#m# is a real number. #(x^2-mx+8) . (x-1) = 0# has two distinct roots. What are all #m# values?

1 Answer
Mar 9, 2016

Answer:

There are three possible #m# values:
#m in {-4sqrt(2), 4sqrt(2), 9}#

Explanation:

(This answer operates under the assumption that there are exactly two distinct roots, rather than at least two distinct roots)

#(x^2-mx+8)(x-1)# has two distinct roots. As #1# is one of those roots, then letting #a# be the other means that, by the fundamental theorem of algebra , one of the following is true:

#(x^2-mx+8)(x-1) = (x-a)^2(x-1)#
or
#(x^2-mx+8)(x-1) = (x-a)(x-1)^2#

If the first is true, then
#x^2-mx+8 = (x-a)^2 = x^2-2ax + a^2#

Equating corresponding coefficients, we have
#8 = a^2 => a = +-sqrt(8)#
#-m = -2a => m = +-2sqrt(8) = +-4sqrt(2)#

Thus, from the first equation, we have two possibilities for #m#.

If the second equation is true, then
#x^2-mx+8 = (x-a)(x-1) = x^2 - (a+1)x + a#

Again, we equate corresponding coefficients to obtain
#8 = a#
#-m = -(a+1) = -9 => m = 9#

Therefore, from the second equation, we have one possibility for #m#.

So, putting them together, there are three possible #m# values:
#m in {-4sqrt(2), 4sqrt(2), 9}#