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:

Question

Accepted Answer

To determine the set of possible values of 'a' for which the difference between the roots of the equation $x^2 + ax + 1 = 0$ is less than $\sqrt{5}$, we start by considering the properties of quadratic equations.

The roots of the quadratic equation $x^2 + ax + 1 = 0$ can be found using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

For our equation, $a = 1$, $b = a$, and $c = 1$. Thus, the roots are:

$$
x = \frac{-a \pm \sqrt{a^2 - 4}}{2}
$$

The difference between the roots is given by:

$$
\Delta = \left| \frac{\sqrt{a^2 - 4}}{1} \right| = \sqrt{a^2 - 4}
$$

We want this difference to be less than $\sqrt{5}$:

$$
\sqrt{a^2 - 4} < \sqrt{5}
$$

Squaring both sides, we get:

$$
a^2 - 4 < 5
$$

$$
a^2 < 9
$$

Taking the square root of both sides, we find:

$$
-3 < a < 3
$$

This inequality describes the interval $(-3, 3)$, which corresponds to Option 1.

**Explanation for why Option 1 is correct:**

Option 1, $(-3, 3)$, is correct because it precisely matches the derived inequality $-3 < a < 3$, which ensures that the difference between the roots is less than $\sqrt{5}$.

**Explanation for why other options are incorrect:**

- **Option 2: $(-3, \infty)$** is incorrect because it includes values of 'a' greater than 3, which would result in $a^2 \geq 9$, violating the condition $a^2 < 9$.

- **Option 3: $(3, \infty)$** is incorrect because it only includes values of 'a' greater than 3, which again would result in $a^2 \geq 9$, violating the condition $a^2 < 9$.

- **Option 4: $(-\infty, -3)$** is incorrect because it only includes values of 'a' less than -3, which would also result in $a^2 \geq 9$, violating the condition $a^2 < 9$.

Thus, Option 1 is the only correct choice that satisfies the condition for the difference between the roots being less than $\sqrt{5}$.