Solve (5x-1)² < 0: Analyzing Negative Values of a Square Function

Quadratic Inequalities with Negative Values

Look at the function below:

y=(5x−1)2 y=\left(5x-1\right)^2

Then determine for which values of x x the following is true:

f(x)<0 f(x) < 0

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Look at the function below:

y=(5x−1)2 y=\left(5x-1\right)^2

Then determine for which values of x x the following is true:

f(x)<0 f(x) < 0

2

Step-by-step solution

To find where f(x)=(5x−1)2<0 f(x) = (5x - 1)^2 < 0 , we start by recognizing a fundamental property of squares:

  • The square of any real number is always non-negative. Therefore, (5x−1)2≥0(5x - 1)^2 \geq 0 for all real x x .

This implies that (5x−1)2(5x - 1)^2 can never be less than zero for any real value of x x .

The inequality (5x−1)2<0 (5x - 1)^2 < 0 has no solution in the real numbers.

Therefore, there are no values of x x for which f(x)<0 f(x) < 0 is true.

So the logical conclusion is: True for no values of x x .

3

Final Answer

True for no values of x x

Key Points to Remember

Essential concepts to master this topic
  • Square Rule: Any real number squared is always non-negative
  • Analysis: (5x−1)2≥0 (5x-1)^2 \geq 0 for all x values
  • Verification: Test any x value: (5(0)−1)2=1≥0 (5(0)-1)^2 = 1 \geq 0 ✓

Common Mistakes

Avoid these frequent errors
  • Solving the equation instead of analyzing the inequality
    Don't set (5x−1)2=0 (5x-1)^2 = 0 and find x=15 x = \frac{1}{5} = wrong focus! The question asks when the expression is negative, not when it equals zero. Always remember that squares of real numbers can never be negative.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of \( x \) where \( f\left(x\right) > 0 \).

AAABBBCCCX

FAQ

Everything you need to know about this question

Why can't a square ever be negative?

+

By definition, when you multiply any real number by itself, the result is always positive or zero. For example: (−3)2=9 (-3)^2 = 9 and 32=9 3^2 = 9 - both positive!

What if I solve (5x−1)2=0 (5x-1)^2 = 0 instead?

+

That gives you x=15 x = \frac{1}{5} , which is where the function equals zero, not where it's negative. The question asks for values where f(x)<0 f(x) < 0 , which never happens!

How do I know there's no solution?

+

Since (5x−1)2≥0 (5x-1)^2 \geq 0 for all real x, and we need (5x−1)2<0 (5x-1)^2 < 0 , there's no overlap. It's like asking for a number that's both positive and negative at the same time!

What's the difference between < 0, = 0, and > 0 for this function?

+
  • < 0: Never happens (no solution)
  • = 0: Only when x=15 x = \frac{1}{5}
  • > 0: For all x≠15 x \neq \frac{1}{5}

Could this work with complex numbers?

+

In complex numbers, squares can be negative, but this problem deals with real numbers only. In real number mathematics, squares are always non-negative.

🌟 Unlock Your Math Potential

Get unlimited access to all 18 The Quadratic Function questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime

More Questions

Click on any question to see the complete solution with step-by-step explanations

Positive and Negative Domains

Solve (3x+30)² > 0: Finding Positive Values of a Quadratic Function
Click to solve
Solve (2x+30)² < 0: Finding Values When a Square is Negative
Click to solve

Product Representation

Solve (2x-16)² > 0: Finding Values Where Function is Positive
Click to solve
Solve: When is -(2x-2¼)² Less Than Zero? Complete Solution Guide
Click to solve