Final answer:
To ensure that the functions (x²+2x, 3x²+kx) are linearly independent, we compute their Wronskian and set it not equal to zero, resulting in the conclusion that the scalar constant k must not be equal to 1.
Step-by-step explanation:
To compute the scalar constant (k) so that the functions (x²+2x, 3x²+kx) are linearly independent, we need to ensure that their Wronskian is not equal to zero. The Wronskian is a determinant used to test whether a set of functions are linearly independent.
The Wronskian for two functions f(x) and g(x) is defined as:
W(f,g) = |f(x) g(x)|
|f'(x) g'(x)|
For our functions f(x) = x²+2x and g(x) = 3x²+kx, these derivatives are f'(x) = 2x+2 and g'(x) = 6x+k.
Thus, the Wronskian is:
W(f,g) = |(x²+2x) (3x²+kx)|
|(2x+2) (6x+k)|
The determinant of this matrix gives us:
W(f,g) = (x²+2x)(6x+k) - (3x²+kx)(2x+2)
Simplifying, we get:
W(f,g) = (6kx² + kx³) - (6x³ + kx²)
For the functions to be linearly independent, the Wronskian must not equal zero. Therefore:
W(f,g) = kx²(6-x) - x²(6-x)
This simplifies to:
W(f,g) = (k-1)x²(6-x)
For W(f,g) ≠ 0 when x ≠ 0, k must not be equal to 1.
Therefore, for the functions to be linearly independent, k ≠ 1.