Final answer:
The values of k for which the line y = kx + 1 is a tangent to the curve y = 2x² + x + 3 are 1 + sqrt(8) and 1 - sqrt(8).
Step-by-step explanation:
To find the values of k for which the line y = kx + 1 is a tangent to the curve y = 2x² + x + 3, we need to find the point(s) of intersection between the line and the curve.
Let's equate the two equations and solve for x:
2x² + x + 3 = kx + 1
2x² + (1-k)x + 2 = 0
For the line to be tangent to the curve, this quadratic equation should have exactly one solution.
This means the discriminant (b² - 4ac) should be equal to 0.
Using the discriminant formula, (1 - k)² - 4(2)(2) = 0
Simplifying, we get k² - 2k - 7 = 0
Using the quadratic formula,
k = (-b ± sqrt(b² - 4ac))/(2a)
k = (2 ± sqrt((-2)² - 4(1)(-7)))/(2(1))
k = (2 ± sqrt(32))/(2)
k = 1 ± sqrt(8)
Therefore, the values of k for which the line is a tangent to the curve are 1 + sqrt(8) and 1 - sqrt(8).