Question

به ازای چه مقادیری از \\(m\\)، سهمی \\(y = mx^2 - mx - 1\\) همواره پایین محور \\(x\\) هاست؟

Explanation:

🆕 New Concept Discovered: Quadratic Sign Analysis
Determining when a parabola lies entirely on one side of the x-axis.

Step 1: Translate the Problem

The Persian text asks:
"For what values of \( m \), is the parabola \( y = mx^2 - mx - 1 \) always below the x-axis?"

For a quadratic function \( y = ax^2 + bx + c \) to be always below the x-axis (meaning \( y < 0 \) for all real \( x \)), two conditions must be met:

1. The parabola must open downwards: \( a < 0 \)
2. The parabola must not intersect or touch the x-axis (no real roots): \( \Delta < 0 \)

Step 2: Apply the First Condition (\( a < 0 \))

Identify the coefficients of the quadratic equation \( y = mx^2 - mx - 1 \):

• \( a = m \)
• \( b = -m \)
• \( c = -1 \)

For the parabola to open downwards:

$$ m < 0 $$

Step 3: Apply the Second Condition (\( \Delta < 0 \))

Calculate the discriminant \( \Delta = b^2 - 4ac \):

$$ \Delta = (-m)^2 - 4(m)(-1) $$

$$ \Delta = m^2 + 4m $$

Set the discriminant to be strictly less than zero:

$$ m^2 + 4m < 0 $$

$$ m(m + 4) < 0 $$

The roots of the equation \( m(m + 4) = 0 \) are \( m = 0 \) and \( m = -4 \).
For the product to be negative, \( m \) must lie between these two roots:

$$ -4 < m < 0 $$

Step 4: Find the Intersection of Both Conditions

We must satisfy both conditions simultaneously:

1. \( m < 0 \)
2. \( -4 < m < 0 \)

The intersection of these two intervals is:

$$ m \in (-4, 0) $$

Answer:

The values of \( m \) for which the parabola is always below the x-axis are:

$$ -4 < m < 0 $$