medals and love :D solve for k pls
$$16\sqrt{k}-5k^{4}=-8723$$

Question

medals and love :D solve for k pls
$$16\sqrt{k}-5k^{4}=-8723$$

anonymous · Answer

what method?   i used wolfram, but no way i could do this by hand

anonymous · Answer

lol any, there's no way of doing it by hand? thought so :P

anonymous · Answer

don't think quadratics work here...

tkhunny · Answer

Try to relax.

There are 2 Real solutions.  Unfortunately, they are quite close together, so this may prove a little confusing.  A slight substitution give a better arrangement:

$M = \sqrt{k}$

This leads to $16M - 5M^{8} = -8723$

In this form, the two Real solutions for M are of opposite signs.  This is encouraging.

Now, we must decide how to proceed.  There is NO general formula for the solution of an 8th degree polynomial.  Unless we are REALLY LUCKY, we will need to resort to any of quite a few numerical methods.

You pick where to go next.

anonymous · Answer

sry but now im confused, u can't take M out, u can't divide,...hmmm

tkhunny · Answer

Who said anything about taking out or dividing.  I simple turned it into a polynomial (Whole Number exponents only), so that all applicable techniques could be applied and we could gain a little knowledge about it.  This also provided the unintended and fortunate result that the two Real solutions are not right next to each other.  This makes them more easily distinguished when we go hunting for them.

$5M^{8} - 16M - 8723 = 0$  Has EXACTLY one Real POSITIVE zero and EXACTLY one Real Negative zero.  That is wonderful information we did not have about k.  This tells us that there must be six (6) Complex zeros (in 3 conjugate pairs).

Now, how do we go about finding the two Real zeros?

anonymous · Answer

break it down into 3 sub-expression and set each equal to 0?

tkhunny · Answer

??  Please demonstrate.  I'm not sure what you are planning.

Personally, I would first want to know WHERE each zero resided.  It can't be very far from M = 0.  That 8 is huge!

M = 0 produces -8723 (Important that it is negative.)
M = 1 produces -8734
M = 2 produces -7475
M = 3 produces +2403 -- There we go.  There is one between M = 2 and M = 3

Since this thing is nearly symmetric outside $-2<M<2$, let's just assume the other Real zero is between M = -2 and M = -3.

Okay, now we know where to look.

tkhunny · Answer

Break it down to sub expressions?  Does that mean factor?  That doesn't seem likely to succeed.

anonymous · Answer

the same way u would do it for the function $$x ^{3}-3x ^{2}-16x-12$$ gives u (x+2)(x+1)(x-6) = 0 and u get x = -2, 6 and -1

anonymous · Answer

yes i know factoring wouldn't work...

tkhunny · Answer

That's factoring.  It cannot be done.  You'll need a better plan.

anonymous · Answer

how else would u get zeroes...

anonymous · Answer

and no pls dont start graphing this thing

anonymous · Answer

oh wtf i can just graph it...LOL

tkhunny · Answer

Bisection.
Whoops!  3 provides 24030, not 2403
We know it's between 2 and 3.  2.5 provides -1134
We know it's between 2.5 and 3.  2.75 provides +7587
We know it's between 2.5 and 2.75.  2.625 provides +2507
We know it's between 2.5 and 2.625.  2.5625 provides +532
Jut keep doing this until you're happy with it.

Graphing might be close enough.

There are many methods for this sort of thing.  Some are better than others.  Some don't work on some problems.

anonymous · Answer

that doesn't seem like an efficient solution...lol but it's cool