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Background: This problem is from my textbook: Let $x^{2}-(m-3) x+m=0 (m \in R)$ be a quadratic equation. Find the values of $m$ for which at leat one one root lies in $(1,\ 2).$

The solution too this problem given in the book is like this:

Case I: Exactly one root lies in $(1,\ 2)$. So, $$ f(1) f(2)<0 \Rightarrow m>10 $$

Case II: Both the roots lie in the interval $(1,\ 2)$. Then, $ D \geq 0 \Rightarrow(m-1)(m-9) \geq 0 $ $$\Rightarrow \quad m \leq 1 \text{ or } m \geq 9.$$ $\text{also } f(1)>0$ and $f(2)>0$ $$\Rightarrow 10>m$$ $\text{and }1<-\frac{b}{2 a}<2$ $$\Rightarrow 5<m<7$$ Thus, no such $m$ exists.

Hence, $m \in(10, \infty)$ .

Question: But how did author decide which "properties" to use to solve such problems? In case I, exactly one roots lies in (1, \ 2), so the followings are the properties which are satisfied by such quadratic:

  • The vertex of the quadratic is larger than $1$ but less than $2$.
  • and Sum and products of the roots will be positive ie let.
  • and the graph of the quadratic is concave up.
  • and $f(1)f(2)\lt 0$
  • and $af(2)\gt 0$
  • and $af(1)\gt 0$

and so on... but the author just chose the last 3rd property, why? Surely we will get many other inequalities and equations, after "mathematicalizing" those properties, involving the term "$m$", shouldn't we solve all the inequalities and equation to get the solution; in the end there might be a chance that other equations/inequalities could have "limited" the set of possible values of $m$, that is we may get extraneous solutions to the problem.

[ Note the word "and", all these properties are satisfied, so suppose that after solving, say 1st property we get something like $m\lt 0$, then we will have to take intersection of this set and $m \in(10, \infty)$. ]

Similarly I can list many properties for second cases also and again the same problem.

So why is the author so sure that solving that specific inequality will give him solution set without any extraneous solutions? Indeed he did got it right. Surely there must be a way to know that after solving a specific equation/inequality generating from certain property will led you to a solution set without any extraneous solutions.

Osmium
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  • @Ryan G Yes I know that, but my point was why author solved it only and still got right answer? – Osmium Jul 17 '21 at 09:58
  • Same polynomial, related question: https://math.stackexchange.com/questions/4176361/let-x2-%e2%88%92m%e2%88%923xm-0-m-in-r-be-a-quadratic-equation-find-the-value-of-m-fo/4355628#4355628 –  Jan 14 '22 at 10:00

2 Answers2

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You have shown that "there exists a root in $(1,2)$" $\implies m > 10$. But this is not yet an equivalence, as you say there could be extraneous solutions introduced. However, note that the inverse is simple. We have: $m>10 \implies f(1)f(2) < 0 \implies$ "there exists a root in $(1,2)$".

  • Yes, but I don't quite understand how that explains which inequality/equation should one select. – Osmium Jul 17 '21 at 10:00
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    It doesn't matter what method you select as long as it produces the answer. It has been shown that "there exists a root in $(0,1)$" is equivalent to $m>10$. If you can use another method to get the result then go for it. –  Jul 17 '21 at 10:04
  • so am I correct to understand that we cant decide which equation/inequality we should solve first, but if we get a solution which reversely imply our questions condition then there is no need to solve other eq/ieq? – Osmium Jul 17 '21 at 10:35
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    @IDKWTD: There is no deterministic procedure by which you can figure out how to solve an equation or inequality. For example, suppose you want to find all real $x > 0$ such that $x^x+x^2+1 = 3x$. There is no obvious way to do it. But graphing it quickly suggests that the expression is zero at $x = 1$ and positive elsewhere, and so suggests that we prove that it is convex and has zero derivative at $x = 1$. – user21820 Jul 17 '21 at 18:10
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For case I, the author could have been clearer, and written

  1. exactly one root lies in $(1, 2) \iff f(1) f(2)<0 \iff m>10$

instead of

  1. exactly one root lies in $(1, 2) \implies f(1) f(2)<0 \implies m>10.$

You're right that strictly speaking, statement (2) merely displays a candidate solution $\{m\in\mathbb R\mid m>10\},$ of which the actual solution is a subset. On the other hand, statement (1) assures the reader that “$m>10$” and “exactly one root lies in $(1, 2)$” are indeed equivalent statements, and, as such, $\{m\in\mathbb R\mid m>10\}$ is the actual solution set with no extraneous portion.

It is valid to write statement (1) instead of the weaker statement (2), because we observe that both statement (2) and its converse are true, since the chain of reasoning is correct in both the forward and reverse directions.

Case II

both (distinct or repeated) roots lie on $(1,2)\implies$ [nonnegative discriminant and line of symmetry lies on $(1,2)]\implies [D\geq0\;$ and $\;1<-\frac{b}{2a}<2] \implies m$ has no possible value

ryang
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  • The author is very bad. Do you have recommendation for a book(s), which includes topics like Rolles theorem, location of roots, nature of roots, inequalities? – Osmium Jul 17 '21 at 11:07
  • Thanks, I found this book by Hung-Hsi Wu, his(her?) book is very rigorous and explains everything well unlike my current book. Thanks again. – Osmium Jul 17 '21 at 11:54
  • @IDKWTD Happy to have helped! – ryang Jul 17 '21 at 12:00
  • One last question; if I was asked to show that a quadratic equation has both roots btwn two numbers, say $k_1$ and $k_2$, then do I just need to shows that $D\geq 0$ and $k_1\lt \frac{-b}{2a}\lt k_2$? My book also mentions that $af(k_1)\lt 0$ and $af(k_2)\lt 0$ are also necessary. – Osmium Jul 17 '21 at 12:19
  • @IDKWTD Yes, that's right: I've just expanded my answer in agreement with your suggestion. (Incidentally, for case II, $“\iff”$ is unnecessary since there is no extraneous solution to worry about.) – ryang Jul 17 '21 at 14:23