Question

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

Answer 1

You are on the right track.
the discriminant = 2²-4(k-1)[-(k-3)] greater than 0

It becomes 4 + 4(k-1)(k-3) .>= 0
4(k-1)(k-3) .>= -4
(k-1)(k-3) .>= -1
k²-4k+3 >= -1
k²-4k+ 4 >= 0
(k-2)² >= 0
So k can be any real # for (k-2)² >= 0

Answer 2

OK, will try and process it! Thx, Rohangrr. (:

Answer 3

u call mr ron rohan or eric

Answer 4

Δ = (1 - k)² -4(-k)(1)
= 1-2k +k² +4k
= 1+2k+k²
= (1+k)² ≥ 0 for any value of k
so the roots are real

Answer 5

Thank youuuu Hoblos (:

Answer 6

yw :)