Find the real values of $m$ such that $a \leq b \leq c$ .

Question

Given a system of equations (where $m$ is a real number),

$(a + b) = 2m^2$ , $(b+ c) = 6m$ , $(c + a) = 2$ , find the real values of $m$ such that $a \leq b \leq c.$

What I Tried :- I tried to do it like this.

Suppose the equations are marked as $x,y,z$ in that order. Then by doing $(x - y + z)$ I get :-

$$2a = 2m^2 - 6m + 2$$ $$\rightarrow a = m^2 - 3m + 1$$

Now I find that $m = \frac{3}{2}$ is a solution iff $a = -\frac{5}{4}$ . From $a$ and $m$, I can deduce that $b = \frac{23}{4}$ and $c = \frac{13}{4}$ , which is not possible as it is given that $a \leq b \leq c$. So we can say that $m = \frac{3}{2}$ is not a solution.

I guess my work is actually fine, but the problem is that this is not the only answer at all. I have got some more cases to check with, [such as what would have happened if we had done $(x + y + z)$ or $(-x + -y + -z)$ $?$] and it is seeming like a very tiring work dealing with a lot of cases like that and finding what $m$ satisfies and what $m$ does not, I believe there is some more easier solution.

Can anyone help?

Answer 1

We have that

• $(a + b) = 2m^2 \implies a=2m^2-b$
• $(c + a) = 2 \implies c=2-a$
• $(b+ c) = 6m \implies b=6m-c=6m-2+a$

therefore

• $a=m^2-3m+1$
• $b=m^2+3m-1$
• $c=-m^2+3m+1$

and

$$a \leq b \leq c \iff m^2-3m+1\le m^2+3m-1\le -m^2+3m+1$$

$$ m^2-3m\le m^2-1\le 0 \iff \frac13 \le m \le 1$$