Finding X: Determine When y=5x² is Greater Than Zero

Q: Why isn't the answer "all x" if the parabola opens upward?

While the parabola $ y = 5x^2 $ does open upward, it touches the x-axis at x = 0 where y = 0. Since we need greater than zero, not greater than or equal to zero, we must exclude this point.

Q: How do I know when a quadratic is always positive or negative?

Look at the coefficient of $ x^2 $! If it's positive (like +5), the parabola opens upward. If negative, it opens downward. Then check where it equals zero to find exclusions.

Q: What's the difference between > 0 and ≥ 0?

> 0 means strictly greater than zero (excludes zero), while ≥ 0 means greater than or equal to zero (includes zero). For this problem, we exclude x = 0 because we need strictly greater than.

Q: Do I need to consider both positive and negative x values?

Yes! Since $ x^2 $ makes any negative number positive, both positive and negative x values (except zero) will make $ 5x^2 > 0 $. For example: $ 5(-3)^2 = 5(9) = 45 > 0 $.

Quadratic Inequalities with Non-Zero Solutions

Given the function:

$y=5x^2$

Determine for which values of x $f\left(x\right) > 0$ holds

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Given the function:

$y=5x^2$

Determine for which values of x $f\left(x\right) > 0$ holds

Step-by-step solution

The given function is $y = 5x^2$ . This is a quadratic function where the coefficient of $x^2$ is positive, making the parabolic shape open upwards.

The expression $5x^2$ is always non-negative. For this function to be greater than zero, it should not be equal to zero. We find the expression equals zero when $x = 0$ .

When $x \neq 0$ , $5x^2$ is positive. Therefore, $y = 5x^2 > 0$ whenever $x \neq 0$ . This applies to both positive and negative $x$ values, except for zero.

Thus, the correct answer is: $x \neq 0$ .

Final Answer

$x\ne0$

Key Points to Remember

Essential concepts to master this topic

Rule: Quadratic expressions $ax^2$ with positive coefficient are always non-negative
Technique: Find when $5x^2 = 0$ gives x = 0, exclude this value
Check: Test x = 2: $5(2)^2 = 20 > 0$ and x = -1: $5(-1)^2 = 5 > 0$ ✓

Common Mistakes

Avoid these frequent errors

Thinking the answer is "all x" since the parabola opens upward
Don't assume $5x^2 > 0$ for all x values = includes x = 0 where the function equals zero! At x = 0, we have $5(0)^2 = 0$ , which is not greater than zero. Always check where the expression equals zero first, then exclude those points.

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 $f\left(x\right) > 0$ .

FAQ

Everything you need to know about this question

Why isn't the answer "all x" if the parabola opens upward?

While the parabola $y = 5x^2$ does open upward, it touches the x-axis at x = 0 where y = 0. Since we need greater than zero, not greater than or equal to zero, we must exclude this point.

How do I know when a quadratic is always positive or negative?

Look at the coefficient of $x^2$ ! If it's positive (like +5), the parabola opens upward. If negative, it opens downward. Then check where it equals zero to find exclusions.

What's the difference between > 0 and ≥ 0?

> 0 means strictly greater than zero (excludes zero), while ≥ 0 means greater than or equal to zero (includes zero). For this problem, we exclude x = 0 because we need strictly greater than.

Do I need to consider both positive and negative x values?

Yes! Since $x^2$ makes any negative number positive, both positive and negative x values (except zero) will make $5x^2 > 0$ . For example: $5(-3)^2 = 5(9) = 45 > 0$ .

How can I check my answer quickly?

Pick a positive number (like x = 1) and a negative number (like x = -2), then substitute: $5(1)^2 = 5 > 0$ ✓ and $5(-2)^2 = 20 > 0$ ✓. Also verify that x = 0 gives zero, not positive.

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