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The number of integral values of a for which $ax^2 - (4-2a)x - 8<0$ for exactly three integral value of x .what i did was find the roots of this equation lets say $x_1$ ,$x_2$ ,will this condition satisfy the criteria: $2<|x_1 -x_2|<4$ and a>0? Or is there some more elegant way to get the required condition in terms of roots?

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You're kind of on the right track. Let's suppose without loss of generality that $x_1 < x_2$ are the roots of $$f(x;a) = ax^2 - (4-2a)x - 8.$$ Then it is clear that we must have $a > 0$; otherwise, there will be infinitely many $x$ such that $f(x;a) < 0$. Moreover, since we require exactly three integer solutions, there must be an integer $k$ such that $$k-2 \le x_1 < k-1 < k < k+1 < x_2 \le k+2.$$ Now write $$f(x;a) = a(x-x_1)(x-x_2) = ax^2 - a(x_1 + x_2)x + ax_1 x_2,$$ hence $$a(x_1 + x_2) = 4-2a, \\ a x_1 x_2 = -8.$$ Solving this system in terms of $a$, we obtain $$\{x_1, x_2\} \in \{-2, 4/a\}.$$ Since $a > 0$, we must have $x_1 = -2$, $x_2 = 4/a$, and we require $3 < x_2 - x_1 \le 4$. This is because one root is already an integer. It follows that $a \in \{2, 3\}$, and $k = 0$.

heropup
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  • Sir it should be 4>(x2-x1)>=3 , hence answer is 2<a<=4 , and also Sir what will we do in case both roots r like function of a , like in this case we r lucky to have one independent of a (root) . –  Mar 24 '21 at 00:24
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    @WayBig Let me revise my answer; I was solving for $f(x;a) \le 0$ rather than $<0$. – heropup Mar 24 '21 at 00:31
  • Yeah Sir , One edit in the above comment by me , it should be 3<(x2-x1)<=4 , hence answer 2<= a<4 . And now Sir can u tell the method if both roots r function of a ? –  Mar 24 '21 at 00:38
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    @WayBig If both roots are a function of $a$, then the question becomes more complicated; e.g., if the roots are $g(a), h(a)$ in some order, then we need to determine for which values of $a$ that $h > g$, and when $g > h$, in order to solve the inequality $3 < |h(a) - g(a)| \le 4$. – heropup Mar 24 '21 at 00:38
  • Agree Sir , this may also happen right Sir if both r function of a and dont give roots as integers at no value of a, then the inequality can be reduced to 2? That is 2<|h(a)-g(a)|<=4 , u told for one integer case right Sir that mod condition ? –  Mar 24 '21 at 00:44
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    @WayBig That's right; the lower bound can change if neither $h$ nor $g$ are integers. – heropup Mar 24 '21 at 00:51
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The general quadratic $Ax^2 + Bx + C = 0$

can be converted to $\left(x + \frac{B}{2A}\right)^2 - \frac{B^2 - 4AC}{(2A)^2} = 0.$

So, a given value of $x$ will satisfy $Ax^2 + Bx + C < 0 \iff \left(x + \frac{B}{2A}\right)^2 < \frac{B^2 - 4AC}{(2A)^2}.$

With this query's quadratic, you have that $A,B,C = a, (2a-4), (-8)$ respectively.

This means that the values of $x$ that will satisfy the inequality are $$\left(x + \frac{2a-4}{2a}\right)^2 < \frac{(2a-4)^2 + 32a}{(2a)^2} = \frac{(2a+4)^2}{(2a)^2}.\tag1$$

Since it is given that $a > 0$, (1) Can be re-expressed as

$$- \frac{2a+4}{2a} < x + \frac{2a-4}{2a} < \frac{2a+4}{2a}$$

which implies that

$$-1 - \frac{2}{a} < x + 1 - \frac{2}{a} < 1 + \frac{2}{a} $$

which implies that

$$-2 < x < \frac{4}{a}. \tag2$$

This means that the positive integer values of $a$ must be found such that inequality (2) above will have exactly 3 satisfying integer values of $x$.

It is immediate that inequality (2) above will be satisfied by $\{-1,0\}$, regardless of the value of $a$. Therefore, the problem reduces to identifying for which values of $a$, $\frac{4}{a} > 1.$

Edit
Thanks to heropup for noticing a mistake of mine.
The actual constraint should be
$1 < \frac{4}{a} \leq 2$, since $x \geq 2$ must be prevented.

Noting that $a$ is required to be a positive integer, this means that $a \in \{2,3\}.$ Therefore, there are exactly 2 satisfying positive integer values for $a$.

user2661923
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The polynomial $ \ ax^2 \ + \ 2·(a-2 )·x \ - \ 8 \ $ turns out to be factorable as $ \ (x + 2)·(ax - 4) \ $ either by inspection and a check applying synthetic division, or by observing that its zeroes "obey" $$ r \ + \ s \ \ = \ \ -\frac{2·(a-2)}{a} \ \ = \ \ -2 \ + \ \frac{4}{a} \ \ , $$ $$ \Delta \ \ = \ \ 4·(a - 2)^2 \ - \ 4·a·(-8) \ \ = \ \ 4a^2 \ - \ 16a \ + \ 16 \ + \ 32a \ \ = \ \ 4·(a + 2)^2 $$ $$ \Rightarrow \ \ r \ - \ s \ \ = \ \ \frac{\sqrt{4·(a + 2)^2}}{a} \ \ = \ \ \frac{2·(a+2)}{a} \ \ = \ \ 2 \ + \ \frac{4}{a}$$ $$ \Rightarrow \ \ 2r \ \ = \ \ \frac{8}{a} \ \ \ , \ \ \ 2s \ \ = \ \ -4 \ \ . $$

Since we wish $ \ ax^2 \ + \ 2·(a-2 )·x \ - \ 8 \ < \ 0 $ to have only three integers within its solution interval, we must have $ \ a > 0 \ $ (an "upward-opening" parabola), as $ \ a < 0 \ $ would have a solution set of two "semi-infinite" disjoint intervals (as heropup observes). This tells us that $ \ r \ = \ \frac{4}{a} \ > \ 0 \ \ . $ We also have a proper polynomial inequality, so $ \ x = -2 \ $ can only be an "open endpoint" of the solution interval. Hence, the integers in the desired interval are $ \ -1 \ , \ 0 \ , \ 1 \ \ , $ forcing the endpoint restriction $$ \ 1 \ < \ \frac{4}{a} \ \le \ 2 \ \ \Rightarrow \ \ 1 \ > \ \frac{a}{4} \ \ge \ \frac12 \ \ \Rightarrow \ \ 4 \ > \ a \ \ge \ 2 \ \ . $$ [We may include $ \ 2 \ $ as an endpoint here, since it will also be excluded from the open solution interval for the polynomial inequality.]

Hence, the only permissible integer values for $ \ a \ $ are

• $ \ \ \mathbf{a \ = \ 2 \ \ :} \quad 2x^2 \ - \ 8 \ < \ 0 \ \ , \ $ solution interval $ \ (-2 \ , \ 2) \ \ \ $ and

• $ \ \ \mathbf{a \ = \ 3 \ \ :} \quad 3x^2 \ - \ 2x \ - \ 8 \ < \ 0 \ \ , \ $ solution interval $ \ \left(-2 \ , \ \frac43 \right) \ \ \ . $

[There seems to be a contradiction between the condition stated in the title and that given in the problem statement. I believe the word "negative" should not be present in the title.]