Show that the roots of the equation x^2 + (1-k)x = k are real for all values of k.?

Question

Show that the roots of the equation x^2 + (1-k)x = k are real for all values of k.

Please it would be nice if you explained the steps a bit. But its ok if you don't want to.

Answer 1

Can't show what just ain't so.

x² + (1-k)x = k
x² + (1-k)x - k = 0
(x + 1)(x – k) = 0

The roots are
x + 1 = 0 and x – k = 0
x = ‾1 and x = k

There are no constraints on k that require it to be real. For instance, if k = i = √‾1, the equation still holds.
(x + 1)(x – i) = 0

This gives a root x = i which is not real.

Answer 2

I will prove it for all real values of k.
Write the equation as
x² + (1-k)x - k = 0.
To show the roots are always real, the discriminant
of the equation must be nonnegative.
So we must show
(1-k)² +4k >= 0.
This is clear if k >= 0, so we may assume k = -m, m >0,
and we must show
(1+m)² -4m >= 0.
Write it out
m² + 2m + 1 - 4m >= 0
m² -2m+ 1 >= 0
(m-1)² >= 0.
Since this last inequality is always true, we are done!

Answer 3

x^2 + (1-k)x = k
=> x^2 +(1 - k)x - k = 0
For roots to be real, Δ ≥ 0
=> (1 - k)^2 + 4k ≥ 0
=> (1 + k)^2 ≥ 0
which is true for all values of k

Answer 4

Long time since I did these and not sure if I could now but I would assume non-zero included negative