The Role of Square Roots in Defining the Quotient

Oct 13, 2025 by sorusgentrification.com 120 views

When dealing with mathematical expressions involving square roots, it's essential to consider the conditions under which the expression is defined. In this case, we're given the quotient $\frac{\sqrt{6x^2}}{\sqrt{4x}}$ and asked to determine the inequality that represents all values of $x$ for which this quotient is defined.

The Role of Square Roots in Defining the Quotient

To begin, let's examine the square roots present in the quotient. The expression $\sqrt{6x^2}$ is defined as long as the radicand, $6x^2$ , is non-negative. This means that $6x^2 \geq 0$ . Since $6$ is a positive constant, the only requirement for the expression to be defined is that $x^2 \geq 0$ . As the square of any real number is non-negative, this condition is always satisfied, and the expression $\sqrt{6x^2}$ is defined for all real values of $x$ .

The Denominator of the Quotient

The denominator of the quotient is $\sqrt{4x}$ . For this expression to be defined, the radicand, $4x$ , must be non-negative. This means that $4x \geq 0$ . Since $4$ is a positive constant, the only requirement for the expression to be defined is that $x \geq 0$ . If $x$ is negative, the expression $\sqrt{4x}$ would involve the square root of a negative number, which is undefined in the real number system.

Combining the Conditions

To determine the inequality that represents all values of $x$ for which the quotient is defined, we need to combine the conditions for the numerator and the denominator. As we've established, the expression $\sqrt{6x^2}$ is defined for all real values of $x$ , and the expression $\sqrt{4x}$ is defined only when $x \geq 0$ . Therefore, the quotient is defined only when $x \geq 0$ .

Conclusion

In conclusion, the inequality that represents all values of $x$ for which the quotient $\frac{\sqrt{6x^2}}{\sqrt{4x}}$ is defined is $x \geq 0$ . This is because the expression $\sqrt{6x^2}$ is defined for all real values of $x$ , and the expression $\sqrt{4x}$ is defined only when $x \geq 0$ .

Answer

The correct answer is:

A. $x \geq 0$

Explanation

The explanation for the correct answer is as follows:

The expression $\sqrt{6x^2}$ is defined for all real values of $x$ .
The expression $\sqrt{4x}$ is defined only when $x \geq 0$ .
Therefore, the quotient is defined only when $x \geq 0$ .

Comparison with Other Options

Let's compare the correct answer with the other options:

Option B: $x > 1$ . This option is incorrect because the quotient is defined for all values of $x$ greater than or equal to 0, not just values greater than 1.
Option C: $x > -1$ . This option is incorrect because the quotient is defined for all values of $x$ greater than or equal to 0, not just values greater than -1.
Option D: $x < 0$ . This option is incorrect because the quotient is not defined for any values of $x$ less than 0.

Final Answer

The final answer is:

$A. $x \geq 0$<br/> **Quotient Definition Inequality: Frequently Asked Questions**$

In the previous article, we explored the conditions under which the quotient $\frac{\sqrt{6x^2}}{\sqrt{4x}}$ is defined. We determined that the inequality representing all values of $x$ for which the quotient is defined is $x \geq 0$ . In this article, we'll address some frequently asked questions related to this topic.

Q: Why is the expression $\sqrt{6x^2}$ defined for all real values of $x$ ?

A: The expression $\sqrt{6x^2}$ is defined for all real values of $x$ because the square of any real number is non-negative. This means that $6x^2 \geq 0$ for all real values of $x$ . As a result, the expression $\sqrt{6x^2}$ is defined for all real values of $x$ .

Q: Why is the expression $\sqrt{4x}$ defined only when $x \geq 0$ ?

A: The expression $\sqrt{4x}$ is defined only when $x \geq 0$ because the radicand, $4x$ , must be non-negative. This means that $4x \geq 0$ . Since $4$ is a positive constant, the only requirement for the expression to be defined is that $x \geq 0$ . If $x$ is negative, the expression $\sqrt{4x}$ would involve the square root of a negative number, which is undefined in the real number system.

Q: What happens if $x$ is negative?

A: If $x$ is negative, the expression $\sqrt{4x}$ would involve the square root of a negative number, which is undefined in the real number system. Therefore, the quotient $\frac{\sqrt{6x^2}}{\sqrt{4x}}$ is not defined for any negative values of $x$ .

Q: Can the quotient be defined for any value of $x$ ?

A: No, the quotient $\frac{\sqrt{6x^2}}{\sqrt{4x}}$ cannot be defined for any value of $x$ . The expression $\sqrt{4x}$ is defined only when $x \geq 0$ , and the expression $\sqrt{6x^2}$ is defined for all real values of $x$ . Therefore, the quotient is defined only when $x \geq 0$ .

Q: What is the significance of the inequality $x \geq 0$ ?

A: The inequality $x \geq 0$ represents all values of $x$ for which the quotient $\frac{\sqrt{6x^2}}{\sqrt{4x}}$ is defined. This means that the quotient is defined for all non-negative values of $x$ .

Q: Can the quotient be defined for any non-negative value of $x$ ?

A: Yes, the quotient $\frac{\sqrt{6x^2}}{\sqrt{4x}}$ can be defined for any non-negative value of $x$ . This means that the quotient is defined for all values of $x$ greater than or equal to 0.

Q: What is the final answer?

A: The final answer is:

A. $x \geq 0$

Conclusion

In conclusion, the quotient $\frac{\sqrt{6x^2}}{\sqrt{4x}}$ is defined only when $x \geq 0$ . This is because the expression $\sqrt{6x^2}$ is defined for all real values of $x$ , and the expression $\sqrt{4x}$ is defined only when $x \geq 0$ . We hope this article has helped to clarify any questions you may have had about this topic.