1
$\begingroup$

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 $−5/3<m≤−1$ and 2 solutions for $−2≤m≤−5/3$

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – dgeyfman
    Commented Sep 5, 2023 at 17:29
  • $\begingroup$ 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. $\endgroup$
    – coffeemath
    Commented Sep 5, 2023 at 17:33
  • $\begingroup$ Quick beginner guide for asking a well-received question + please avoid "no clue" questions. $\endgroup$ Commented Sep 5, 2023 at 18:09
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Sep 5, 2023 at 18:10
  • $\begingroup$ 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 $\endgroup$
    – MadBlack
    Commented Sep 5, 2023 at 18:37

2 Answers 2

1
$\begingroup$

For what values of $m$ does

\begin{equation} f(x)=x^2+(m-4)x-(5m+1)\tag{1} \end{equation} 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.

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

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

and

\begin{equation} -m-4\le\sqrt{(m+2)(m+10)}\le -m\tag{4} \end{equation}

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 $[2,4]$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

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