Solve y=x²: Finding Values Where Function is Negative

Quadratic Functions with Sign Analysis

Given the function:

y=x2 y=x^2

Determine for which values of x f(x)<0 f(x) < 0 holds

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Step-by-step written solution

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1

Understand the problem

Given the function:

y=x2 y=x^2

Determine for which values of x f(x)<0 f(x) < 0 holds

2

Step-by-step solution

To determine for which values of x x the function y=x2 y = x^2 satisfies f(x)<0 f(x) < 0 , we need to analyze the nature of the quadratic function.

Step 1: Recognize the function y=x2 y = x^2 is parabolic and opens upwards. For any real number x x , x2 x^2 is always non-negative, i.e., x20 x^2 \geq 0 .

Step 2: Since squaring any real number results in a value greater than or equal to zero, it is not possible for x2 x^2 to be less than zero.

Conclusion: Therefore, there are no real values of x x for which x2<0 x^2 < 0 . The correct conclusion is that no x x satisfies x2<0 x^2 < 0 .

3

Final Answer

x0 x\ne0

Key Points to Remember

Essential concepts to master this topic
  • Rule: Squaring any real number always gives non-negative result
  • Technique: Analyze x20 x^2 \geq 0 for all real x values
  • Check: Test specific values: (3)2=9>0 (-3)^2 = 9 > 0 and 32=9>0 3^2 = 9 > 0

Common Mistakes

Avoid these frequent errors
  • Thinking negative inputs give negative outputs
    Don't assume x2<0 x^2 < 0 when x is negative = wrong conclusion! Squaring removes the negative sign, so (5)2=25>0 (-5)^2 = 25 > 0 . Always remember that squaring any real number produces a positive result or zero.

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 x2 x^2 ever be negative?

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When you square any number, you multiply it by itself. A negative times a negative gives a positive, and a positive times a positive stays positive. So x20 x^2 \geq 0 always!

What about when x is negative like x = -4?

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Even when x is negative, x2 x^2 becomes positive! For example: (4)2=(4)×(4)=16>0 (-4)^2 = (-4) \times (-4) = 16 > 0 . The square removes the negative sign.

So the answer is 'No x' then?

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Exactly! There are no real values of x where x2<0 x^2 < 0 . The function y=x2 y = x^2 is always greater than or equal to zero.

What's the minimum value of x2 x^2 ?

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The minimum value is 0, which occurs when x=0 x = 0 . At this point, 02=0 0^2 = 0 , and for any other x value, x2>0 x^2 > 0 .

How do I remember this for tests?

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Think: "Squares are never negative!" Whether you start with positive or negative numbers, squaring always gives you a positive result or zero. It's a fundamental property of real numbers.

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