The Conditions for a Defined Quotient

Oct 13, 2025 by magicalceylontours.com 57 views

When dealing with mathematical expressions, it's essential to consider the conditions under which the expression is defined. In the case of a quotient, such as the one given in the problem, we need to ensure that the denominator is not equal to zero and that the expression inside the square root is non-negative. In this article, we will explore the conditions for which the quotient $$\frac{\sqrt{15(x-1)}}{\sqrt{2x^2}}$$ is defined.

The Conditions for a Defined Quotient

To determine the conditions for which the quotient is defined, we need to consider the following:

The expression inside the square root in the numerator, $$15(x-1)$$, must be non-negative.
The expression inside the square root in the denominator, $$2x^2$$, must be positive.
The denominator, $$\sqrt{2x^2}$$, cannot be equal to zero.

Analyzing the Conditions

Let's analyze each of the conditions in more detail:

The Expression Inside the Square Root in the Numerator

The expression $$15(x-1)$$ must be non-negative. This means that:

15(x-1) \geq 0

Simplifying the inequality, we get:

x-1 \geq 0

Adding 1 to both sides, we get:

x \geq 1

The Expression Inside the Square Root in the Denominator

The expression $$2x^2$$ must be positive. This means that:

2x^2 > 0

Since $$x^2$$ is always non-negative, we can divide both sides by 2:

x^2 > 0

Taking the square root of both sides, we get:

x \neq 0

The Denominator Cannot Be Equal to Zero

The denominator, $$\sqrt{2x^2}$$, cannot be equal to zero. This means that:

\sqrt{2x^2} \neq 0

Squaring both sides, we get:

2x^2 \neq 0

Since $$x^2$$ is always non-negative, we can divide both sides by 2:

x^2 \neq 0

Taking the square root of both sides, we get:

x \neq 0

Combining the Conditions

Now that we have analyzed each of the conditions, we can combine them to determine the conditions for which the quotient is defined. The expression inside the square root in the numerator must be non-negative, which means that $$x \geq 1$$. The expression inside the square root in the denominator must be positive, which means that $$x \neq 0$$. The denominator cannot be equal to zero, which also means that $$x \neq 0$$.

Therefore, the conditions for which the quotient is defined are:

x \geq 1 \text{ and } x \neq 0

Simplifying the Conditions

We can simplify the conditions by combining the two inequalities:

x \geq 1 \text{ and } x \neq 0

This can be written as:

x > 0

Conclusion

In conclusion, the conditions for which the quotient $$\frac{\sqrt{15(x-1)}}{\sqrt{2x^2}}$$ is defined are $$x > 0$$. This means that the quotient is defined for all values of $$x$$ that are greater than zero.

Answer

The correct answer is:

x > 0

In this article, we will address some of the most frequently asked questions related to the conditions for which the quotient $$\frac{\sqrt{15(x-1)}}{\sqrt{2x^2}}$$ is defined.

Q: What is the condition for the expression inside the square root in the numerator?

A: The expression inside the square root in the numerator, $$15(x-1)$$, must be non-negative. This means that $$x \geq 1$$.

Q: What is the condition for the expression inside the square root in the denominator?

A: The expression inside the square root in the denominator, $$2x^2$$, must be positive. This means that $$x \neq 0$$.

Q: What is the condition for the denominator to be non-zero?

A: The denominator, $$\sqrt{2x^2}$$, cannot be equal to zero. This means that $$x \neq 0$$.

Q: How do I combine the conditions to determine the conditions for which the quotient is defined?

A: To combine the conditions, we need to consider the intersection of the two inequalities. In this case, we have $$x \geq 1$$ and $$x \neq 0$$. The intersection of these two inequalities is $$x > 0$$.

Q: What is the final answer?

A: The final answer is $$x > 0$$.

Q: Why is the answer $$x > 0$$ and not $$x \geq 1$$?

A: The answer $$x > 0$$ is correct because it satisfies both conditions: $$x \geq 1$$ and $$x \neq 0$$. The condition $$x \geq 1$$ is a subset of the condition $$x > 0$$, so $$x > 0$$ is a more general and correct answer.

Q: What if $$x = 0$$?

A: If $$x = 0$$, then the denominator $$\sqrt{2x^2}$$ is equal to zero, which means that the quotient is undefined.

Q: What if $$x < 0$$?

A: If $$x < 0$$, then the expression inside the square root in the numerator, $$15(x-1)$$, is negative, which means that the quotient is undefined.

Q: What if $$x = 1$$?

A: If $$x = 1$$, then the expression inside the square root in the numerator, $$15(x-1)$$, is equal to zero, which means that the quotient is undefined.

Conclusion

Answer

The correct answer is:

x > 0

This is option C.