Problem 46 - Precalculus : Building Concepts and Connections

Question

For what value(s) of \(c\) will the inequality \(x^{2}+c>0\) have all real numbers as its solution? Explain.

Step-by-Step Solution

Answer

The inequality \(x^{2}+c > 0\) will have all real numbers as its solution for any \(c\) that is greater than 0.

Step 1: Reason about the inequality

Let's first observe the inequality. The expression \(x^{2}+c>0\) involves a squared term. Squaring a real number, whether it's positive or negative, always gives a non-negative result. Therefore, \(x^{2}\) is always non-negative, i.e., \(x^{2} \geq 0\).

Step 2: Find a condition for c

To make \(x^{2}+c > 0\), the value of \(c\) must be such that the entire expression is always greater than zero. Since \(x^{2}\) is always non-negative, the smallest it can be is 0. That only happens when \(x\) itself is 0. Therefore, to ensure \(x^{2}+c>0\), we need \(c > 0\). Any value of \(c > 0\) will make the expression \(x^{2}+c\) positive, regardless of the value of \(x\). This means that for every real number \(x\), \(x^{2}+c > 0\) holds true.

Step 3: Formulate the conclusion

Based on the analysis above, we conclude that for all real numbers to be solutions to the inequality \(x^{2}+c > 0\), \(c\) must be any real number greater than 0.

Key Concepts

real numbers

When we talk about real numbers, we're discussing a broad set of numbers that include both rational and irrational numbers. These are the numbers that can be found on the number line.
Real numbers include:

• Positive and negative integers (e.g., -3, 0, 5)
• Fractions and decimals (e.g., 1/2, -0.75)
• Irrational numbers like \pi and the square root of 2

In the context of inequalities, specifying that all real numbers are solutions means any number that fits this broad inclusive definition can satisfy the given inequality. So when you solve an inequality and say it holds for all real numbers, you're asserting that regardless of the number picked, it will satisfy the conditions of the inequality. This makes the solution very broad.
In the exercise discussed here, we claim that the inequality holds true for all these numbers when a certain condition on \(c\) is met.

squared term

Squared terms are fascinating in algebra because they always yield a non-negative result, regardless of the input.
The reason for this is rooted in the operation itself: when you square a real number, you multiply it by itself. This process removes any original negative sign:

• If the original number is positive, squaring it keeps it positive.
• If it's negative, squaring it turns it positive.
• And if it's zero, it remains zero after squaring.

Therefore, any expression of the form \(x^2\) will \underline{always}\ be non-negative, ranging from zero to positive values.
In the original exercise, the squared term \(x^2\) is crucial to understand why \(c\) must be greater than 0 for the inequality \(x^{2}+c>0\) to hold for all real numbers. Because the lowest possible value of \(x^2\) is 0, ensuring that \(c\) remains strictly positive guarantees that the entire expression \(x^2 + c\) remains positive no matter what real number \(x\) is chosen.

non-negative

The term 'non-negative' might sound complex, but it's straightforward – it refers to any number that is zero or greater.
In other words, non-negative numbers include all positive numbers and zero. Therefore:

• Zero is non-negative.
• Positive numbers are non-negative.
• Negative numbers are obviously not non-negative.

This concept is essential when dealing with squared numbers. The reason lies in that squaring any real number results in a non-negative outcome.
In the problem we are discussing, knowing that \(x^2\) is non-negative allows us to determine that for the inequality \(x^2+c>0\), \(c\) must also be non-negative to ensure the inequality holds true. However, since the expression \(x^2\) can equal zero, \(c\) needs to be strictly positive to ensure the entire expression remains strictly positive, hence satisfying the inequality with every real number \(x\).