Let $m \in \mathbb{R}$
Let $f:\mathbb{R} \to \mathbb{R} \space\space |\space \space f(x) = x^2+(m+1)x-((3/4)m+5/4)\space\space \forall \space\space x \space \in \space \mathbb{R}$
Let $\mathbb{A} = \{x\space|\space f(x)=0\}$
We are to find the range of values of $m$ such that
$||\mathbb{A}||=2$
$r>3 \space \forall\space r \in \mathbb{A}$
The first condition is basically that the given quadratic polynomial should have two distinct real roots, for which it's necessary and sufficient that the discriminant of the quadratic polynomial is strictly greater than zero, that is
$$(m+1)^2 - 4((3/4)m+5/4) > 0 $$ $$\iff m^2+5m+6 >0$$ $$\iff (m+2)(m+3)>0$$ $$\iff m \in (-\infty, -3)\cup(-2,\infty)$$
The second condition however is related to restricting both the zeroes of the quadratic polynomial in a certain bound(that is, both the zeroes of the quadratic polynomial should be strictly greater than 3). I am unable to think of a condition on $m$ to achieve this restriction.
Any hint or help will be appreciated. Thank You
You can write formula for roots as a function of $m$. $$ x_1 = \frac { -(m + 1) - \sqrt{ (m+2)(m+3) } } {2},\qquad x_2 = \frac { -(m + 1) + \sqrt{ (m+2)(m+3) } } {2}. $$ If you want both of them to be greater than 3, it is necessary and sufficient to have minimum greater than 3, that is $x_1 > 3$. This condition gives us inequality with square root $$ -(m + 1)- \sqrt{ m^2 + 5m + 6 } > 6 \quad \iff \quad \sqrt{ m^2 + 5m + 6 } < -(m +7). $$ Since RHS needs to be positive, we assume $m < -7$ and then square inequality to obtain $$ m^2 + 5m + 6 < m^2 + 14 m + 49 \quad \iff \quad -43 < 9m. $$ But $m > -\frac {43} 9$ and $m < -7$ contradict with each other, so there are no parameters $m\in \mathbb R$ such that both conditions 1 & 2 hold.