To solve these problems, we need to analyze the given mathematical expressions.
Part 1: Find the set of values for \( m \) such that \( x^2 + 3m + m \) is positive for all real values of \( x \).
First, simplify the expression:
\[ x^2 + 3m + m = x^2 + 4m \]
To ensure \( x^2 + 4m > 0 \) for all real \( x \), we need the quadratic function \( x^2 + 4m \) to be positive definite.
- The leading coefficient of the quadratic term (which is \( x^2 \)) is positive.
- The quadratic will be positive for all \( x \) if the constant term \( 4m \) is positive, meaning:
\[ 4m > 0 \]
This simplifies to:
\[ m > 0 \]
Thus, the set of values for \( m \) is:
\[ m \in (0, \infty) \]
Part 2: Find the value of \( k \) for which the equation \( (m-3)x^2 - 8x + 4 = 0 \) has real solutions.
For a quadratic equation \( ax^2 + bx + c = 0 \) to have real solutions, the discriminant must be non-negative. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
First, identify the coefficients from the equation:
- \( a = m - 3 \)
- \( b = -8 \)
- \( c = 4 \)
Now calculate the discriminant:
\[ D = (-8)^2 - 4(m - 3)(4) = 64 - 16(m - 3) \]
Simplifying this gives:
\[ D = 64 - 16m + 48 = 112 - 16m \]
For real solutions, we need \( D \geq 0 \):
\[ 112 - 16m \geq 0 \]
Solving for \( m \):
\[ 112 \geq 16m \]
\[ \frac{112}{16} \geq m \]
\[ 7 \geq m \]
Thus, we find:
\[ m \leq 7 \]
Summary
- The set of values for \( m \) such that \( x^2 + 4m > 0 \) for all real \( x \) is \( m \in (0, \infty) \).
- For the quadratic equation to have real solutions, \( m \) must satisfy \( m \leq 7 \).
Conclusively:
- The solution set for \( m \) from Part 1 is \( (0, \infty) \).
- The solution set for \( m \) from Part 2 is \( (-\infty, 7] \).
Values that satisfy both conditions:
The intersection of the two sets where \( 0 < m \leq 7 \) gives:
\[ m \in (0, 7] \]