No not at all, I just wanted to point out that using the calculus may not be much help to a student posting in the Algebra forum. But there's no need to remove your post on my account.

I worked it out to be c = -13.

On to part 2!

Would I work it out the same as part 1? (i.e. solve for m when the discriminant = 0)

So far I've done:

mx + 6 = -2x2 - 6x +2

-2x2 - 6x - 4 - mx = 0

(I have a feeling this isn't correct.)

Yes, you want to equate the discriminant to zero, but I would write the quadratic as:

$\displaystyle 2x^2+(m+6)x+4=0$

Now, what is the discriminant here?

The discriminant there would be (m + 6)2 - 32 which, solved for x, gives us -6 ±4√2 -- the correct answer.

Thanks for the help, Mark.

One final question: How do you properly type equations? I tried pasting them from Microsoft Equation 3.0 (via Word) but to no avail.

Originally Posted by Fratricide
The discriminant there would be (m + 6)2 - 32 which, solved for x, gives us -6 ±4√2 -- the correct answer.

Thanks for the help, Mark.

One final question: How do you properly type equations? I tried pasting them from Microsoft Equation 3.0 (via Word) but to no avail.
Good work!

To express mathematical expression, many forums, including MHF, use $\displaystyle \LaTeX$. A search here and online will yield information and tutorials on its usage.

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