Algebra question

Does anyone see a way to do this problem algebraicly(sp)?


I can solve it via graphing...but not with algebra.


The area of a rectangle is 15960 and it has a diagonal equal to 193. Use a system of non-linear equations to find the dimensions.


eqn 1: L*W=15960
eqn 2: L^2 + W^2 = 193^2

Answer is 168x95.

 

Well, normally I'd just use a calculator, but my cousin's instructor wants the work shown. Aside from graphing it...I dont see how.

Elimination isn't possible, and substitution doesn't work.

 

The method you seek is based on what used to be called completing the square.

(L+W)^2 = L^2+2*L*W+W^2

Thus adding twice the first equation to the second yields

37249 + 15960 +15960 = 69169

take the square root of 69169 = 263

but this is = to L+W

So by substitution into equation 1

W(263-L)=15960
263*W-W^2=15960
W^2-263*W+15960=0
(W-168)(W-95)=0

W=168 or 95
L= 263 - 168 = 95
0r
L=263 -95=168

Hope this helps

 

Its actually still called that.

Nifty. I didnt even think of that. I am curious though...where are you deriving the (L+W)^2 equation?

 

Actually, I am lost as to the reasoning behind all of this:

None of my equations are quadratic, which is why I never considered completing the square.

Im sure its something simple and hard to convey across a forum.

 

There are Completing the Square Calculators like found here:-

http://www.algebrahelp.com/calculators/equation/completingthesquare/

Now, the neat thing about this, is that it doesn't merely solve the equation, but gives the full working. May be of some use in understanding the formulae derivations.

 

It doesn't help, as my equations deal with two variables, not one?

 

here is a far better solution;

your 2 equations

L^2 + W^2 = 193^2
LW = 15960

from the second equation you get W = 15960/L

substitute back into the first equation (and I am going to call 15960 = c cause I am lazy)

L^2 + (c/L) = 193^2
L^2 + (c^2)/L^2 = 193^2 (multiply through by L^2 and call 193^2 b)
L^4 + c^2 = (b^2)(L^2) (move everything to one side)
L^4 - (b^2)(L^2) + c^2 = 0

solve for L there will be 4 solutions of which only 2 will make sense and only one will actually give a proper value for W

 

Two equations, two unknowns you youngsters have life so easy.

......................

I notice BM didn't actually work through his solution which ended in a quartic equation.
Is that because there are no formulaeic solutions to equations higher than quadratic?
They can sometimes be solved by reduction or factorisation, if as in this case, the arithmetic works out neatly.

........................

I'm sorry the limits of the typography, which are more beyond me than the maths, failed to show that I started with a standard result, as taught to 12 year old schoolchildren, in my (Queen Victoria's) day.

(a+b)^2 is a perfect square which can be expanded to be

(a+b)^2 = (a+b)*(a+b) by definition

thus (a+b)^2 = (a*a)+(a*b)+(a*b)+(b*b)

or (a^2)+(2*a*b)+(b^2)

or {a^2+b^2} + {2ab}

or {2ab} + {a^2+b^2}

Notice that the first quantity in braces is the twice the first of Addys equations and the second quantity is the second.

This is why I added twice the first equation to the second.
After a little rearranging that leads to a quadratic which can be solved by formulae or by factorisation.

...........

I don't know if this was junior's homework or someone looking for an algorithm to program?

 

This picture might help.

 

This is for my cousin in Her College Algebra class. I am actually in Calculus II, but a problem like this, we just simple graphed for the answer, which is considerably quicker.

 

Brilliant!

This is why I suck at math, I can never see how to manipulate things.

I bow to you, sir.

 

I'll leave tidying up the presentation as an exercise for the student then.
It looks so much better, written out longhand.

 

Already taken care of. Thanks

Once I saw you change it to 2ab +a^2+b^2, it all clicked.

 

To solve that particular "quartic", you note that it is actually a quadratic in L^2, therefore you use the standard solutions to a quadratic equation and take the square root of the result.

If px^2 + qx + r = 0 then x = ( -q +- sqrt ( q^2 - 4pr ) ) / 2p

Thus with p = 1, q = -(b^2) and r = c^2:

L^2 = ( b^2 +- sqrt ( b^4 - 4c^2 ) ) / 2

Now c = 15960 and b = 193^2 = 37249 hence

L^2 = ( 37249 + sqrt ( 37249 * 37249 - 4 * 15960 * 15960 ) ) / 2 = +-168

and

L^2 = ( 37249 - sqrt ( 37249 * 37249 - 4 * 15960 * 15960 ) ) / 2 = +-95

Remember the negative roots are also a valid answer here.

PS my text book gives formula solutions for cubics and quartics neither of which are needed here, see also http://en.wikipedia.org/wiki/Quartic_equation

 

Soryy I meant you can't automatically guarantee to obtain solutions to higher equations without reverting to complex number arithmetic. Using these equations of all degree are soluble.

But thanks for working out the 'reduction of degree' solution, although I have never come across rectangles with negative sides before, perhaps you could post a picture alongside mine?

 

OMG!! I think my brain is leaking out my ears now.... I thought this would be rather straightforward..... boy was I wrong.. too many years since hardcore math.

~C

 

This isn't hardcore, not really. See, I have an issue with manipulating equations to my advantage, hence I didn't see it till studiot pointed it out. This is pretty basic algebra, just requires a little unconventional thinking to solve.

Graphing it is an easier approach, to be honest.

Thought you guys might find it funny that I showed my cousin how to solve it, she took it to class, and nobody has it solved yet. Further more, the instructor hasn't either. Its suspected he can't.

 

Attached is a picture of a rectangle with negative sides . The original rectangle is shown solid (with positive W and L) and a rotated or translated rectangle is shown (with negative W and L).

Sorry, I didn't read the original read post stating the the equations were derived from a rectangle and was focussing on the solution to the quartic and so wasn't concerned on the interpretation of the solutions, which is "an exercise left to the reader" as they say in textbooks . The two equations are simply a hyperbola and a circle to which the solutions are the intersections of these curves.

 

Math like a lot of things has multiple solutions. My way probably would have given the correct answer (I hope, I didn't actually finish to the end answers ). this is not to say that studiot's answer was wrong or right, just different.

and an equation in the form that I had it was a square of a square. you just complete the square twice or run it through a quadratic twice and you are done.