We are given the quadratic equation $x^2 + 2ax + (3a + 10) = 0$. Let the roots be $\alpha$ and $\beta$. We are given the condition $\alpha < 1 < \beta$.
Analyzing the Roots Condition
The condition $\alpha < 1 < \beta$ means that the number 1 lies strictly between the two roots of the quadratic equation. This implies that the quadratic equation must have two distinct real roots.
The condition for a quadratic equation $Ax^2 + Bx + C = 0$ to have two distinct real roots is that its discriminant, $\Delta = B^2 - 4AC$, must be positive ($\Delta > 0$).
Calculating the Discriminant
For the given equation $x^2 + 2ax + (3a + 10) = 0$, we have:
- $A = 1$
- $B = 2a$
- $C = 3a + 10$
The discriminant is:
$\Delta = (2a)^2 - 4(1)(3a + 10)$
$\Delta = 4a^2 - 12a - 40$
Solving the Discriminant Inequality
We require $\Delta > 0$ for distinct real roots:
$4a^2 - 12a - 40 > 0$
Divide the inequality by 4:
$a^2 - 3a - 10 > 0$
Factor the quadratic expression:
$(a - 5)(a + 2) > 0$
This inequality holds true when $a$ is less than the smaller root ($-2$) or greater than the larger root ($5$).
Therefore, the possible values of $a$ are:
$a < -2$ or $a > 5$
Conclusion
The set of all possible values of $a$ that satisfy the condition is the union of the two intervals:
$(-\infty, -2) \cup (5, \infty)$
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