Question:

Step 1: Apply Vieta's formulas to the first equation $x^2 + px + q = 0$. The roots are $\alpha$ and $\beta$. $$ \alpha + \beta = -p $$ $$ \alpha\beta = q $$ Step 2: Determine the sum of roots for the second equation $x^2 + mx + n = 0$. The roots are $\alpha + \frac{1}{\alpha}$ and $\beta + \frac{1}{\beta}$. The sum of these roots is $-m$: $$ -m = \left(\alpha + \frac{1}{\alpha}\right) + \left(\beta + \frac{1}{\beta}\right) $$ $$ -m = (\alpha+\beta) + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) $$ $$ -m = (\alpha+\beta) + \frac{\alpha+\beta}{\alpha\beta} $$ Substitute the expressions from Step 1: $$ -m = -p + \frac{-p}{q} = -p\left(1 + \frac{1}{q}\right) = -p\left(\frac{q+1}{q}\right) $$ Therefore, $$ m = \frac{p(q+1)}{q} $$ Step 3: Determine the product of roots for the second equation $x^2 + mx + n = 0$. The product of these roots is $n$: $$ n = \left(\alpha+\frac{1}{\alpha}\right)\left(\beta+\frac{1}{\beta}\right) $$ $$ n = \alpha\beta + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + \frac{1}{\alpha\beta} $$ $$ n = \alpha\beta + \frac{\alpha^2+\beta^2}{\alpha\beta} + \frac{1}{\alpha\beta} $$ We know that $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$. Substitute from Step 1: $$ \alpha^2+\beta^2 = (-p)^2 - 2q = p^2 - 2q $$ Substitute this back into the expression for $n$: $$ n = q + \frac{p^2-2q}{q} + \frac{1}{q} $$ $$ n = q + \frac{p^2-2q+1}{q} $$ Step 4: Apply the given condition $m+n=0$. Substitute the expressions for $m$ and $n$ from Step 2 and Step 3: $$ \frac{p(q+1)}{q} + q + \frac{p^2-2q+1}{q} = 0 $$ Since $q \neq 0$, multiply the entire equation by $q$: $$ p(q+1) + q^2 + p^2 - 2q + 1 = 0 $$ Rearrange the terms to form a quadratic equation in $p$: $$ p^2 + p(q+1) + (q^2 - 2q + 1) = 0 $$ Recognize that $q^2 - 2q + 1 = (q-1)^2$: $$ p^2 + p(q+1) + (q-1)^2 = 0 $$ Step 5: Determine the range of $q$ for which real values of $p$ exist. For $p$ to be a real number, the discriminant of the quadratic equation in $p$ must be non-negative. The discriminant $D$ is given by $D = (q+1)^2 - 4(1)(q-1)^2$. $$ D \geq 0 $$ $$ (q+1)^2 - 4(q-1)^2 \geq 0 $$ Expand the terms: $$ (q^2+2q+1) - 4(q^2-2q+1) \geq 0 $$ $$ q^2+2q+1 - 4q^2+8q-4 \geq 0 $$ $$ -3q^2+10q-3 \geq 0 $$ Multiply by $-1$ and reverse the inequality sign: $$ 3q^2-10q+3 \leq 0 $$ Factor the quadratic expression: $$ (3q-1)(q-3) \leq 0 $$ This inequality holds when $q$ is between the roots $\frac{1}{3}$ and $3$. $$ \frac{1}{3} \leq q \leq 3 $$ Step 6: Identify the possible integral values of $q$. The problem states that $q$ is a real number and $q \neq 0$. The derived range $\frac{1}{3} \leq q \leq 3$ satisfies $q \neq 0$. The integral values of $q$ within this range are $1, 2, 3$. The problem statement only requires $p$ and $q$ to be real numbers, which is ensured by the discriminant condition. The roots $\alpha$ and $\beta$ of the first equation can be complex, and the roots of the second equation will still be well-defined and satisfy the conditions. The possible integral values in the range of $q$ are $1, 2, 3$.