How many different integers, n, are there such that the difference between 2 x n squared and 7 is less than 1?
The supposed answer is 6 but i have NO idea how that's possible??
The question you were looking for, well, nine years ago...
It is n squared root not n squared for which the answer is 6
You copied the question wrong.
Well, it seems like quite a puzzling question you've got there! Let's unravel it together.
To start, the difference between 2 times n squared and 7 can be expressed as:
2n² - 7
Now, according to the question, this difference needs to be less than 1. So we have the inequality:
2n² - 7 < 1
To solve this inequality, we can add 7 to both sides:
2n² < 8
Then, divide both sides by 2:
n² < 4
Now, taking the square root of both sides (while considering both the positive and negative square root), we get:
n < 2 or n > -2
So, it seems that any integer value of n that is less than 2 or greater than -2 would satisfy the given condition.
Therefore, there are infinitely many integers satisfying the inequality, not just 6. That might explain the confusion! So go ahead and pick any integer value smaller than 2 or larger than -2, and you've got yourself a valid solution.
Remember, numbers can be quite funny sometimes, so don't let them confuse you too much!
To find the number of different integers, n, for which the difference between 2n² and 7 is less than 1, we can use the following steps:
Step 1: Write the inequality based on the given information.
The difference between 2n² and 7 is less than 1 can be written as:
|2n² - 7| < 1
Step 2: Solve the inequality.
We can split the inequality into two separate inequalities:
2n² - 7 < 1 and -(2n² - 7) < 1
Let's solve the first inequality:
2n² - 7 < 1
Adding 7 to both sides:
2n² < 8
Dividing both sides by 2:
n² < 4
Taking the square root of both sides:
-2 < n < 2
Now, let's solve the second inequality:
-(2n² - 7) < 1
Expanding the parentheses and simplifying:
-2n² + 7 < 1
Subtracting 7 from both sides:
-2n² < -6
Dividing both sides by -2 and reversing the inequality:
n² > 3
Taking the square root of both sides:
n > √3 or n < -√3
Step 3: Determine the integers that satisfy the inequalities.
To find the integers that satisfy the inequalities, we need to look for whole number values of n between -2 and 2 (inclusive) and beyond √3 and -√3.
The integers within the range of -2 to 2 are -2, -1, 0, 1, 2 (total 5 integers).
The integers beyond √3 and -√3 are √3 and -√3 (total 2 integers).
So, the total number of different integers, n, is 5 + 2 = 7.
Therefore, the supposed answer of 6 is incorrect. The correct answer is 7.
To find the number of different integers, n, such that the difference between 2n^2 and 7 is less than 1, we can set up an inequality and solve it.
The first step is to write the inequality based on the given conditions:
|2n^2 - 7| < 1
To remove the absolute value, we split the inequality into two separate inequalities:
2n^2 - 7 < 1 and -2n^2 + 7 < 1
Simplifying these inequalities, we get:
2n^2 < 8 and -2n^2 < -6
Dividing both sides of the inequalities by 2 and -2 respectively, we have:
n^2 < 4 and n^2 > 3
Taking the square root of both sides in the second inequality, we get:
n > √3 or n < -√3
Now, since we are looking for integer solutions, we need to find the integers that satisfy these conditions.
For n > √3, we can consider the values n = 2 and n = 3, as their squares are 4 and 9 respectively, satisfying n^2 < 4 condition.
For n < -√3, we can consider the values n = -2 and n = -3, as their squares are also 4 and 9 respectively, satisfying n^2 < 4 condition.
Therefore, there are a total of 4 different integers that satisfy the given inequality: n = -3, -2, 2, and 3.
It seems there has been an error in the answer you were provided with. The correct number of different integers, n, that satisfy the given inequality is 4, not 6.
|2n^2 - 7| < 1
We could go into a formal solution with all the AND and OR operators, but clearly n must be a small integer, so lets just try some values of n
n = 0 ---> |-7| < 1 , no
n = ± 1 --> |2-7| < 1 , no
n = ± 2 ---> |8-7| < 1 , no
n = ± 3 ---> | 18-7| < 1 no
for larger ± n's the situation is getting even worse
I don't see a single integer that would work
by definition |anything| ≥ 0
so
0 < 2n^2 - 7 < 1
7 < 2n^2 < 8
no integer value of n fits that statement.