(a) Given the function $f(x) = x^2 + 5x + p$ where $x \ is \\mathbb{R}$, $-3 \ \leq p \leq 8$, and $p \in \mathbb{Z}$ - Leaving Cert Mathematics - Question 1 - 2020

For the roots of $f(x)$ to differ by 3, we can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 5$, and $c = p$.

Following this, the difference between the roots can be expressed using:

$\sqrt{b^2 - 4ac} = 3$

Substituting in our values, this becomes:

$\sqrt{5^2 - 4(1)(p)} = 3$

Squaring both sides, we get:

$25 - 4p = 9$,

which simplifies to:

$4p = 16 \Rightarrow p = 4.$

For the graph of $f(x)$ to not cross the x-axis, the discriminant must be less than zero:

$b^2 - 4ac < 0$

Substituting for our coefficients gives:

$5^2 - 4(1)(p) < 0$

This leads us to:

$25 - 4p < 0$

Solving for $p$, we have:

$4p > 25 \Rightarrow p > \frac{25}{4} = 6.25.$\

The integer values satisfying this are:

$p \geq 7 \text{ or } p \leq 8.$

To solve for $x$, we start by rearranging the inequality:

$|2x + 5| \leq 1$.

This can be expressed as two linear inequalities:

$-1 \leq 2x + 5 \leq 1.$\

Solving these, we break it down into two parts.

1. Solving $-1 \leq 2x + 5$:

   $-1 - 5 \leq 2x \Rightarrow -6 \leq 2x \Rightarrow -3 \leq x.$

2. Solving $2x + 5 \leq 1$:

   $2x \leq 1 - 5 \Rightarrow 2x \leq -4 \Rightarrow x \leq -2.$\

Combining these results gives the range:

$-3 \leq x \leq -2.$