If x= -5+(45+4k-k^2)^0.5

[AJWILL]

If x= -5+(45+4k-k^2)^0.5 where k is a positive integer. How many values of k exists if x has to be positive?

[A] 9

8

[C] 7

[D] 6

[E] 5

[eagleeye]

If x= -5+(45+4k-k^2)^0.5 where k is a positive integer. How many values of k exists if x has to be positive?

[A] 9

8

[C] 7

[D] 6

[E] 5

If x is positive:
-5+(45+4k-k^2)^0.5 >0
=> (45+4k-k^2)^0.5 > 5
=> 45+4k-k^2 > 25 (squaring both sides)
=> k^2-4k-20 <0
I can't think of any simple factors
I am going to solve the equation k^2-4k-20 = 0 using completing squares
(You can use the quadratic formula as well)

k^2-4k+4 = 24
(k-2)^2 = 4*6
=> k-2 = +/- 2*sqrt(6)
=> k = 2 +/- 2*sqrt(6)
k ~= 2 +/- 2*(1.4*1.7)
k ~= 2 +/- 2.8*1.7 = 2 +/- 4.76

Hence k lies between 2 - 4.76 and 2+4.76
=> k lies between -2.76 and 6.76. But we know that k is a positive integer.

Hence k can be any of 1,2,3, .... 6. Hence k has 6 possible values.

D is correct.

[truplayer256]

Note that in order for x to be positive (45 + 4k - k^2)^(0.5) > 5
=> 45 + 4k - k^2 > 25

=> k^2 - 4k - 45 < -25

=> k^2 - 4k - 20 < 0

=> (k - 2)^2 - 24 < 0

=> (k - 2)^2 < 24

=> (k - 2) > -sqrt(24) or k - 2 < sqrt(24)

k > -sqrt(24) + 2 => Approximates to about k > -4.9 + 2 => k > -2.9

k < 4.9 + 2 => k < 6.9

Only positive integers of k that exist between -2.9 < k < 6.9 are k = 1,2,3,4,5,6

Choose D.