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The question asks to find the product of \(\sqrt{a} - \sqrt{bc}\) and its conjugate.

The conjugate of \(\sqrt{a} - \sqrt{bc}\) is \(\sqrt{a} + \sqrt{bc}\). Now, let's multiply the expression \(\sqrt{a} - \sqrt{bc}\) by its conjugate \(\sqrt{a} + \sqrt{bc}\).

We apply the difference of squares formula:
\[
(\sqrt{a} - \sqrt{bc})(\sqrt{a} + \sqrt{bc}) = (\sqrt{a})^2 - (\sqrt{bc})^2
\]

Simplifying both terms:
\[
(\sqrt{a})^2 = a \quad \text{and} \quad (\sqrt{bc})^2 = bc
\]

Thus, the product is:
\[
a - bc
\]

The correct answer is the first option: \(a - bc\).

Do you have any questions or need further details?

Here are some related questions you might want to explore:
1. What is the importance of conjugates in simplifying expressions?
2. Can the difference of squares formula be applied to non-radical terms?
3. How do we find the conjugate of a binomial expression involving square roots?
4. What other types of expressions benefit from using conjugates?
5. Can we apply the same technique to complex numbers?

**Tip:** Always remember that multiplying a binomial by its conjugate eliminates the square roots, simplifying the expression.

Which of the following is the product of \(\sqrt{a} - \sqrt{bc}\) and its conjugate?

Learn how to find the product of the expression \(\sqrt{a} - \sqrt{bc}\) and its conjugate using the difference of squares formula. This algebraic problem is suitable for students in Grades 9-11, covering key concepts such as radicals, conjugates, and algebraic simplification.

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