Question

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

Answer 1

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}`