Author Topic: exp equation to share (Read 677 times) Tweet Share

evaever · « **on:** April 15, 2011, 08:34:27 pm »

Find the values of k such that the equation k e^(2x) -2e^x + k = 0 has two real solutions.

m@tty · « **Reply #1 on:** April 15, 2011, 09:08:38 pm »

You see this is a quadratic equation in e^x

Thus you can use the discriminant, and for two solutions solve such that it is positive.

So solve $(-2)^2 - 4k^2 >0$

$4k^2<4$

$k^2 < 1$

$|k| < 1$

luffy · « **Reply #2 on:** April 15, 2011, 09:09:14 pm »

let $z = e^x.$

$\therefore kz^2 - 2z + k = 0$

As e^x is one-to-one, two solutions will simply occur when the discriminant is greater than 0.

Note: Technically, as the question didn't say "distinct real solutions", discriminant could equal 0 and there will be two solutions.... They will just be the same xD

a = k
b = -2.
c = k.

$b^2 - 4ac > 0$

$4 - 4k^2 > 0$

$4k^2 < 4$

$k^2 < 1$

$-1 < k < 1$ .

However, reject k = 0 as this would not satisfy two real solutions.

$\therefore -1<k<0 \cup 0<k<1$

EDIT: Reject the negative k-values. Therefore, 0<k<1

evaever · « **Reply #3 on:** April 15, 2011, 09:46:07 pm »

anyone has a different opinion?

luken93 · « **Reply #4 on:** April 15, 2011, 09:54:51 pm »

Quote from: evaever on April 15, 2011, 09:46:07 pm

anyone has a different opinion?

I do, it's only from 0 < k < 1

I admit, I cheated after I found the same answer as you guys above, but I don't know why...

I thought the question was waaayyy too easy for evaever

m@tty · « **Reply #5 on:** April 15, 2011, 10:09:29 pm »

I can see why $k \ne 0$ and $k\not<0$ as the equation clearly fails because $e^x>0$

But how to exclude them I don't know, except of course just by testing after you have the result as above ( $|k|<1$ )..

luffy · « **Reply #6 on:** April 15, 2011, 10:37:07 pm »

Quote from: m@tty on April 15, 2011, 10:09:29 pm

I can see why $k \ne 0$ and $k\not<0$ as the equation clearly fails because $e^x>0$

But how to exclude them I don't know, except of course just by testing after you have the result as above ( $|k|<1$ )..

I can't believe I didn't type to reject k<0, despite me writing it on my piece of paper LOL

evaever · « **Reply #7 on:** April 16, 2011, 07:09:35 am »

Quote from: m@tty on April 15, 2011, 10:09:29 pm

I can see why $k \ne 0$ and $k\not<0$ as the equation clearly fails because $e^x>0$

But how to exclude them I don't know, except of course just by testing after you have the result as above ( $|k|<1$ )..

$e^{x}=\frac{1\pm \sqrt{1-k^{2}}}{k}$

To have two solutions:

$k>0$
and
$1-k^{2}>0$

$\therefore 0<k<1$

m@tty · « **Reply #8 on:** April 16, 2011, 09:34:17 am »

Quote from: evaever on April 16, 2011, 07:09:35 am

Quote from: m@tty on April 15, 2011, 10:09:29 pm
I can see why $k \ne 0$ and $k\not<0$ as the equation clearly fails because $e^x>0$

But how to exclude them I don't know, except of course just by testing after you have the result as above ( $|k|<1$ )..
$e^{x}=\frac{1\pm \sqrt{1-k^{2}}}{k}$

To have two solutions:

$k>0$
and
$1-k^{2}>0$

$\therefore 0<k<1$

Indeed, good work.

Yeah I thought of that after I posted.

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Author Topic: exp equation to share (Read 677 times) Tweet Share

evaever

exp equation to share

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evaever

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