The inequality will be greater than zero
if D is less than zero.
D<0
4(4k−1)2−4.1.(15k2−2k−7)<0
Simplifying this we get
K2−6k+8<0
(k−4)(k−2)<0
∴k lies in (2,4)
In this interval, only one integer is present i.e 3
∴k=3
-Vinculum
Used as reference:
https://www.toppr.com/ask/en-us/question/the-integer-k-for-which-the-inequality-x2-24k-1x-15k2/
