(a) f(x) = x^2 + 5x + p where x ∈ ℝ, -3 ≤ p ≤ 8, and p ∈ ℤ - Leaving Cert Mathematics - Question 1 - 2020

Let the roots of the quadratic be r1 and r2. From the condition, we have:

$r2 - r1 = 3$

Using Vieta's formulas, we know:

• The sum of the roots: r1 + r2 = -rac{b}{a} = -5
• The product of the roots: r1 imes r2 = rac{p}{a} = p

Therefore, we can express r2 as:

$r2 = r1 + 3$

Substituting into the sum of roots gives:

$r1 + (r1 + 3) = -5$

This leads to:

$2r1 + 3 = -5$ $2r1 = -8$ $r1 = -4$

Thus, we have:

$r2 = -1$

Now, substituting the roots into the product gives:

$r1 imes r2 = (-4)(-1) = 4$

This means:

$p = 4$

For the graph of f(x) not to cross the x-axis, the discriminant must be less than 0:

$b^2 - 4ac < 0$

Where for our function, a = 1, b = 5, c = p:

$5^2 - 4(1)(p) < 0$ $25 - 4p < 0$ $4p > 25$ $p > 6.25$

Since p is also constrained to -3 ≤ p ≤ 8, the valid integer values are:

• $p = 7$
• $p = 8$

Start with the inequality:

$|2x + 5| - 1 ≤ 0$

This simplifies to:

$|2x + 5| ≤ 1$

This leads to two inequalities:

1. $2x + 5 ≤ 1$
2. $2x + 5 ≥ -1$

Solving these:

1. From $2x + 5 ≤ 1$:

   • $2x ≤ -4$
   • $x ≤ -2$

2. From $2x + 5 ≥ -1$:

   • $2x ≥ -6$
   • $x ≥ -3$

Thus, the range of values of x is:

$-3 ≤ x ≤ -2$