Key Concepts and Formulas

• For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product of the roots is $\alpha \beta = \frac{c}{a}$.
• The difference of the roots is related to the sum and product by the formula: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.
• For any real number $x$, $\sqrt{x^2} = |x|$. Also, if $a^2 < b^2$ and $a, b > 0$, then $a < b$.

Step-by-Step Solution

Step 1: Identify the coefficients and find the sum and product of the roots.

The given quadratic equation is $x^2 + ax + 1 = 0$. Here, the coefficients are $A = 1$, $B = a$, and $C = 1$. We need to find the sum and product of the roots $\alpha$ and $\beta$.

Using the formulas: $\alpha + \beta = -\frac{B}{A} = -\frac{a}{1} = -a$ $\alpha \beta = \frac{C}{A} = \frac{1}{1} = 1$

Step 2: Express the difference of the roots in terms of $a$.

We are given that the difference between the roots is related to $a$. We can use the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ to relate the difference of the roots to their sum and product, which we found in Step 1.

Substituting the values we found: $(\alpha - \beta)^2 = (-a)^2 - 4(1) = a^2 - 4$

Step 3: Apply the given condition and set up the inequality.

We are given that $|\alpha - \beta| < \sqrt{5}$. To eliminate the absolute value and square root, we can square both sides of the inequality. Since both sides are non-negative, the inequality is preserved.

$(|\alpha - \beta|)^2 < (\sqrt{5})^2$ $(\alpha - \beta)^2 < 5$

Now, substitute the expression for $(\alpha - \beta)^2$ from Step 2: $a^2 - 4 < 5$

Step 4: Solve the inequality for $a$.

We now solve the inequality $a^2 - 4 < 5$ for $a$.

First, add 4 to both sides: $a^2 < 9$

Taking the square root of both sides, we must consider both positive and negative roots: $\sqrt{a^2} < \sqrt{9}$ $|a| < 3$

This inequality is equivalent to: $-3 < a < 3$

Step 5: Write the solution set.

The solution set for $a$ is the open interval $(-3, 3)$.

Common Mistakes & Tips

• Discriminant Consideration: The problem doesn't explicitly state that the roots must be real. If the roots were required to be real, we would need to ensure the discriminant $a^2-4 \ge 0$, which gives $a \le -2$ or $a \ge 2$. In that case, the solution would be the intersection of $(-3, 3)$ and $(-\infty, -2] \cup [2, \infty)$, resulting in $(-3, -2] \cup [2, 3)$. Since the problem does not specify real roots, we proceed with the solution obtained.
• Squaring Inequalities: Remember to be careful when squaring inequalities. Ensure that both sides are non-negative before squaring.
• Absolute Value Interpretation: The inequality $|a| < k$ is equivalent to $-k < a < k$.

Summary

We used the relationships between the roots and coefficients of the quadratic equation, along with the given condition on the difference of the roots, to derive an inequality for $a$. Solving this inequality, we found that the set of possible values for $a$ is the open interval $(-3, 3)$. This means $a$ must lie strictly between $-3$ and $3$.

The final answer is \boxed{(-3, 3)}, which corresponds to option (C).