xy=m\,\,(1)\\x^2+x-y=1-2m\,\,(2) \end{array} \right.$$have experience

My try: From $(1)$ and $(2)\,\Rightarrow $: $4x^3-2\left(y+2\right)x^2+2xy+x^2+x-y=1\\\Leftrightarrow 4x^3-3x^2+x-1=y\left(2x^2-2x+1\right)\\\Leftrightarrow y=\dfrac{4x^3-3x^2+x-1}{2x^2-2x+1}\\\Leftrightarrow y=2x+\dfrac{1}{2}-\dfrac{3}{2\left(2x^2-2x+1\right)}$

From $(2)\,\Rightarrow $ $2m=1+y-x^2-x\\=1+2x+\dfrac{1}{2}-\dfrac{3}{2\left(2x^2-2x+1\right)}-x^2-x\\=-x^2+x+\dfrac{3}{2}-\dfrac{3}{2\left(2x^2-2x+1\right)}$

And I don't know how to contine, plz help me :(

The result is: $m\le 1-\dfrac{\sqrt{3}}{2}$

Answer 1

From your last expression, after division by $2$, the value of $m$ is $$m=\frac{-x^4+2x^3-x}{2x^2-2x+1}.$$ To maximize this we find its derivative and get $$m'(x)=\frac{-(2x-1)(2x^4-4x^3+4x^2-2x-1)}{(2x^2-2x+1)^2}.$$ So there is a critical point at $x=1/2$ and we need to set the fourth degree polynomial in the numerator to zero. [The denominator is positive for all $x$].

It is maybe just luck, but if we substitute $x=1/2+t$ into the fourth degree polynomial it becomes $2t^4+t^2-11/8$ which is quadratic in $t^2$ and leads to $t^2=(-1+\sqrt{12})/4.$ [We cannot use the negative sign choice here since we need $t^2 \ge 0.$ The negative choice corresponds to the pair of nonreal conjugate complex solutions to the fourth degree equation.] So we now have three critical points for $m$, namely $1/2$ and $1/2 \pm \sqrt{-1+\sqrt{12}}/2.$ The value of $m$ is largest among these when $x=1/2+\sqrt{-1+\sqrt{12}}/2,$ and for this $x$ we have $m=1-\sqrt{3}/2.$ Note also that since clearly $m \to -\infty$ as $x \to \pm \infty$ we can be sure the maximum of $m$ must occur at one of the above critical points.

EDIT After a closer look it turns out the maximum $m=1-\sqrt{3}/2$ occurs at both of the critical points $x=1/2 \pm \sqrt{-1+\sqrt{12}}/2.$ This seems unusual since the expression for $m$ is not symmetrical around $x=1/2,$ but somehow the two critical points, and their $m$ values, are symmetrical around $x=1/2.$ I couldn't find a reason for this symmetry from the initial equations.

Answer 2

Hint:

You have $3$ unknowns and $2$ equations. Therefore you shouldnt expect to solve them uniquely. There is one more information that needs to be used. $x$ and $y$ are real. For example when you have $2m=1+y-x^2-x$. You can write it as a polinomial of $x$ as $x^2+x-y-1-2m=0$ when you find the roots of this polinomial you will get $$\frac{-1\pm\sqrt{1-4(-y-1-2m)}}{2}$$ if $x$ is positive you will get an inequality as $1-4(-y-1-2m)>0$. If you apply the same idea to the other equations you should be able to get what you wanna.