Problems with Inequalities

It seems like I am facing some confusion while handling with inequalities,I was doing some tasks where it is asked to find the interval of the variable,after some steps I deduced the the necessary expression but I am facing some problem while trying to deduce the final expression.

Consider for example: $$a^2 \ge 8;\qquad 0 < a < 6; \qquad a < \frac{11}{3}$$

For the start I write $a^2 \ge 8$ as $a \lt -2\sqrt{2} \quad\text{or}\quad a > 2\sqrt{2}$ but the final answer is $2\sqrt{2} < a < \frac{11}{3}$. How can this be deduced?

The original problem is as follows:

If both the roots of the equation $x^2-ax+2 = 0$ lie in the interval $(0,3)$, then find the value of $a$.

Since the two roots lie in the same interval, those three conditions follow.

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Sketching these inequalities on a number line should go a long way. – J. M. Dec 9 '10 at 13:34
That's was a nice hint J.M! – Quixotic Dec 9 '10 at 18:55

What you are forgetting is that you have more than one inequality.

The inequality $a^2\geq 8$, by itself, only gives you $|a|\geq \sqrt{8}=2\sqrt{2}$, or equivalently as you (almost) write: $$a\geq 2\sqrt{2}\qquad\text{or}\qquad a\leq -2\sqrt{2}.$$ Added: Note that you took $a^2\geq 8$, a weak inequality, and "translated" it into "$a\gt 2\sqrt{2}$ or $a\lt -2\sqrt{2}$." Technically, this is incorrect because you have now excluded the possibility of equality. Since you were starting with $\geq$, you should have continued with $\geq$ and $\leq$.

However, you have other inequalities that need to be satisfied as well.The second inequality tells you that $0\lt a \lt 6$ must be satisfied. How does this mesh with the inequalities you already have?

If $0\lt a \lt 6$ and $a\leq -2\sqrt{2}\lt 0$, then that tells you that $a$ must be both greater than and less than zero. This is a very big problem for $a$, which decides to solve this problem by ceasing to exist. There is no solution to the inequality $0\lt a\lt 6$ that also satisfy $a\lt -2\sqrt{2}$.

So putting these two together, you know that you have $a\geq 2\sqrt{2}$, and $0\lt a \lt 6$. Since satisfying $a\geq 2\sqrt{2}$ automatically guarantees you that $0\lt a$, you can combine them all by saying that the first two inequalities you have are equivalent to $$2\sqrt{2} \leq a \lt 6.$$ Finally, we need to add in the condition that $a\lt \frac{11}{3}$. Since $\frac{11}{3}\lt 6$, if $a$ is less than $\frac{11}{3}$ it will also automatically be less than $6$. It still needs to be greater than or equal to $2\sqrt{2}$, though. So combining them all you get that you must have $2\sqrt{2} \leq a \lt 6$ in order to satisfy all three (four, if you count "$0\lt a\lt 6$" as two inequalities) inequalities you have.

Okay, the original problem was: "if the two roots of $x^2-ax+2=0$ lie in the interval $(0,3)$, then find the values of $a$."

The roots are $\frac{a\pm\sqrt{a^2-8}}{2}$. So you assumed that you only needed to ask for $a^2-8\geq 0$ in order for the roots to be real. However, the use of the word "both" suggests that in fact they want there to be two distinct roots, which would require $a^2-8\gt 0$. If you take $a=2\sqrt{2}$, the lower limit I suggested earlier, then you have a single (repeated) root at $\sqrt{2}$ (which satisfies the condition). If they want two roots, then you need the discriminant positive, not merely nonnegative.

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+1,superb explanation,after reading your explanation, I think we can sum up the rules like this $\text { maximum of all condition } \lt \text{ or } \le \text { <variable> } \lt \text{ or } \le \text { minimum of all condition }$. – Quixotic Dec 9 '10 at 18:21
@Debanjan: that's if you have that kind of conditions. Sometimes you have disjunctions rather than conjunctions (like the first inequality of the OP). But yes, if you need to satisfy $x\leq a_1$, $x\leq a_2,\ldots, x\leq a_m$, then this is equivalent to $x\leq \min(a_1,\ldots,a_m)$. – Arturo Magidin Dec 9 '10 at 18:26
Thanks for writing Mr Magidin.But I have two doubts first it seems from the discussion so far for inequalities with "or" condition first step would be to chose the one that satisfies other and then apply the rule stated by Debanjan right ...? – Quixotic Dec 9 '10 at 18:48
The second one is in my module the answer suggested is $2\sqrt{2} \lt a \lt \frac{11}{3}$ – Quixotic Dec 9 '10 at 18:49
@Tretwick Marian: Updated the answer. I think the issue here is that when they say "two roots", they mean "two distinct roots", not "two roots, counted with multiplicity". – Arturo Magidin Dec 9 '10 at 22:03

