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Oct 4, 2005, 12:43:27 AM10/4/05

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

Is it possible to solve for L in this equation?

Thanks,

Aaron

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Oct 11, 2005, 2:26:16 AM10/11/05

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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

equation is quadratic only.

[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

Use the Quadratic Formula,

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] ----***

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?

Oct 11, 2005, 2:26:19 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?

Perhaps second post,sorry...had more to add.

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...

More about what you want/need would be helpful...

Oct 11, 2005, 2:26:22 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?

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

--

But I venture the challenging statement

that if American democracy ceases to

move forward as a living force, seeking

day and night by peaceful means to

better the lot of our citizens, then

Fascism and Communism, aided, unconsciously

perhaps, by old-line Tory Republicanism,

will grow in strength in our land.

--Franklin Delano Roosevelt

Oct 11, 2005, 2:26:24 AM10/11/05

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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

Professor of Biomolecular Engineering, University of California, Santa Cruz

Undergraduate and Graduate Director, Bioinformatics

(Senior member, IEEE) (Board of Directors, ISCB)

life member (LAB, Adventure Cycling, American Youth Hostels)

Effective Cycling Instructor #218-ck (lapsed)

Affiliations for identification only.

Oct 11, 2005, 2:26:26 AM10/11/05

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Yes.

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))

--

I don't know who you are Sir, or where you come from,

but you've done me a power of good.

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

Use the Quadratic Formula,

L = {-(-10) +,-sqrt[(-10)^2 -4(0.0005)(Q)]} / (2*0.0005)

L = {10 +,-sqrt[100 -0.002Q]} / 0.001

L = {10 +,-sqrt[100(1 -0.00002Q)]} / (1/1000)

L = {10 +,-10sqrt[1 -0.00002Q]} * (1000)

L = 10,000 [1 +,-sqrt(1 -0.00002Q)]

That is it.

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