The function $f$ is such that $$f(x) = 2x^3 + 5x^2 - 4x - 3,$$ where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 5 - 2017

For $f(x) + a$ to have only one real root, the discriminant of the equation must be zero:

$D = b^2 - 4ac = 0.$ Here, we can consider the polynomial:

$f(x) + a = 2x^3 + 5x^2 - 4x - 3 + a = 0.$

This leads to a new constant term:

a is modified to $a - 3$ so:

The new cubic polynomial becomes:

$D = -4(2)(a - 3) = 0,$
leading to: $a - 3 = 0\ \implies a = 3$. Also, $f(x)$ has turning points near $x = -2$ and $x = \frac{1}{3}$. Thus:

• The critical values involve maximum and minimum to determine the upward and downward nature of the cubic function.
  So, we need:

a should be placed so that it gets more than maximum $(\frac{-100}{27})$ and less than the minimum $(-2)$ on the limits.
So, $a > \frac{100}{27} \text{ and } a < -9.$