# Discuss the number of solution for a system of parametric equations

I need to discuss the number of solutions of the following parametric system.

$$\begin{cases} x^2+(m-4)x-5m-1=0\\ 2\le x\le4 \end{cases}$$

There is a hint saying to put $$y=x^2$$

So far I've proceded in the following way: $$x^2$$ is not parametric so, unless I'm mistaken, the curve should represent a family of parables. Then the real solutions of the intersection with the $$y=0$$ axis are for $$m^2+12m+20≥0$$ , which gives $$m1≤−10 \vee m2≥−2$$

Unfortunately I'm struggling finding the intervals of $$m$$ where we get one or two solutions complying with the constraint $$2\le x\le4$$

The solution on the book is: 1 solution for $$−5/3 and 2 solutions for $$−2≤m≤−5/3$$

• I'm pretty sure it has infinite solutions for $2 \leq x \leq 4$. Did you mean a parametric form for the solutions? Also, I'm not sure how putting $y=x^2$ is helpful. Commented Sep 5, 2023 at 17:29
• It might be possible to express the solutions in terms of $m,$ by applying the quadratic equation. One would in that case have to say which $m$ values would lead to real solutions for $x$. [However if $x$ is allowed to be complex, nothing much can be said. Commented Sep 5, 2023 at 17:33
• Commented Sep 5, 2023 at 18:09
• If $x$ is real, then the two conditions place restrictions on the value of $m$ which can be found by resolving four cases. Perhaps that is what the exercise is asking for. Commented Sep 5, 2023 at 18:10
• Ok let's see how I've proceded: $x^2$ is not parametric so, unless I'm mistaken, the curve should represent a family of parables. Then the real solutions of the intersection with $y=0$ axis are for $m^2+12m+20 \ge 0$, which gives $m1 \le -10$ or$m2 \ge-2$. Now I shoud find the intervals of $m$ where we get one or two solutions falling in the given interval Commented Sep 5, 2023 at 18:37

For what values of $$m$$ does

$$$$f(x)=x^2+(m-4)x-(5m+1)\tag{1}$$$$ have solution(s) on the interval $$[2,4]$$?

The equation $$x^2+(m-4)x-(5m+1)=0$$ has, for appropriate values of $$m$$, real solution(s)

$$x=\frac{4-m\pm\sqrt{(m+2)(m+10)}}{2}$$

So in order for $$x$$ to be a real number it must be the case that either $$m\le -10$$ or $$m\ge-2$$.

Since the graph of $$y=f(x)$$ is a parabola there are at most two solutions for any given value of $$m$$ so we must find values of $$m$$ for which there are (A) one solution in $$[2,4]$$ or (B) two solutions in $$[2,4]$$.

Let us examine the double inequality.

$$$$2\le\frac{4-m\pm\sqrt{(m+2)(m+10)}}{2}\le4\tag{2}$$$$

This can be broken into the two double inequalities $$$$m\le\sqrt{(m+2)(m+10)}\le m+4\tag{3}$$$$

and

$$$$-m-4\le\sqrt{(m+2)(m+10)}\le -m\tag{4}$$$$

Fron the left inequality of (3) it can be concluded that $$m\ge-2$$ and from the right inequality that $$-2\le m\le-1$$ so $$-2\le m\le-1$$.

From the left inequality of (4) it can also be concluded that $$m\ge-2$$ and from the right inequality that $$-2\le m\le -\frac{5}{3}$$.

So (2) is telling us that either $$-2\le m\le-1$$ or $$-2\le m\le -\frac{5}{3}$$. Therefore we should expect solutions for solutions of $$f(x)=0$$ to occur for values of $$m$$ lying in the interval $$[-2,-1]$$ and that perhaps there is something interesting happening at $$m=-\frac{5}{3}$$.

So we should first find what solutions of equation (1) lie in $$[2,4]$$ for the three critical values $$m=-2,-\frac{5}{3},-1$$.

When $$m=-2$$ equation (1) has one double solution, $$x=3$$ lying in the interval $$[2,4]$$ which by convention counts as two solutions.

When $$m=-1$$, equation (1) has two solutions but only one, $$x=4$$, which lies in the interval $$[2,4]$$.

When $$m=-\frac{5}{3}$$, equation (1) has two solutions, $$x=2$$ and $$x=\frac{11}{3}$$, both of which lie in the interval $$[2,4]$$.

To find out what is happening for values of $$m$$ in the interiors of intervals $$\left[-2,-\frac{5}{3}\right]$$ and $$\left(-\frac{5}{3},-1\right]$$.

For $$m=-\frac{11}{6}$$ halfway between $$-2$$ and $$-\frac{5}{3}$$ we find two solutions of (1) lying in the interval $$[2,4]$$ so we conclude that for $$-2\le m\le -\frac{5}{3}$$ there will be two solutions.

For $$m=-\frac{4}{3}$$ halfway between $$-\frac{5}{3}$$ and $$-1$$ we find two solutions of (1), but only one of the two solutions lies in the interval $$[2,4]$$, so we conclude that for $$m\in\left(-\frac{5}{3},-1\right]$$ only one solution of (1) lies in the interval $$[2,4]$$.

Hints:

https://www.desmos.com/calculator/oxkjalmcnj

• express $$m=f(x)$$, this gives a rational function, find its range over the interval $$x\in[2,4]$$ (i.e. boundary values $$A$$ and $$B$$ and minimum in $$C$$).

• since $$f$$ is $$\searrow$$ then $$\nearrow$$ and continuous on $$[2,4]$$ you can study initial problem on intervals $$m\in[-2,-\frac 53]$$ and $$m\in[-\frac 53,-1]$$

• express the two solutions $$x_{min}$$ and $$x_{max}$$ with quadratic formula and find their range for the two intervals considered (hint: $$x_{min}\searrow$$ and $$x_{max}\nearrow$$)

• conclude