# Help solving equation

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### HoMoon115

Oct 4, 2005, 12:43:27 AM10/4/05
to

Q = 10L - 1/2000 (L)squared

Is it possible to solve for L in this equation?

Thanks,
Aaron

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### ticbol

Oct 11, 2005, 2:26:16 AM10/11/05
to

Sure it is possible, but L will be in terms of Q only.

Q = 10L - 1/2000 (L)squared

I assume the -1/2000 (L)squared is -(1/2000)(L^2) so that the given
[If it were -1/(2000 L^2), then the given equation would be cubic.]

Q = 10L -(1/2000)(L^2)
Q = 10L -0.0005(L^2)
Put them all to the lefthand side,
0.0005(L^2) -10L +Q = 0
L = {-(-10) +,-sqrt[(-10)^2 -4(0.0005)(Q)]} / (2* 0.0005)
L = {10 +,-sqrt[100 -(0.002)Q]} / 0.001
L = 10,000 +,-1000sqrt[100 -0.002Q]

That means,
L = 10,000 +1000sqrt[100 -0.002Q] ----***
or,
L = 10,000 -1000sqrt[100 -0.002Q] ----***

### Duane Bozarth

Oct 11, 2005, 2:26:18 AM10/11/05
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HoMoon115 wrote:
>
> Q = 10L - 1/2000 (L)squared
>
> Is it possible to solve for L in this equation?

Q = 10L - 0.0005L^2 ?

If that's what you mean, ever hear of the quadratic formula?

### Duane Bozarth

Oct 11, 2005, 2:26:19 AM10/11/05
to

HoMoon115 wrote:
>
> Q = 10L - 1/2000 (L)squared
>
> Is it possible to solve for L in this equation?

Ever hear of quadratic formula? (The old b and +/- sqrt(4ac) thingy...)

Yes in terms of Q but not numerically unless you select a value for Q.
Otherwise, you have only one equation w/ two unknowns...

And, of course, there are limits on what Q can be if you wish to
restrict the solution to the real line rather than the complex plane...

### Jeffrey Turner

Oct 11, 2005, 2:26:22 AM10/11/05
to

HoMoon115 wrote:

> Q = 10L - 1/2000 (L)squared
>
> Is it possible to solve for L in this equation?

Sure is. You now know how many Ls make a Q, how do you think
you'd find out how many Qs you'd need for an L?

--Jeff

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### Kevin Karplus

Oct 11, 2005, 2:26:24 AM10/11/05
to

On 2005-10-04, HoMoon115 <homo...@yahoo.com> wrote:
>
> Q = 10L - 1/2000 (L)squared
>
> Is it possible to solve for L in this equation?

Yes.

Assuming that the equation you meant to write was
Q = 10 L - (1/2000) L^2 ,
this is a standard quadratic formula and can be solved with any of the
techniques taught in Algebra 1 for solving quadratics.

------------------------------------------------------------
Kevin Karplus kar...@soe.ucsc.edu http://www.soe.ucsc.edu/~karplus
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### MA Math Teacher

Oct 11, 2005, 2:26:26 AM10/11/05
to

Yes.

### Jim Spriggs

Oct 11, 2005, 2:26:25 AM10/11/05
to

HoMoon115 wrote:
>
> Q = 10L - 1/2000 (L)squared
>
> Is it possible to solve for L in this equation?

L = 2000(5 +/- sqrt(25 + Q/2000))

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### ticbol

Oct 11, 2005, 2:26:35 AM10/11/05
to

Let me try answering this again. Looks like my answer 3 days ago would
not be not posted.anymore.

Q = 10L -(1/2000)L^2
Q = 10L -0.0005L^2
0.0005L^2 -10L +Q = 0