Solve the Quadratic Inequality: x² + 9 > 0

Quadratic Inequalities with Always-Positive Expressions

Solve the following equation:

x2+9>0 x^2+9>0

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1

Understand the problem

Solve the following equation:

x2+9>0 x^2+9>0

2

Step-by-step solution

Let's explore this problem step-by-step:

The inequality given is x2+9>0 x^2 + 9 > 0 .

1. To understand this inequality, we start by considering the expression x2 x^2 . We know that for any real number x x , x20 x^2 \geq 0 . This means x2 x^2 is always non-negative.

2. Since x20 x^2 \geq 0 for every real number, adding 9 to x2 x^2 will necessarily make the expression greater than zero, because a non-negative number plus a positive number gives a positive result: x2+99>0 x^2 + 9 \geq 9 > 0 .

3. Therefore, the inequality x2+9>0 x^2 + 9 > 0 holds true for all real numbers x x . There is no value of x x that makes the left side equal to or less than zero.

4. Thus, the solution to the inequality is that it holds for all values of x x .

Consequently, the correct choice from the options provided is:

  • All values of x x

Therefore, the solution is that the inequality x2+9>0 x^2 + 9 > 0 is true for all values of x x .

3

Final Answer

All values of x x

Key Points to Remember

Essential concepts to master this topic
  • Rule: x20 x^2 \geq 0 for all real numbers x
  • Technique: Since x2+99>0 x^2 + 9 \geq 9 > 0 , inequality holds everywhere
  • Check: Test any value: (5)2+9=34>0 (-5)^2 + 9 = 34 > 0

Common Mistakes

Avoid these frequent errors
  • Setting the expression equal to zero to find boundary points
    Don't solve x2+9=0 x^2 + 9 = 0 for real solutions = no real roots exist! This leads to confusion about intervals. Always recognize that x2+9 x^2 + 9 is always positive since the minimum value is 9.

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 find where x² + 9 = 0?

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Great question! The equation x2+9=0 x^2 + 9 = 0 has no real solutions because x2 x^2 is never negative. The smallest value of x2+9 x^2 + 9 is 9 (when x = 0).

How do I know this is true for ALL values of x?

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Since x20 x^2 \geq 0 always, adding 9 gives us x2+99 x^2 + 9 \geq 9 . Because 9 > 0, the expression is always greater than zero no matter what x equals!

Should I test specific values to be sure?

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Absolutely! Try x = 0: 02+9=9>0 0^2 + 9 = 9 > 0
Try x = -3: (3)2+9=18>0 (-3)^2 + 9 = 18 > 0
Every value you test will work!

What if the problem was x² + 9 < 0 instead?

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Then there would be no solution! Since x2+9 x^2 + 9 is always positive, it can never be less than zero. The answer would be 'no real solutions' or the empty set.

Is this different from solving x² - 9 > 0?

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Yes, very different! For x29>0 x^2 - 9 > 0 , you'd factor to get (x3)(x+3)>0 (x-3)(x+3) > 0 , which has specific intervals as solutions. But x2+9 x^2 + 9 is always positive!

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