4

Find the range of values of $m$ in the quadratic equation $x^2+mx+6=0$ such that both the roots of the equation $\alpha,\beta<1$.

My attempt - it is given that

$\alpha<1$ and $\beta<1$

$\rightarrow \alpha+\beta<2$

But $\alpha+\beta=-m$

Thus $m>-2$.

But this solution doesn't involve the coefficient term i.e. $6$.

Any solution to the above question is appreciated.

Snehil Sinha
  • 1,179

5 Answers5

4

You made a logical fallacy as old as time.

What you have proven is:

If $\alpha, \beta<1$, then $m>-2$

What you HAVE NOT PROVEN is:

If $m>-2$, then $\alpha, \beta < 1$.

You can easily see that the second statement is false, since if $m=0>-2$, the equation has no real solutions.

5xum
  • 123,496
  • 6
  • 128
  • 204
  • Sir so how do we find the range of m.please can u post ur answer. – Snehil Sinha Aug 19 '15 at 09:59
  • @AxelBlaze Try to write the explicit formulas for $\alpha$ and $\beta$ – 5xum Aug 19 '15 at 10:03
  • @gebra But the statement "for all $m>-2$, $a,b<1$" is false. That's my whole point. If $m=0$ is a true statement, then $a,b<1$ is not a true statement. – 5xum Aug 29 '15 at 12:46
  • it depends on how you interpret the question, I would consider the bound $a,b < 1$ only a statement about the real part of a and b. If the imaginary part must be 0, then they should state that, or otherwise state that both roots of the equation a,b, < 1 are in the set of reals. – jg mr chapb Aug 29 '15 at 18:57
  • @gebra Look, however you want to spin this, the answer to the question "when are both roots of $x^2 + mx + 6$ smaller than $1$" is not "when $m>-2$". Also, the standard way of interpreting $x<1$ is "$x$ is real and smaller than $1$". – 5xum Aug 29 '15 at 21:59
2

This is a step-by-step description of how I would do it. There might be more clever ways out there, and there might be unanticipated problems with this approach, but it is the first thing I would try:

  1. Write an explicit formula for $\alpha$ and $\beta$ (it should use both the constant coefficient $6$ and the linear coefficient $m$)
  2. Decide which of $\alpha$ and $\beta$ is the biggest
  3. Set that root to be less than $1$
  4. Solve
Arthur
  • 199,419
2

As you see solving for $x$, one can start from the system $$\begin {cases} y=mx \\ \\ y=-x^2-6 \end {cases}$$ studying the intersections of a generic line passing through the origin ($m$ is just its slope) with a (fixed) parabola.

After checking the lines of tangency with $m=\pm \sqrt {24}$ and the line passing through the point $(1,-7)$, it is easy to deduce that both the roots are less than $1$ for every (and only) $m \ge \sqrt {24}$ .

Tony Piccolo
  • 4,426
1

Find the range of values of $m$ in the quadratic equation $x^2+mx+6=0$ such that both the roots of the equation $\alpha,\beta<1$.

Our graph has its wings pointing upwards. When both roots are less than one, both wings of the function's graph transect the OX axis to the left of the point $x=1$.

In order to be sure that both transection points are to the left of $x=1$, we need to know:

  1. There exist two transection points: hence, $D>0$
    (If $D=0$, there will only be one transection point, meaning only one root; if $D<0$, there will exist no transection points: no roots).
  2. If we draw a vertical line from $x=1$, it will strike the graph above the OX axis: hence $f(1)>0$
    (Indeed, if it strikes the graph below the OX axis, that would mean that the point $x=1$ lies between the two roots)
  3. But what if it will strike the left wing of the graph, not the right one? What if the whole shebang lies to the right of $x=1$? In order to exclude this possibility, let's say that the lowest point of the graph should also lie to the left of $x=1$. Hence, $x_0<1$. The formula for $x_0$ is $x_0=\frac{-B}{2A}$

Then I would unite these three conditions in a system and solve it.

enter image description here

CopperKettle
  • 603
  • 1
    If you are interested, your method is described by A.Tartinville in his book (1886). By the way, x0=-B/A or -B/2A ? – Tony Piccolo Aug 19 '15 at 18:42
  • @TonyPiccolo - thank you for the correction! It's $\frac{-B}{2A}$ of course! – CopperKettle Aug 19 '15 at 18:55
  • 2
    @TonyPiccolo - thanks for the mention of Tartinville. I took the method from a website run by Igor Yakovlev, a Moscow Physics Institute graduate and a math teacher. – CopperKettle Aug 19 '15 at 18:59
  • @CopperKettle Can u please also send the link for the translated page to english?I am an Indian and I can't understand Russian! – Snehil Sinha Aug 20 '15 at 12:27
  • @AxelBlaze - sorry, I have no information on similar self-study pages in English. The webpage is in Russian, of course. – CopperKettle Aug 20 '15 at 12:30
  • @AxelBlaze - hey, I've got an idea: you might ask a "soft question" at StackExchange, providing the link to my answer and requesting weblinks to similar pages in English (i mean with explanations on quadratics) – CopperKettle Aug 20 '15 at 12:32
  • @CopperKettle A nice idea.But unfortunately I have no idea about soft questions as I have never posted them.Still I will try to post one.I thus request you to please edit my question if there's any mistake in it or if the language might be improved.And should I place my soft question on maths stackexchange or any other stack exchange community? – Snehil Sinha Aug 20 '15 at 12:39
  • @AxelBlaze - place it here, on maths SE. I've seen such questions. Here's an example – CopperKettle Aug 20 '15 at 12:44
  • @CopperKettle CAn you please go through my soft question and suggest corrections (if any). (http://math.stackexchange.com/questions/1403858/soft-question-weblinks-to-pages-with-explanation-on-quadratics) – Snehil Sinha Aug 20 '15 at 13:02
0

Just use the pq-formula:
$$x_{1/2} = -\frac{m}{2} \pm \sqrt{\frac{m^2}{4}-6}$$ from that we ca deduce:
1) $m^2 \geq 24$, because the radiant under the root can't be negative
2) $$-\frac{m}{2} \pm \sqrt{\frac{m^2}{4}-6} < 1$$ just solve that and you have your range

Inuyaki
  • 226