Solve x² + 4 > 0: Understanding Quadratic Inequalities

Quadratic Inequalities with Always-Positive Expressions

Solve the following equation:

x2+4>0 x^2+4>0

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1

Understand the problem

Solve the following equation:

x2+4>0 x^2+4>0

2

Step-by-step solution

To solve this problem, let's examine the inequality x2+4>0 x^2 + 4 > 0 .

The expression x2+4 x^2 + 4 consists of two terms: x2 x^2 and 4 4 . Notice that:

  • The term x2 x^2 is always non-negative, which means x20 x^2 \geq 0 for any real number x x .
  • The constant term 4 4 is positive.

Combining these observations, we see that:

  • Since x2 x^2 is non-negative, x2+44 x^2 + 4 \geq 4 .
  • Therefore, x2+4 x^2 + 4 is always greater than zero, as adding 4 to a non-negative number will always yield a positive result.

Thus, there are no values of x x for which the expression x2+4 x^2 + 4 is zero or negative. Instead, the expression is always positive for all real numbers x x .

Therefore, the solution to the inequality x2+4>0 x^2 + 4 > 0 is all values of x x .

3

Final Answer

All values of x x

Key Points to Remember

Essential concepts to master this topic
  • Property: x² is always non-negative for any real number x
  • Analysis: Since x² ≥ 0, then x² + 4 ≥ 4 > 0
  • Verification: Test any value like x = 0: 0² + 4 = 4 > 0 ✓

Common Mistakes

Avoid these frequent errors
  • Trying to solve x² + 4 = 0 first
    Don't attempt to find roots by setting x² + 4 = 0 since this has no real solutions! This wastes time and confuses the analysis. Always recognize that x² + positive constant is always positive for all real numbers.

Practice Quiz

Test your knowledge with interactive questions

Solve the following equation:

\( x^2+4>0 \)

FAQ

Everything you need to know about this question

Why can't I solve x² + 4 = 0 to find boundary points?

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Because x2+4=0 x^2 + 4 = 0 means x2=4 x^2 = -4 , but squares of real numbers cannot be negative! This equation has no real solutions, so there are no boundary points to consider.

How do I know when a quadratic expression is always positive?

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Look for expressions like x2+positive number x^2 + \text{positive number} . Since x20 x^2 \geq 0 always, adding a positive constant makes the entire expression always positive.

What if the problem was x² - 4 > 0 instead?

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Then you would solve x24=0 x^2 - 4 = 0 to find boundary points x = ±2, then test intervals. The key difference is subtracting vs. adding the constant!

Do I need to graph this inequality?

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Not necessary here! Since x2+4 x^2 + 4 is always positive, the graph of y = x² + 4 is always above the x-axis. The solution is simply all real numbers.

How can I be sure my answer covers all real numbers?

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Test several values: when x = 0, we get 4 > 0 ✓. When x = -3, we get 9 + 4 = 13 > 0 ✓. When x = 100, we get 10,004 > 0 ✓. Every test confirms the pattern!

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