Math Help..

[xenon200]

Does anyone know how to even BEGIN on this?

1) A quadratic equation is given by kx² - (k-6)x + 2 = 0. For each relationship between m and n given, find the corresponding value of k.

a) m + n = 6
b) m²n² = 10

(I don't need both of these solved, just some insight on the procedure to be followed..)

It looked similiar to some discriminant problems we had done before, so I worked this out - but don't see how it helps, if at all;

0 = b² - 4ac
0 = (k-6)² - (4)(k)(2)
0 = k² - 12k + 36 - 8k
0 = k² - 20k + 36
0 = (k - 18) (k - 2)

k = 18, k = 2

This is supposed to tie in directly with the unit we're working now (Sums and Products of Quadratic Roots), but I really don't understand what is supposed to be done here.

~x2

[RADAR1797]

Dang, this one is tough.

I used the second equation to solve for the first and got:

M^2 -6m+10^(1/2) = 0

I used the quadratic equation a determined that

M= 5.416 and n = 0.584

I do not know how this ties into k though.

-RADAR

[RADAR1797]

Xenon200,

I think you miss used the quadratic equation on the primary problem. The correct formula is:

Solving for ax^2 + bx + c = 0

x = (-b +or- (b^2 - 4ac)^(0.5))/2a

-RADAR

[xenon200]

:S

What about this?

Since m + n = -b / a, wouldn't this be true?

(k-6) / k = 6

At which point you would solve for k;

(k-6) / k = 6
k - 6 = 6k
-6 = 5k
-6/5 = k

Whaddya think?

[RADAR1797]

I don't understand how you arrived at m + n = -b/a

-RADAR

[RADAR1797]

Oh I think I see how you arrived at it, but it coincidence. Don't use constants like you do variables.

-RADAR

[xenon200]

m + n is the only thing we've been "taught" so far.

~x2

[xenon200]

Newest dilemma - I don't know how to use;

m + n = -b / a

and

mn = c / a

and apply to solve the second question...

m²n² = 10



~x2