Verified Answer :

  • Identify the conditions for the equation to be defined: the radicands \(x^2 - 2x - 2\) and \(2x - 3\) must both be non-negative.
  • Solve \(2x - 3 \geq 0\) to get \(x \geq 1.5\).
  • Solve \(x^2 - 2x - 2 \geq 0\) to find the roots \(1 \pm \sqrt{3}\) and determine that the expression is non-negative for \(x \leq 1 - \sqrt{3}\) or \(x \geq 1 + \sqrt{3}\).
  • Find the intersection of these sets, resulting in the domain \(x \geq 1 + \sqrt{3}\).
  • The domain is \(\boxed{[1 + \sqrt{3}, \infty)}\).

Explanation

  1. Defining the Conditions for the Domain

To determine the domain of the equation \(\sqrt{x^2 - 2x - 2} + \sqrt{2x - 3} = 5,\) we must ensure that all terms are defined. Square roots of real numbers are defined only when their radicands are non-negative. Therefore, we have two conditions that must be simultaneously satisfied:

  1. \(x^2 - 2x - 2 \geq 0\)
  2. \(2x - 3 \geq 0\)
  1. Solving the Linear Inequality

First, we solve the inequality \(2x - 3 \geq 0\):
\[ 2x \geq 3 \]
\[ x \geq \frac{3}{2} \]
Thus, \(x \geq 1.5\). This condition specifies that any solution must be at least \(1.5\).

  1. Solving the Quadratic Inequality

Next, we solve the quadratic inequality \(x^2 - 2x - 2 \geq 0\). We find the roots of the corresponding quadratic equation \(x^2 - 2x - 2 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3} \]
The parabola \(y = x^2 - 2x - 2\) opens upwards, so the expression is non-negative outside the roots. Thus, the inequality holds for \(x \leq 1 - \sqrt{3}\) or \(x \geq 1 + \sqrt{3}\).
Approximating these values, \(1 - \sqrt{3} \approx -0.732\) and \(1 + \sqrt{3} \approx 2.732\). So, the condition is \(x \leq 1 - \sqrt{3}\) or \(x \geq 1 + \sqrt{3}\).

  1. Determining the Final Domain

Finally, we find the intersection of the two conditions: \(x \geq 1.5\) and (\(x \leq 1 - \sqrt{3}\) or \(x \geq 1 + \sqrt{3}\)).
Since \(1.5 > 1 - \sqrt{3}\), the interval \(x \leq 1 - \sqrt{3}\) is excluded.
Comparing \(1.5\) and \(1 + \sqrt{3}\), we note that \(1.5 < 2.732\), so the intersection is the set of values \(x \geq 1 + \sqrt{3}\).
Thus, the domain is \([1 + \sqrt{3}, \infty)\).

Examples

In engineering, determining the domain of an equation is crucial when analyzing the stability of a physical system. For example, if you are modeling the tension in cables with specific geometric constraints, the domain of the resulting equation tells you which physical configurations are actually possible. Restricting variables to values where calculations are well-defined ensures that your model does not predict physically impossible or mathematically undefined scenarios.

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