Show that \( 2 x^{2}+k x+k^{2}+1 \), where \( k \) is a constant, is always positive.

To show that the expression \( 2x^2 + kx + k^2 + 1 \) is always positive for all \( x \), we can analyze it as a quadratic function in \( x \). The general form of a quadratic function is given by: \[ ax^2 + bx + c \] where \( a = 2 \), \( b = k \), and \( c = k^2 + 1 \). ### Step 1: Check the leading coefficient The leading coefficient \( a = 2 \) is positive. This means that the parabola opens upwards. ### Step 2: Find the discriminant The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c \) is given by: \[ D = b^2 - 4ac \] Substituting our values for \( a \), \( b \), and \( c \): \[ D = k^2 - 4(2)(k^2 + 1) = k^2 - 8k^2 - 8 = -7k^2 - 8 \] ### Step 3: Analyze the discriminant The discriminant \( D \) must be non-positive for the quadratic to have no real roots (which means it does not cross the x-axis). We can see that: \[ D = -7k^2 - 8 \] Since \( -7k^2 \) is always non-positive (as \( k^2 \geq 0 \)), and \( -8 \) is negative, we have: \[ D \leq -8 < 0 \] This means that \( D < 0 \) for all values of \( k \). Therefore, the quadratic \( 2x^2 + kx + (k^2 + 1) \) has no real roots. ### Step 4: Conclusion Since the quadratic opens upwards (because \( a > 0 \)) and has no real roots (because \( D < 0 \)), it must be positive for all \( x \). Thus, we conclude that: \[ 2x^2 + kx + k^2 + 1 > 0 \quad \text{for all } x \in \mathbb{R} \] This shows that the expression \( 2x^2 + kx + k^2 + 1 \) is always positive.