2x(square) - (k-2)x + 1 where there exist equal roots?

Question

2x(square) - (k-2)x + 1 where there exist equal roots

Answer 1

(k - 2) ² = (4)(2)(1) for equal roots.
(k - 2) = ± √8
(k - 2) = ± 2√2
k = 2 ± 2√2
k = 2 ( 1 ± √2 )

Answer 2

to get equal roots, the discriminant must be zero!

write
2x^2 - (k-2)x + 1, then a = 2, b = -(k-2), c = 1,

and the roots are x = (-b +- sqrt(b^2 - 4ac))/2a

note the roots are equal when the expression in the sqrt sign, b^2 - 4ac = 0, or
(-(k-2))^2 - 4*2*1 = 0, or

(2-k)^2 - 8 = 0 or
4 - 4k + k^2 - 8 = 0 or
k^2 - 4k - 4 = 0, solve by quad formula (above)

k = (4 +- sqrt(16 - 4*1*(-4)))/2*1
k = (4 +- sqrt(32))/2
k = (4 +- 4sqrt(2))/2
k = (4 + 4sqrt(2))/2 | k = (4 - 4sqrt(2))/2
k = 2 + 2sqrt(2) | k = 2 - 2sqrt(2)

There are the values of k which gives equal roots to original equation.

Answer 3

x1 = (k-2)/4 + root{ (k-2)^-4 }/8
x2 = (k-2)/4 - root{ (k-2)^-4 }/8