1. Discriminants 2

Show that mx² - (m + 1)x + 1 = 0 will have rational roots if m is an element of Q (I think Q means quotient?).

Could someone please show me how to do this?

2. Originally Posted by xwrathbringerx
Show that mx² - (m + 1)x + 1 = 0 will have rational roots if m is an element of Q (I think Q means quotient?).

Could someone please show me how to do this?
Have you calculated the discriminant? What do you notice? Therefore, what are the roots ....? By the way, Q is the symbol for rational numbers .....

3. Can the solution be like this:

For rational roots, discriminant = square no.

Discriminant = (- m - 1)^2 - 4(m)(1)

... (working out) ...

= (m -1)^2

m -1 = (/q + 3)^2
= (p+3q)^2 / q^2
= rational square no.

4. Originally Posted by xwrathbringerx
Can the solution be like this:

For rational roots, discriminant = square no.

Discriminant = (- m - 1)^2 - 4(m)(1)

... (working out) ...

= (m -1)^2

m -1 = (/q + 3)^2
= (p+3q)^2 / q^2
= rational square no.
$x = \frac{m+1 \pm |m - 1|}{2} \in Q$ if $m \in Q$.

5. Errr sori but what exactly do you mean? Which part of my solution am I meant to fix?

6. Originally Posted by xwrathbringerx
Discriminant = (- m - 1)^2 - 4(m)(1)

... (working out) ...

= (m -1)^2

m -1 = (/q + 3)^2
= (p+3q)^2 / q^2
= rational square no.
Originally Posted by xwrathbringerx
Errr sori but what exactly do you mean? Which part of my solution am I meant to fix?
What I mean is that I have no idea where the stuff in red has come from or what it means.

7. Oops I meant:

Assume that m is rational.
Therefore, m can be written in the form p/q where p and q are positive integers with no common factor.
Hence, (m-1)^2 = (p/q -1)^2
= (p-q)^2 / q^2
= rational square no.

8. Originally Posted by xwrathbringerx
Oops I meant:

Assume that m is rational.
Therefore, m can be written in the form p/q where p and q are positive integers with no common factor.
Hence, (m-1)^2 = (p/q -1)^2
= (p-q)^2 / q^2
= rational square no.
But how is that relevant to answering the question?

9. Don't I have to prove (m-1)^2 is rational?

The only way I could think of doing that was to prove m was rational.

10. Originally Posted by xwrathbringerx
Don't I have to prove (m-1)^2 is rational? Mr F says: Where does the question ask this?

The only way I could think of doing that was to prove m was rational.
Here is a reminder of the question and a reply I gave:

Originally Posted by xwrathbringerx
Show that mx² - (m + 1)x + 1 = 0 will have rational roots if m is an element of Q (I think Q means quotient?).

[snip]
Originally Posted by mr fantastic
$x = \frac{m+1 \pm |m - 1|}{2} \in Q$ if $m \in Q$.