Any polynomial inequality over the reals can be reduced to testing it on the intervals between its roots - where the polynomial has constant sign. This will yield a finite number of intervals where the inequality holds true. When you have multiple such inequalities you simply intersect the intervals for each polynomial to obtain the finite number of intervals where all of the inequalities are true. In fact the same idea generalizes to arbitrary first order multivariate statements about the reals - as Tarski has shown by his celebrated method of quantifier elimination for real-closed fields. It has an algorithmic form that is essentially a higher-dimensional cylindrical version of the finite interval decomposition for the above one dimensional case - google "cylindrical algebraic decomposition" or CAD algorithm.

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Everything you have done so far is correct. Once you get down to linear constraints, you essentially have to intersect all the permitted ranges. E.g. since $0<a<6$, the negative solutions $a<-2\sqrt{2}$ are out. Thus the only possibility is $a>2\sqrt{2}$. Now the remaining inequalities also give you an upper bound.

A good strategy is, as I said at the beginnig, to sketch all the ranges and then intersect them to find the range in which all inequalities are satisfied. If you had some "or"-conditions, you would take unions of ranges, rather than intersections.

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Could you explain If you had some "or"-conditions, you would take unions of ranges, rather than intersections. ? – Quixotic Dec 9 '10 at 18:42
@Tretwick Marian: Suppose you have a bunch of inequalities, $I_1$, $I_2,\ldots,I_k$, each with a particular solution set $S_1$, $S_2,\ldots,S_k$. If you want to find all values of $a$ that satisfy all the inequalities, then you need to take $S_1\cap S_2\cap\cdots\cap S_k$, because you need your $a$ to be a solution to each inequality, so you need it to be in each solution set. But if you want to find all $a$ that satisfy at least one of the inequalities, then your solution set would be $S_1\cup S_2\cup\cdots\cup S_k$... (cont) – Arturo Magidin Dec 9 '10 at 22:07
@Tretwick Marian: (cont)... For example: in order to satisfy "$0\lt a$ and $a\lt 6$", you take the solution set to $0\lt a$ (which is $(0,\infty)$), and the solution set of $a\lt 6$ (which is $(-\infty,6)$), and you intersect them $(0,\infty)\cap(-\infty,6) = (0,6)$ to get the solution. But in order to find the solution to "$a\lt -2\sqrt{2}$ or $a\gt 2\sqrt{2}$", you take the solution set to the first, $(-\infty,-2\sqrt{2})$, the solution set to the secont, $(2\sqrt{2},\infty)$, and you take their union $(-\infty,-2\sqrt{2})\cup(2\sqrt{2},\infty)$ as the solution set of the "system." – Arturo Magidin Dec 9 '10 at 22:09

For your original problem, here's the approach I'd take:

I know what the set of graphs of $y=x^2-ax+2$ looks like as $a$ varies:

Okay, the picture in my head isn't that clear, but I know that they all have y-intercept 2 and move in more or less that fashion as $a$ varies (the locus of the vertices of those parabolas is itself a parabola).

I know that I want the interval from when the equation starts to have real roots up to when the greater of its two roots is 3. Starting with the latter condition, $3=\frac{a+\sqrt{a^2-8}}{2}\implies(6-a)^2=a^2-8\implies 36-12a=-8\implies a=\frac{11}{3}$.

The earlier condition happens when $x^2-ax+2$ is a perfect square, so when $\frac{a}{2}=\pm\sqrt{2}$. The negative value of $a$ leads to the parabola that touches the x-axis on the negative side, so the relevant value is $a=2\sqrt{2}$.

From these two bounding values and the animated image in my head, the desired values of $a$ are $2\sqrt{2}<a<3$.

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No computer power during exam. – Quixotic Dec 10 '10 at 20:08
@Tretwick Marian: No computer needed. As I said, I already know what happens to the graph as $a$ changes (it's useful to know how varying each coefficient of a quadratic changes the graph); I put the computer graphic in to illustrate the picture in my mind. – Isaac Dec 10 '10 at 20:23