The exercises involve solving quadratic and factorizable equations. I'll address one problem step-by-step and explain the process. Let’s work through exercise 73 (a) first:

Problem:

$x^2 - 36 = 0$

Solution:

1. Recognize this as a difference of squares: $x^2 - 36 = (x - 6)(x + 6)$

2. Set each factor equal to 0 (product zero property): $x - 6 = 0 \quad \text{or} \quad x + 6 = 0$

3. Solve for $x$ in each case:

   • $x = 6$
   • $x = -6$

Final Answer:

$x = 6 \quad \text{or} \quad x = -6$

Would you like me to solve any other problem, or explain further?

Here are related questions to expand understanding:

1. How can we recognize a difference of squares quickly?
2. How would the solution differ if there was a linear term (e.g., $x^2 - 36x = 0$)?
3. What other factorization methods can solve quadratic equations?
4. Can we solve $x^2 - 36 = 0$ graphically? What does the graph look like?
5. How do we verify solutions to quadratic equations?

Tip: Always simplify the equation first and look for common patterns like factorization or completing the square.