For what value(s) of k is the inequation x^2 – kx + 2 >(or equal) 0 true for all values of x?

Think graphs. The graph of y=x^2-kx+2 is a U parabola and for y>0 for all x
the equation x^2-kx+2=0 must have no real solutions so b^2-4ac<0
giving k^2-8<0, k^2<8 giving -sqrt8<k<sqrt8

b) You can do this in a similar way to a)
x^2+2x-2n>0 for all x so 2^2+8n<=0 giving n <=-1/2

x^2 - kx + 2 ≥
=> (x - k/2)^2 + 2 - k^2/4 ≥
=> 2 - k^2/4 ≥ 0 ... [because (x - k/2)^2 is ≥ 0 for all the values of x
=> 2 ≥ k^2/4
=> k^2 ≤ 8
=> l k l ≤ 2√2
=> - 2√2 ≤ k ≤ 2√2
=> k ∈[- 2√2, 2√2].

i got -3 because when you put them into brackets 2 is the multiple of the numbers K is added together... the only numbers 2 can be is -1 and -2 which added together equals -3


n= 1/2 because as i said before about the multiple and the added numbers....in this case the added numbers is 2 ..which is 1 + 1 =2 and their multiple would be 1, so 1 (divided by) 2 is 1/2 or 0.5

I can think of two ways of approaching these problems.
* Complete the square on the left; there are tutorials online, I'm sure, on how to do this process.
* Solve the corresponding equality (so, x^2 - kx + 2 = 0 in the first case) by using the quadratic formula: the solution to your problem is reasonably clear depending on whether the equation has a solution (which, in turn, is dependent on whether the 'determinant' is positive).

I would consider the first solution the more elegant.