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.

breadysetgo Dec 12, 2023

#3**0 **

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.

Boseo Dec 18, 2023