$x^2 - (m - 2)x + 6 = 0$ : Possible values of $m$.

Question

One of the roots of the equation $x^2 - (m - 2)x + 6 = 0$ lies between $(0, 2)$ and the other root lies between $(3, 4)$. It is known that $m$ is an integer. What is the sum of all the possible values of $m$?

Let $a$ be the root that lies between $(0, 2)$ and let $b$ be the root that lies between $(3, 4)$. I can write it mathematically as $0 \lt a \lt2$ and $3 \lt b \lt4$ and also we can say that $3 \lt a+b \lt6$.

$a+b=m-2$ and this means that $3 \lt m-2 \lt6 \Rightarrow 5 \lt m \lt 8$. So the possible integer values of $m$ should be $6$ and $7$ and the possible sum should be $13$.

But as per the solution this is not the correct answer. What have I done wrong? Please help !!!

Thanks in advance !!!

Answer 1

No real root is possible for integral $m$ in (0,2) and (3,4)

Let $f(x)=x^2-(m-2)x+6$. $f(0)=6>0$, then for a root in (0,2) we should have $f(2))<0 \implies 14-2m<0 \implies m>7$.

Next for a root in (3,4), $f(3)=9-3(m-2)+6=21-3m<0$, next we demand $f(4)>0 \implies 16-4(m-2)+6>0\implies m<7.5$. Hence no integral value of $m$ is possible.

For $m=7$ the roots are 2 and 3. For $7 <m<7.5$, roots lie in (0,2) and (3,4).Hence, no real root is possible for integral m in (0,2) and (3,4).

Answer 2

The problem with using Viete's relations the way you did for the inequality is that they apply even when the roots are complex. So $ \ x^2 - (6-2)x + 6 \ = \ 0 \ $ has the roots $ \ 2 \ \pm \ i\sqrt2 \ \ , $ which sum to $ \ 4 \ \ \ , $ satisfying $ \ 3 \ < \ a+b \ < \ 6 \ \ . $

One way we can look at the quadratic polynomial is to consider its "vertex form" describing an "upward-opening" parabola with $ \ y-$intercept $ \ (0 \ , \ 6 ) \ \ . $ "Completing the square" produces $$ x^2 \ - \ (m-2)x \ + \ 6 \ \ = \ \ \left(x \ - \ \frac{m-2}{2} \right)^2 \ + \ \left(6 \ - \ \frac{(m-2)^2}{4} \right) \ \ , $$ which tells us that the vertex is the $ \ x-$intercept of the parabola for $$ 6 \ - \ \frac{(m-2)^2}{4} \ \ = \ \ 0 \ \ \Rightarrow \ \ m \ \ = \ \ 2 \ \pm \ 2·\sqrt6 \ \ . $$

Our quadratic equation then has real roots for $ \ m \ < \ 2 \ - \ 2·\sqrt6 \ $ and $ \ m \ > \ 2 \ + \ 2·\sqrt6 \ \ . $ But the $ \ y-$intercept at $ \ (0 \ , \ 6 ) $ means that we cannot has positive real roots for $ \ m \ < \ 2 \ - \ 2·\sqrt6 \ \ , $ since the vertex is "to the left" of the $ \ y-$axis, leaving us with $ \ m \ > \ \approx \ 6.899 \ \ . $

With $ \ m \ $ required to be an integer, we can assess cases directly. For $ \ m \ = \ 7 \ \ , $ the equation is $ \ x^2 - (7-2)x + 6 \ = \ (x-2)·(x-3) \ = \ 0 \ \ ; $ unfortunately, neither of the roots are in the specified open intervals. The next integer case leads to $ \ x^2 - (8-2)x + 6 \ = \ 0 \ \ , $ with roots $ \ 3 - \sqrt3 \ \approx \ 1.27 \ $ and $ \ 3 + \sqrt3 \ \approx \ 4.73 \ \ , $ the larger of these being "to the right" of the interval $ \ (3 \ , \ 4 ) \ \ . $ Using larger integers simply shifts the vertex "lower and to the right", making the larger of the roots even larger. Thus, we confirm Z Ahmed's conclusion that there are no possible integer values of $ \ m \ $ to sum. (Trick question...)