For what value(s) of n is the inequation x^2 + 2x ≥ 2n true for all values of x?

We need to find the value(s) of \( n \) for which the inequality \( x^2 + 2x \geq 2n \) holds for all values of \( x \).

### Step 1: Rewrite the Inequality
Start by rearranging the inequality to bring all terms to one side:
\[
x^2 + 2x - 2n \geq 0
\]
This can be considered as a quadratic inequality of the form:
\[
x^2 + 2x - 2n \geq 0
\]
We need to find the values of \( n \) such that this inequality is satisfied for all \( x \).

### Step 2: Analyze the Quadratic Expression
For the quadratic inequality \( x^2 + 2x - 2n \geq 0 \) to hold for all \( x \), the quadratic expression must either have no real roots (i.e., the discriminant must be non-positive) or it must have a double root where the quadratic expression is non-negative.

### Step 3: Calculate the Discriminant
The discriminant \( \Delta \) for the quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[
\Delta = b^2 - 4ac
\]
For our quadratic expression \( x^2 + 2x - 2n \), we have:
- \( a = 1 \)
- \( b = 2 \)
- \( c = -2n \)

The discriminant is:
\[
\Delta = 2^2 - 4(1)(-2n) = 4 + 8n = 4(1 + 2n)
\]

### Step 4: Ensure Non-Negativity of the Quadratic
For the quadratic expression to be non-negative for all \( x \), the discriminant must be non-positive:
\[
4(1 + 2n) \leq 0
\]
Simplifying, we get:
\[
1 + 2n \leq 0
\]
\[
2n \leq -1
\]
\[
n \leq -\frac{1}{2}
\]

### Conclusion
The inequality \( x^2 + 2x \geq 2n \) is true for all values of \( x \) if and only if \( n \leq -\frac{1}{2} \).

Would you like a more detailed explanation or have any questions?

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Here are 5 related questions:
1. How do you determine if a quadratic expression is always non-negative?
2. What is the effect of the discriminant on the sign of a quadratic expression?
3.

Discover how to find values of n for which the quadratic inequality x^2 + 2x ≥ 2n holds true for all values of x. This problem involves analyzing quadratic inequalities and calculating discriminants. Suitable for high school students.

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