Which Choice Is Equivalent To The Quotient Below When $-3 \leq X \ \textless \ 3$?$\frac{\sqrt{9-x^2}}{\sqrt{3-x}}$A. $\sqrt{3+x}$B. $\sqrt{6+x-x^2}$C. 1D. $\sqrt{3-x}$

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Understanding the Problem

When dealing with mathematical expressions involving square roots, it's essential to understand the conditions under which the expressions are defined. In this case, we have the quotient $\frac{\sqrt{9-x^2}}{\sqrt{3-x}}$, and we need to find an equivalent expression when $-3 \leq x \ \textless \ 3$. To begin, let's analyze the square roots in the given expression.

Analyzing the Square Roots

The expression $\sqrt{9-x^2}$ is defined when $-3 \leq x \leq 3$, as the square root of a non-negative number is always non-negative. However, the expression $\sqrt{3-x}$ is defined when $x \leq 3$, as the square root of a non-negative number is always non-negative. Therefore, the given expression is defined when $-3 \leq x \leq 3$.

Rationalizing the Denominator

To simplify the given expression, we can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of $\sqrt{3-x}$ is $\sqrt{3-x}$ itself. Therefore, we have:

$$\frac{\sqrt{9-x^2}}{\sqrt{3-x}} \cdot \frac{\sqrt{3-x}}{\sqrt{3-x}} = \frac{\sqrt{9-x^2} \cdot \sqrt{3-x}}{3-x}$$

Simplifying the Expression

Now, let's simplify the numerator of the expression. We can use the fact that $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ to rewrite the numerator as:

$$\sqrt{9-x^2} \cdot \sqrt{3-x} = \sqrt{(9-x^2)(3-x)}$$

Expanding the Product

To expand the product in the numerator, we can use the distributive property:

$$(9-x^2)(3-x) = 27 - 9x - 3x^2 + x^3$$

Simplifying the Expression Further

Now, let's simplify the expression further by combining like terms:

$$\sqrt{27 - 9x - 3x^2 + x^3} = \sqrt{(3-x)(9-x^2)}$$

Using the Difference of Squares Formula

We can use the difference of squares formula to rewrite the expression as:

$$\sqrt{(3-x)(9-x^2)} = \sqrt{(3-x)(3+x)(3-x)}$$

Simplifying the Expression Further

Now, let's simplify the expression further by combining like terms:

$$\sqrt{(3-x)(3+x)(3-x)} = \sqrt{(3-x)^2(3+x)}$$

Simplifying the Expression Further

Now, let's simplify the expression further by taking the square root of the perfect square:

$$\sqrt{(3-x)^2(3+x)} = (3-x)\sqrt{3+x}$$

Simplifying the Expression Further

Now, let's simplify the expression further by dividing both the numerator and the denominator by $3-x$:

$$\frac{(3-x)\sqrt{3+x}}{3-x} = \sqrt{3+x}$$

Conclusion

Therefore, the equivalent expression to the quotient $\frac{\sqrt{9-x^2}}{\sqrt{3-x}}$ when $-3 \leq x \ \textless \ 3$ is $\boxed{\sqrt{3+x}}$.

Comparison with the Given Options

Let's compare the equivalent expression $\sqrt{3+x}$ with the given options:

• Option A: $\sqrt{3+x}$ is equivalent to the given expression.
• Option B: $\sqrt{6+x-x^2}$ is not equivalent to the given expression.
• Option C: 1 is not equivalent to the given expression.
• Option D: $\sqrt{3-x}$ is not equivalent to the given expression.

Final Answer

Therefore, the final answer is $\boxed{\sqrt{3+x}}$.

Introduction

In our previous article, we explored the equivalent expression to the quotient $\frac{\sqrt{9-x^2}}{\sqrt{3-x}}$ when $-3 \leq x \ \textless \ 3$. In this article, we'll delve deeper into the world of mathematical expressions and provide a Q&A guide to help you understand the concept better.

Q: What is the condition for the given expression to be defined?

A: The given expression is defined when $-3 \leq x \leq 3$, as the square root of a non-negative number is always non-negative.

Q: Why do we need to rationalize the denominator?

A: We need to rationalize the denominator to simplify the expression and make it easier to work with. Rationalizing the denominator involves multiplying both the numerator and the denominator by the conjugate of the denominator.

Q: What is the conjugate of the denominator?

A: The conjugate of the denominator $\sqrt{3-x}$ is $\sqrt{3-x}$ itself.

Q: How do we simplify the numerator?

A: We can use the fact that $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ to rewrite the numerator as $\sqrt{(9-x^2)(3-x)}$.

Q: How do we expand the product in the numerator?

A: We can use the distributive property to expand the product in the numerator: $(9-x^2)(3-x) = 27 - 9x - 3x^2 + x^3$.

Q: How do we simplify the expression further?

A: We can use the difference of squares formula to rewrite the expression as $\sqrt{(3-x)(9-x^2)} = \sqrt{(3-x)(3+x)(3-x)}$.

Q: How do we simplify the expression further?

A: We can simplify the expression further by combining like terms: $\sqrt{(3-x)(3+x)(3-x)} = \sqrt{(3-x)^2(3+x)}$.

Q: How do we simplify the expression further?

A: We can simplify the expression further by taking the square root of the perfect square: $\sqrt{(3-x)^2(3+x)} = (3-x)\sqrt{3+x}$.

Q: How do we simplify the expression further?

A: We can simplify the expression further by dividing both the numerator and the denominator by $3-x$: $\frac{(3-x)\sqrt{3+x}}{3-x} = \sqrt{3+x}$.

Q: What is the equivalent expression to the quotient $\frac{\sqrt{9-x^2}}{\sqrt{3-x}}$ when $-3 \leq x \ \textless \ 3$?

A: The equivalent expression is $\boxed{\sqrt{3+x}}$.

Q: How do we compare the equivalent expression with the given options?

A: We can compare the equivalent expression $\sqrt{3+x}$ with the given options A, B, C, and D.

Q: What is the final answer?

A: The final answer is $\boxed{\sqrt{3+x}}$.

Conclusion

In this Q&A guide, we've explored the concept of equivalent expressions and how to simplify mathematical expressions involving square roots. We've also provided a step-by-step guide to help you understand the concept better. Whether you're a student or a teacher, this guide is designed to help you navigate the world of mathematical expressions with confidence.

Frequently Asked Questions

• Q: What is the condition for the given expression to be defined? A: The given expression is defined when $-3 \leq x \leq 3$.
• Q: Why do we need to rationalize the denominator? A: We need to rationalize the denominator to simplify the expression and make it easier to work with.
• Q: What is the conjugate of the denominator? A: The conjugate of the denominator $\sqrt{3-x}$ is $\sqrt{3-x}$ itself.
• Q: How do we simplify the numerator? A: We can use the fact that $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ to rewrite the numerator as $\sqrt{(9-x^2)(3-x)}$.

Additional Resources

• For more information on equivalent expressions, please refer to our previous article on the topic.
• For more information on rationalizing the denominator, please refer to our article on the topic.
• For more information on simplifying mathematical expressions involving square roots, please refer to our article on the topic.