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

Mar 9, 2016

There are three possible $m$ values:
$m \in \left\{- 4 \sqrt{2} , 4 \sqrt{2} , 9\right\}$

#### Explanation:

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

$\left({x}^{2} - m x + 8\right) \left(x - 1\right)$ 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:

$\left({x}^{2} - m x + 8\right) \left(x - 1\right) = {\left(x - a\right)}^{2} \left(x - 1\right)$
or
$\left({x}^{2} - m x + 8\right) \left(x - 1\right) = \left(x - a\right) {\left(x - 1\right)}^{2}$

If the first is true, then
${x}^{2} - m x + 8 = {\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$

Equating corresponding coefficients, we have
$8 = {a}^{2} \implies a = \pm \sqrt{8}$
$- m = - 2 a \implies m = \pm 2 \sqrt{8} = \pm 4 \sqrt{2}$

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

If the second equation is true, then
${x}^{2} - m x + 8 = \left(x - a\right) \left(x - 1\right) = {x}^{2} - \left(a + 1\right) x + a$

Again, we equate corresponding coefficients to obtain
$8 = a$
$- m = - \left(a + 1\right) = - 9 \implies m = 9$

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

So, putting them together, there are three possible $m$ values:
$m \in \left\{- 4 \sqrt{2} , 4 \sqrt{2} , 9\right\}$