How many different integers, n, are there such that the difference between
2 \sqrt{n} and 7 is less than 1?
A. 0
B. 2
C. 4
D. 6
E. 8
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We construct the inequality:
|2 \sqrt {n} -7| <1
And we know that if we apply the knowledge of
if |X|< D , then -D < X < D
We can reassume the inequality to become:
-1 < 2 \sqrt{n} - 7 < 1
And so we begin simplifying-
Let’s add 7 to each respective side, it becomes:
6 < 2 \sqrt {n} < 8
Next we can divide by 2:
3 < \sqrt{n} < 4
We can raise to the square of 2:
3^2 < n < 4^2 which we can quickly re-write as 9 < n < 16
With this simple inequality, we can list out our possible values, and count how many they are:
n = { 10, 11, 12, 13, 14, 15 } which make for 6 values
Our answer is, therefore, D.
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