Let me give you a clue about where you went wrong.
As long as the the minimum point of the equation is > 0, the
equation will be >0 for all values of x.
TS already knows that as long as the the minimum point of the
equation is > 0, the equation will be >0 for all values of
x as TS has said that for the quadratic function to be > 0,
the discriminant must be < 0.
Question
Find the range of values of k for
which x^2 − 2kx − k +6 >0 for all values of k.
Solution
x^2
– 2kx –k + 6 > 0
x^2
− 2kx + k^2 – k^2 – k + 6 >
0
(x –
k)^2 – k^2 – k + 6 > 0
Since (x – k)^2 is always positive
(or zero),
–
k^2 – k + 6 > 0
(− k
+ 2) (k + 3) > 0
Since the quadratic function in k
> 0and it has a maximum point ie
coefficient of k^2 is − 1 ,
so
− 3
< k < 2
PS :
I made a mistake in the earlier answer. The answer is now
corrected. Sorry for the
mistake.
