Any help is appreciated :) I don't know how to start this problem!
For an integer y, the inequality x^2+yx+36<0 has no real solutions in x. Find the number of different possible values of y.
We can solve this problem similarly to the previous one by analyzing the discriminant of the quadratic:
Discriminant and no real solutions: Recall that for the quadratic ax^2 + bx + c = 0 to have no real solutions, the discriminant b^2 - 4ac must be negative. In this case, a = 1, b = y, and c = 1.
Therefore, the condition for no real solutions becomes:
y^2 - 4 * 1 * 1 < 0
Simplifying the inequality:
y^2 - 4 < 0
Adding 4 to both sides:
y^2 < 4
Taking the square root of both sides (remembering that we need both positive and negative solutions):
-2 < y < 2
Integer solutions: Now, we need to count the number of integer values of y within this range. Since -2 and 2 are not integers, we consider the closed interval [-1, 1]. However, we need to be careful because the original inequality involved x^2 instead of just x.
Special cases: The original inequality x^2 + yx + 1 < 0 becomes equal to zero for x = -1 and x = 1 when y = -1 and y = 1, respectively. These values of y cannot satisfy the condition of no real solutions because they introduce real roots through x = -1 and x = 1.
Therefore, we exclude -1 and 1 from the possible values of y.
Final count: After excluding -1 and 1, we are left with the closed interval [-1, 0) and (0, 1]. Both intervals contain 1 integer each: 0 for the first and 1 for the second.
Therefore, there are 1 + 1 = 2 possible values of y for which the inequality has no real solutions (excluding y = -1 and y = 1).
Note: This case is different from the previous one because the original inequality involves x^2, which can produce double roots at x = 0. This necessitates excluding y = -1 and y = 1, which would otherwise satisfy the condition but introduce double roots that violate the requirement of no real solutions.