Answered

Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

For how many integers $n$ does the inequality $x^2 + nx + 15 < 0$ have no real solutions $x$?.

Sagot :

facundo3141592

By making the determinant equal to or smaller than zero, we will see that there are 15 values of n such that the inequality has no real solution.

Solving the inequality:

Here we have the inequality:

x^2 + n*x + 15 < 0

We want to see for which values of n we don't have real solutions.

Now, we can write the quadratic equation:

y = x^2 + nx + 15.

If the determinant of this quadratic equation is negative or 0, then we have a single root or we don't have roots, and because the leading coefficient is positive, that would mean that the function never becomes negative. Then for these cases, the inequality has no real solutions.

Then we must have:

n^2 - 4*15*1  ≤ 0

Now we can solve this for n.

n^2 ≤ 60

Then if n^2  is smaller than 60, we don't have real solutions, and we know that:

√60 = 7.75

Then we have:

-7.75 ≤ n ≤ 7.75

But n can only be an integer, so we can write:

-7  ≤ n ≤ 7

From -7 to 7 there are 15 integers, then there are 15 values of n such that the inequality has no real solutions.

If you want to learn more about inequalities, you can read:

https://brainly.com/question/11234618

We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.