+0  
 
+1
3
2
avatar+5 

For an integer n, the inequality x^2+nx+15<-21
has no real solutions in x. Find the number of different possible values of n.

 Dec 13, 2023
 #1
avatar+222 
+1

We want to find the number of different possible values of n for which the quadratic x^2 + nx + 15 < -21has no real solutions.

 

For a quadratic ax^2 + bx + c = 0 to have no real solutions, the discriminant b^2 - 4ac must be negative. In our case, a = 1, b = n, and c = 15. Therefore, we need n^2 - 4 * 1 * 15 < 0.

 

Simplifying the inequality, we get:

n^2 - 60 < 0 n^2 < 60

 

Since n is an integer, the possible values for n are -7, -6, ..., 6, 7. However, we need to consider the endpoints as well. For n = -7 or n = 7, the quadratic becomes a perfect square, which means it has a double root, not no real solutions. Therefore, the only valid values of n are -6, -5, ..., 4, 5, which gives us a total of 12 possibilities.

 Dec 13, 2023
 #2
avatar+25 
0

great job! I was thinking move the -21 to the other side and say b^2-4ac<0

Iwillsueyou69420  Dec 15, 2023

2 Online Users

avatar