Final answer:
The value of n for which the equation 6v² + 3v - (n + 3) = 0 has one real solution is n = -27/8. This is determined by setting the discriminant of the quadratic equation, b² - 4ac, equal to zero and solving for n.
Step-by-step explanation:
To find the values of n for which the equation 6v² + 3v - (n + 3) = 0 has one real solution, we look at the discriminant of the quadratic equation, which is given by b² - 4ac. A quadratic equation has one real solution (is a perfect square) when the discriminant is equal to zero. In our equation, a = 6, b = 3, and c = -(n + 3).
The discriminant is therefore 3² - 4(6)(-(n + 3)). Setting this to zero for one real solution, we get 9 + 24(n + 3) = 0. Simplifying and solving for n, we have 9 + 24n + 72 = 0, which leads to 24n = -81, and then n = -81/24, which simplifies to n = -27/8.
To check if this solution is reasonable, we substitute n = -27/8 back into the discriminant and find that 3² - 4(6)(-27/8 - 3) does indeed equal zero. Therefore, n = -27/8 is the correct value for which the quadratic equation has one real solution.