To solve the equation $1 + 1.5\left(\sqrt{x^2 - 1} - \frac{1.5x^2}{\sqrt{x^2 - 1}}\right) = 0$, we can first simplify the expression inside the parentheses.

1. Rewrite the equation: $1 + 1.5\left(\sqrt{x^2 - 1} - \frac{1.5x^2}{\sqrt{x^2 - 1}}\right) = 0$

2. Simplify the expression: $\sqrt{x^2 - 1} - \frac{1.5x^2}{\sqrt{x^2 - 1}} = \frac{(\sqrt{x^2 - 1})^2 - 1.5x^2}{\sqrt{x^2 - 1}} = \frac{x^2 - 1 - 1.5x^2}{\sqrt{x^2 - 1}} = \frac{-0.5x^2 - 1}{\sqrt{x^2 - 1}}$

3. Substitute back into the equation: $1 + 1.5 \cdot \frac{-0.5x^2 - 1}{\sqrt{x^2 - 1}} = 0$

4. Multiply through by $\sqrt{x^2 - 1}$ to eliminate the fraction: $\sqrt{x^2 - 1} + 1.5(-0.5x^2 - 1) = 0$

5. Distribute: $\sqrt{x^2 - 1} - 0.75x^2 - 1.5 = 0$

6. Rearranging gives: $\sqrt{x^2 - 1} = 0.75x^2 + 1.5$

7. Square both sides: $x^2 - 1 = (0.75x^2 + 1.5)^2$

8. Expand the right-hand side: $(0.75x^2 + 1.5)(0.75x^2 + 1.5) = 0.5625x^4 + 2.25x^2 + 2.25$

9. Set the equation: $x^2 - 1 = 0.5625x^4 + 2.25x^2 + 2.25$

10. Rearranging this gives: $0.5625x^4 + 1.25x^2 + 3.25 = 0$

This is a quadratic in terms of $y = x^2$: $0.5625y^2 + 1.25y + 3.25 = 0$

Now you can use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find $y$.

Calculation of $y$:

1. $a = 0.5625$
2. $b = 1.25$
3. $c = 3.25$

Calculating the discriminant: $b^2 - 4ac = (1.25)^2 - 4(0.5625)(3.25)$

Calculating further gives: $= 1.5625 - 7.35 = -5.7875$

Since the discriminant is negative, there are no real solutions for $x^2$, which means there are no real values for $x$.

If you need further details or have any questions, feel free to ask! Here are some related questions:

1. How do you interpret complex solutions in equations?
2. Can you provide examples of similar equations with real solutions?
3. What steps can you take when faced with a negative discriminant?
4. How does squaring both sides affect the solutions of an equation?
5. What are the implications of having no real solutions in a mathematical model?

Tip: Always check for extraneous solutions after squaring both sides of an equation, as this can introduce false solutions.