For how many different negative values of x is √x+2001 an integer?
A) 42
B) 43
C) 44
D) 45
A lot to list out, any way to find it faster??
clearly, -2001 <= x <= 0
Thus, 0 <= x+2001 <= 2001
So, how many numbers between 0 and 2001 are squares?
√2001 = 44.7
So, There are 45 numbers such that 0 <= x <= 44, meaning that 0 <= x^2 <= 2001
Yes, there is a faster way to solve this problem.
First, let's rewrite the expression as √x + 2001 = k, where k is an integer.
Simplifying the equation, we have √x = k - 2001.
To find the values of x, we need to find the values of k that make k - 2001 a perfect square.
Let's solve for the smallest negative value of x. Assume k - 2001 = -1, then k = 2000.
To find the largest negative value, assume k - 2001 = -45 (since option D) is the maximum). Solving for k, we have k = 1956.
So, the range of values for k is 2000 to 1956.
To find the number of negative values in this range, we subtract 1956 - 2000 and add 1 (to include both endpoints).
So, there are 45 different negative values of x.
Therefore, the answer is D) 45.
To find the number of different negative values of x for which √x + 2001 is an integer, we need to consider the range of values for x.
Since we are looking for negative values of x, we can rewrite the expression as √(-x) + 2001.
We know that the square root of a negative number is not a real number, so √(-x) is not an integer.
So, we solve the equation √(-x) + 2001 = 0 to find the value of x for which the expression becomes an integer.
√(-x) + 2001 = 0
√(-x) = -2001
Since we are working with negative values of x, we can rewrite -2001 as -2001^2 to eliminate the negative sign:
√(-x) = √(2001^2)
Taking the square root of both sides, we get:
(-x) = 2001
Multiply both sides by -1 to eliminate the negative sign:
x = -2001
It is given that √x + 2001 is an integer when x = -2001.
Therefore, there is only one negative value of x for which √x + 2001 is an integer.
Therefore, the answer is: A) 1