Question

If the difference between the roots of the equation $x^2 + ax + 1 = 0$ is less than $\sqrt{5}$, then the set of possible values of $a$ is

Answer 1

(-3, -2] ∪ [2, 3)

Answer 2

Let the roots of the quadratic equation $x^2 + ax + 1 = 0$ be $\alpha$ and $\beta$. According to Vieta's formulas:

1. Sum of the roots: $\alpha + \beta = -a$
2. Product of the roots: $\alpha \beta = 1$

The difference between the roots, $|\alpha - \beta|$, can be expressed using the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$. Substituting the values from Vieta's formulas: $(\alpha - \beta)^2 = (-a)^2 - 4(1)$ $(\alpha - \beta)^2 = a^2 - 4$

Therefore, $|\alpha - \beta| = \sqrt{a^2 - 4}$.

The problem states that the difference between the roots is less than $\sqrt{5}$: $|\alpha - \beta| < \sqrt{5}$ Substitute the expression for $|\alpha - \beta|$: $\sqrt{a^2 - 4} < \sqrt{5}$

To solve this inequality, we square both sides (both sides are non-negative): $(\sqrt{a^2 - 4})^2 < (\sqrt{5})^2$ $a^2 - 4 < 5$ $a^2 < 9$

This inequality implies $-3 < a < 3$.

Additionally, for the roots to be real (which is implicitly assumed when talking about their difference), the discriminant of the quadratic equation must be non-negative. The discriminant $D$ for the equation $Ax^2 + Bx + C = 0$ is $D = B^2 - 4AC$. For $x^2 + ax + 1 = 0$, $A=1$, $B=a$, $C=1$. So, $D = a^2 - 4(1)(1) = a^2 - 4$.

For real roots, $D \ge 0$: $a^2 - 4 \ge 0$ $a^2 \ge 4$

This inequality implies $a \le -2$ or $a \ge 2$.

Now, we need to find the values of $a$ that satisfy both conditions:

1. $-3 < a < 3$
2. $a \le -2$ or $a \ge 2$

Let's visualize these conditions on a number line: Condition 1: $a \in (-3, 3)$ Condition 2: $a \in (-\infty, -2] \cup [2, \infty)$

The intersection of these two sets is: $(-3, -2] \cup [2, 3)$

This means the possible values of $a$ are such that $-3 < a \le -2$ or $2 \le a < 3$.