+0

# Find the number of integers n that satisfy

+1
104
2
+69

Find the number of integers n that satisfy

$$7\sqrt{-n^2+22n-21}≤n+39$$

Apr 7, 2021
edited by imbadatmath  Apr 7, 2021

#1
+420
0

The expression $$-n^2+22n-21$$ will only be positive if $$1\leq x\leq21$$. (If it's negative then the left-hand side is undefined), which means that both sides of the expression must be positive, so we can square both sides without worrying about changing signs.

$$49(-n^2+22n-21)\leq n^2+78n+1521\\-49n^2+1078n-1029\leq n^2+78n+1521\\ 0\leq50n^2-1000n+2550\\0\leq n^2-20n+51\\0\leq (x+3)(x+17)$$

The zeroes of the quadratic of the left-hand side are -3 and -17, which means that the answer is $$\boxed{15}$$

Apr 7, 2021
#2
+1

$$-n^2 + 22n - 21 = -(n^2 - 22n + 21) = - (n-21)(n-1)$$

For the above to be non-negative, as stated in the previous solution:

$$1 \le n \le 21$$

Therefore we are starting with a possibility of 21 integers.

From the previous answer, the following was derived:

$$0 \le n^2 - 20n + 51$$

Factoring this yields:

$$0 \le (n-3)(n-17)$$

For this to be true:

$$3 \le n \le 17$$

The number of integers satisfying this is 15.

Apr 7, 2021
edited by Guest  Apr 7, 2021