Finding the Domain: Analyze the Function (3x+12)/√(5x-10)

Q: Why can't x equal 2 if the square root equals zero?

When x = 2, we get $ \sqrt{5(2)-10} = \sqrt{0} = 0 $. This makes the denominator zero, and division by zero is undefined in mathematics!

Q: What's the difference between ≥ and > for domain problems?

Use ≥ when the square root is not in a denominator (zero is okay). Use > when it is in a denominator (zero would cause division by zero).

Q: How do I solve 5x - 10 > 0 step by step?

Start with $ 5x - 10 > 0 $. Add 10: $ 5x > 10 $. Divide by 5: $ x > 2 $. Always keep the inequality direction when dividing by positive numbers!

Q: What if I had a negative number under the square root?

Square roots of negative numbers are not real. That's why we need the expression under the square root to be positive (greater than zero).

Q: Does the numerator 3x + 12 affect the domain?

No! The numerator $ 3x + 12 $ can be any real number. Only the denominator creates restrictions because it cannot equal zero.

Q: How can I check if x = 3 is in the domain?

Substitute: $ \sqrt{5(3)-10} = \sqrt{15-10} = \sqrt{5} $. Since √5 is positive and real, x = 3 is in the domain!

Domain Restrictions with Square Root Denominators

Look at the following function:

$\frac{3x+12}{\sqrt{5x-10}}$

What is the domain of the function?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:09 Does the function have a domain? If it does, let's find out what it is.

00:14 Remember, the root must be of a positive number. It should be greater than zero.

00:22 Now, let's go step by step to isolate X.

00:34 And that's how we find the solution to the question!

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the following function:

$\frac{3x+12}{\sqrt{5x-10}}$

What is the domain of the function?

Step-by-step solution

The function given is $\frac{3x+12}{\sqrt{5x-10}}$ .

To find the domain, we focus on the expression within the denominator's square root: $\sqrt{5x-10}$ .

The expression $5x-10$ must be greater than $0$ for the square root to be defined and the denominator to be non-zero.

Let's solve the inequality:

Set $5x-10 > 0$ .
Add $10$ to both sides: $5x > 10$ .
Divide both sides by $5$ : $x > 2$ .

This means the domain of the function is all $x$ such that $x > 2$ .

The domain is, therefore, correctly expressed as $x > 2$ .

Final Answer

$x > 2$

Key Points to Remember

Essential concepts to master this topic

Rule: Square root expressions must be positive and non-zero
Technique: Set 5x - 10 > 0 and solve inequality
Check: Test x = 3: √(5(3) - 10) = √5 works ✓

Common Mistakes

Avoid these frequent errors

Using ≥ instead of > for square root denominators
Don't set 5x - 10 ≥ 0 = including x = 2 in domain! This makes the denominator zero and creates division by zero. Always use > to ensure the denominator stays positive and non-zero.

Practice Quiz

Test your knowledge with interactive questions

$\frac{6}{x+5}=1$

What is the field of application of the equation?

FAQ

Everything you need to know about this question

Why can't x equal 2 if the square root equals zero?

When x = 2, we get $\sqrt{5(2)-10} = \sqrt{0} = 0$ . This makes the denominator zero, and division by zero is undefined in mathematics!

What's the difference between ≥ and > for domain problems?

Use ≥ when the square root is not in a denominator (zero is okay). Use > when it is in a denominator (zero would cause division by zero).

How do I solve 5x - 10 > 0 step by step?

Start with $5x - 10 > 0$ . Add 10: $5x > 10$ . Divide by 5: $x > 2$ . Always keep the inequality direction when dividing by positive numbers!

What if I had a negative number under the square root?

Square roots of negative numbers are not real. That's why we need the expression under the square root to be positive (greater than zero).

Does the numerator 3x + 12 affect the domain?

No! The numerator $3x + 12$ can be any real number. Only the denominator creates restrictions because it cannot equal zero.

How can I check if x = 3 is in the domain?

Substitute: $\sqrt{5(3)-10} = \sqrt{15-10} = \sqrt{5}$ . Since √5 is positive and real, x = 3 is in the domain!

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