Question

Prove that if x^2>k(x-2) for all real x, the 0 13 years ago

Answer 1

x^2>kx-2k
x^2-kx+2k>0
so the Discriminant of quadratic x^2-kx+2k must be less than or equal to zero...

Answer 2

because if x^2-kx+2k>0 then there is no roots for x^2-kx+2k=0
am i right?

Answer 3

i dunno

Answer 4

I didn't think to involve the discriminant, i was trying to solve it as a quadratic inequality.

Answer 5

ok

Answer 6

I would say, if
x^2>k(x-2)
then x^2-kx+2k>0
this is a parabola opening upward (cup shaped). So we want its minimum value to be bigger than 0. The min value occurs at the vertex of the parabola
the vertex is at x= -b/(2a). In this case, k/2
the value of the parabola at the vertex is
(k/2)^2-k(k/2)+2k>0
you can simplify this down to
8k-k^2>0
(8-k)k>0

this is true if both terms are positive or both terms are negative.
Let's do both negative:
(8-k)<0 and (at the same time) k<0 --> 8<k and k<0 (can not happen)

Both terms positive:
(8-k)>0 and k>0 --> 8>k and (at the same time) k>0 yes. we have
0<k<8

Answer 7

totally lost mayne

Answer 8

Step by step. The condition given
x^2>k(x-2)

is the same as
x^2-kx+2k>0

agree?

Answer 9

If you don't agree, we have a problem!

Answer 10

yes I whole heartily agree.

Answer 11

and you agree that is the equation of a cup shaped parabola with a minimum value