Thread: Simultaneous Equation

Hiya,

I need to find the range of values of k which give distinct roots, given the following 2 equations:

kx + y = 3
x^2 + y^2 = 5

Thanks guys.

Kermola

The angles B and B' are 90°, but angle BAB' is not 90°. You can find , from where you can find the slope of AB' (recall that ).

Thanks for that.
Any way to solve it without trigonometry and without drawing out the graphs? i.e. solve it algebraically?
Thanks

You can express y from the first equation and substitute it into the second one. You'll get a quadratic equation in x with coefficients depending on k. Determine for which k the discriminant is positive.

Oh I've managed to figure it out!
Correct me if I'm wrong :

from (1) - y = 3 - kx
therefore x^2 + (3-kx)^2 = 5
x^2 + 9 - 6kx + 9 = 5
(k^2 + 1)x^2 - 6kx + 4 = 0
using b2 - 4ac to find real roots
(-6k)^2 - 16k^2 - 16 > 0
36k^2 - 16k^2 - 16 > 0
20k^2 > 16
k^2 > 4/5
-2/(5^(1/2)) < k > 2/(5^(1/2))


Hmm.. what program do people use to write proper it in proper mathematical form to post on these forums? That'd be helpful for the future.

Kermola

Ha-ha, I was early by 5 seconds!

Originally Posted by kermola

Oh I've managed to figure it out!
Correct me if I'm wrong :

from (1) - y = 3 - kx
therefore x^2 + (3-kx)^2 = 5
x^2 + 9 - 6kx + 9 = 5
(k^2 + 1)x^2 - 6kx + 4 = 0
using b2 - 4ac to find real roots
(-6k)^2 - 16k^2 - 16 > 0
36k^2 - 16k^2 - 16 > 0
20k^2 > 16
k^2 > 4/5
-2/(5^(1/2)) < k < 2/(5^(1/2))

The last line is incorrect.

Originally Posted by kermola

Hmm.. what program do people use to write proper it in proper mathematical form to post on these forums? That'd be helpful for the future.

See the LaTeX Help subforum. Wrap LaTeX code in [TEX]...[/TEX] tags.

I've since editted my post - we keep missing each other!

Thanks for the latex pointer - i'll check it out.
Is the editted post correct?
Thanks so much for help.

Kerm

Originally Posted by kermola

-2/(5^(1/2)) < k > 2/(5^(1/2))

It should be k > 2/(5^(1/2)) or k < -2/(5^(1/2)). When you want to abbreviate k > x and k < y, it's OK to write x < k < y. However, it is not customary to abbreviate k < x or k > y. If you write y < k < x, it would be confused with k < x and k > y.

I was just being a bit lazy when I edited the post. Thanks a lot for the help!

K