Question #0e900

1 Answer
Jun 22, 2015

Answer:

In case A: Either #k=-4# with repeated root #x = 1# or #k=0# with roots #x = -1+-sqrt(2)#.

In case B: #k = -3/5# with roots #x = 1/3# and #x = -1#

In case C: #x = -k +-sqrt(2-3k)#

Explanation:

In case A, since the roots are reciprocals of one another, we must have:

#(2k+2) = +-(k-2)#

If #(2k+2) = (k-2)# then #k = -4# and #x = 1#

If #(2k+2) = -(k-2)# then #k = 0# and #x = -1+-sqrt(2)#

In case B, first divide through by #k# to get:

#x^2-(1+k)/kx +(3k+2)/k = 0#

The sum of the roots is #(1+k)/k#

The product of the roots is #(3k+2)/k#

So we are told:

#(1+k)/k = 2 xx (3k+2)/k#

Multiply through by #k# to get:

#1+k = 2(3k+2) = 6k+4#

So #5k = -3# and #k = -3/5#

So the quadratic is:

#-3/5x^2-2/5x+1/5 = 0#

Multiply by #-5# to get:

#3x^2+2x-1 = 0#

That is:

#(3x-1)(x+1) = 0#

So #x=1/3# or #x=-1#

In case C,

#x = -k +-sqrt(2-3k)#

We seem to be missing some additional constraint.