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
[A] 9
8
[C] 7
[D] 6
[E] 5
AJWILL wrote: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.
