# Integer values of $m$ for which $x^2-(m-3)x+m = 0$ has greater then $2$

Integer values of $m$ for which both the roots of the equation $x^2-(m-3)x+m = 0$ has greater then $2$

$\underline{\bf{My\;\; Try}}:$ Let $\alpha\;\;,\beta>2$ be the roots of the equation.

and here roots are real and equal which is $>2$

So $D\geq 0$ and $\alpha+\beta >4$ and $\alpha.\beta >4$

$(m-3)^2-4m\geq 0$ and $m-3>4$ and $m>4$

$m\leq 1 \cup m\geq 9$ and $m>7$ and $m>4$

So $m\leq 1 \cup m\geq 9$

So we get Infinite no. of integer values which satisfy the above conditin.

Now I did not understand How can I get finite no. of integer values.

Help Required.

Thanks

We can write $x^2 -(m-3)x + m = (x-\alpha)(x-\beta)$, and so $\alpha \beta = m$ and $\alpha + \beta = m - 3$. We already see that $m = \alpha \beta > 4$. Combined with the fact that $D = (m-3)^2 - 4m \geq 0$ this gives $m \geq 9$. Now let $\alpha = 2 + \epsilon$, then $\beta = m - 5 - \epsilon$, and we find $m = \alpha \beta = 2m - 10 + \epsilon (m - 7 - \epsilon) = 2m - 10 + \epsilon( \beta - 2) > 2m - 10$, and so $m < 10$. So $m$ is an integer satisfying $9 \leq m < 10$, which implies $m = 9$.
$$\begin{cases} \alpha, \beta \in \mathbb{R}, \\ \alpha, \beta \ge 2; \end{cases} \begin{cases} D \ge 0, \\ \alpha + \beta > 4, \\ \alpha \cdot \beta > 4, \\ \alpha, \beta \ge 2; \end{cases} \begin{cases} (m-3)^2 - 4m \ge 0, \\ (m-3) \ge 4, \\ m \ge 4, \\ \alpha, \beta \ge 2; \end{cases} \begin{cases} (m-9)(m-1)\ge 0, \\ m \ge 7, \\ m \ge 4, \\ \alpha, \beta \ge 2; \end{cases} \begin{cases} (m-9)(m-1)\ge 0, \\ m \ge 7, \\ m \ge 4, \\ \alpha, \beta \ge 2; \end{cases} \begin{cases} m \ge 9, \\ \alpha, \beta \ge 2; \end{cases} \begin{cases} m \ge 9, \\ \alpha = \left((m-3) - \sqrt{(m-3)^2 - 4m}\right) / 2, \\ \beta = \left((m-3) + \sqrt{(m-3)^2 - 4m}\right) / 2, \\ \alpha, \beta \ge 2; \end{cases} \begin{cases} m \ge 9, \\ \left((m-3) - \sqrt{m^2-10m+9}\right) / 2 \ge 2; \\ \end{cases} \begin{cases} m \ge 9, \\ m - 7 \ge \sqrt{m^2-10m+9}; \\ \end{cases} \begin{cases} m \ge 9, \\ (m - 7)^2 \ge m^2-10m+9; \\ \end{cases} \begin{cases} m \ge 9, \\ 40 \ge 4m; \\ \end{cases} \begin{cases} m \ge 9, \\ m \le 10. \\ \end{cases}$$ So $m$ should be 9 or 10. Both are coreect: $m = 9$ gives $\alpha = \beta = 3$ and $m = 10$ gives $\alpha = 2$, $\beta = 5$.
you need the equation $(x+2) ^2-(m-3)(x+2)+m = 0$ has positive roots. Simplifying this you get $x^2+(7-m)x+10-m=0$ . now sum and product of roots must be $>0$. then $-(7-m)>0, 10-m>0$. so $7<m<10$. also you need $D\ge 0$ its found above: $m\ge9$
So $m=9$