Solve x² + 4 > 0: Quadratic Inequality Practice

Q: Why doesn't this quadratic have any roots?

The expression $ x^2 + 4 $ can never equal zero because squares are never negative. Since $ x^2 \geq 0 $, the smallest possible value is $ 0 + 4 = 4 $.

Q: How do I know the answer is 'all real numbers'?

Since $ x^2 + 4 $ is always at least 4, it's always greater than zero. Try any number: $ x = 5 $ gives $ 25 + 4 = 29 > 0 $, $ x = -3 $ gives $ 9 + 4 = 13 > 0 $.

Q: What if the inequality was x² + 4 < 0 instead?

Then there would be no solution! Since $ x^2 + 4 $ is always positive, it can never be less than zero. The answer would be 'no real solutions'.

Q: Do I need to factor or use the quadratic formula?

No! This type of problem doesn't require factoring or formulas. Just analyze the expression: since squares are non-negative and we're adding a positive constant, the result is always positive.

Q: How is this different from solving x² - 4 > 0?

With $ x^2 - 4 > 0 $, you can find roots by setting $ x^2 - 4 = 0 $ (giving $ x = ±2 $). But $ x^2 + 4 = 0 $ has no real solutions, so the expression is always positive.

Quadratic Inequalities with Always-Positive Expressions

Solve the following equation:

$x^2+4>0$

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Understand the problem

Solve the following equation:

$x^2+4>0$

Step-by-step solution

The inequality we are solving is $x^2 + 4 > 0$ . Let's analyze this expression:

Consider $x^2$ , which is always non-negative for any real number $x$ . Therefore, $x^2 \geq 0$ .

When we add 4 to $x^2$ , the result is $x^2 + 4$ . Because $x^2 \geq 0$ , adding 4 ensures that $x^2 + 4$ is always greater than 4.

Thus, for any real value of $x$ , the expression $x^2 + 4$ will always satisfy the inequality $x^2 + 4 > 0$ .

In conclusion, the inequality holds true for all values of $x$ . So, the answer is: All values of $x$ .

Final Answer

All values of $x$

Key Points to Remember

Essential concepts to master this topic

Property: $x^2$ is always non-negative for any real number
Technique: Since $x^2 \geq 0$ , adding 4 gives $x^2 + 4 \geq 4 > 0$
Check: Test any value like $x = 0$ : $0^2 + 4 = 4 > 0$ ✓

Common Mistakes

Avoid these frequent errors

Setting the expression equal to zero and solving
Don't solve $x^2 + 4 = 0$ and look for roots = no real solutions exist! This leads students to think there's no answer. Always recognize that $x^2 + 4$ is always positive since it's at least 4 for any real $x$ .

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 doesn't this quadratic have any roots?

The expression $x^2 + 4$ can never equal zero because squares are never negative. Since $x^2 \geq 0$ , the smallest possible value is $0 + 4 = 4$ .

How do I know the answer is 'all real numbers'?

Since $x^2 + 4$ is always at least 4, it's always greater than zero. Try any number: $x = 5$ gives $25 + 4 = 29 > 0$ , $x = -3$ gives $9 + 4 = 13 > 0$ .

What if the inequality was x² + 4 < 0 instead?

Then there would be no solution! Since $x^2 + 4$ is always positive, it can never be less than zero. The answer would be 'no real solutions'.

Do I need to factor or use the quadratic formula?

No! This type of problem doesn't require factoring or formulas. Just analyze the expression: since squares are non-negative and we're adding a positive constant, the result is always positive.

How is this different from solving x² - 4 > 0?

With $x^2 - 4 > 0$ , you can find roots by setting $x^2 - 4 = 0$ (giving $x = ±2$ ). But $x^2 + 4 = 0$ has no real solutions, so the expression is always positive.

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