Solving Algebra Problem: a = SQRT(b/x)-cx

[beeresearch]

Hi Guys,

My algebra is a little rusty, and I have become stuck with this problem, would anyone mind showing me the right path..

The problem is..

a = SQRT(b/x)-cx

And I need to solve for x

Thanks in advance..

Steven

 

[Gear300]

Adding cx to both sides and squaring gives a^2 +2acx + c^2*x^2 = b/x. Multiplying by x and subtracting by b gives c^2*x^3 - 2acx^2 +a^2*x - b = 0. Now you just have to find a root for this third degree polynomial.

 

[beeresearch]

Now you just have to find a root for this third degree polynomial.

That's the hard part =(

Steven

 

[Gear300]

For a polynomial ax^3 + bx^2 + cx + d = 0, the roots may be given by these:
http://www.josechu.com/ecuaciones_polinomicas/cubica_solucion.htm
It might be simpler to use Newton's method to come up with an expression that is approximately close. Even still, there is most likely a simpler way...I just can't see it right off the back.

 

[Freddy_Turnip]

c^2*x^3 - 2acx^2 +a^2*x - b = 0

shouldn't that be:

+ 2acx^2

 

[srijithju]

For a polynomial ax^3 + bx^2 + cx + d = 0, the roots may be given by these:
http://www.josechu.com/ecuaciones_polinomicas/cubica_solucion.htm
It might be simpler to use Newton's method to come up with an expression that is approximately close. Even still, there is most likely a simpler way...I just can't see it right off the back.

Hey I never knew that there was an exact expression for the roots of a cubic polynomial . I infact never thought of a method apart from trying the factor theorem ( or using numerical technique) .

How was it show that this is the solution ( I mean apart from back substitution) ?
I thought it was in fact impossible to find the roots for a general polynomial equation of order > 2.

 

[Elucidus]

Solution formulae for polynomials of up to fourth order were known by mid-sixteenth century. I believe Gerolamo Cardano compiled them all in his Ars Magna of 1545. See

http://www-math.cudenver.edu/~wcherowi/courses/m4010/polynom2.pdf

for some background of how this all came about.

Solution formulae for higher order polynomials were sought thereafter, until things came to a head in the early 19th century. In particular Paulo Ruffini, Niels Abel, and Evariste Galois separatelly discovered (and proved to varying degrees, Galois being the deepest) that no general solution formulae exist for polynomials of order 5 or higher. This is not say one cannot solve any particular such polynomial (and many can be) but rather that there does not exist a formulae that works on all polynomials of a certain degree (> 4).

The theorem in particular that states the impossibility of finding such a formula is called the Abel-Ruffini Theorem. Ruffini's first attempt at the proof in 1799 was slightly flawed, but Abel fixed it not long after. His proof suffered from exploring unknown territory and was consequently clunky and arcane. Galois developed a rich and powerful body of theory to clearly attack this particular puzzle. Galois' work has lead to the development of the body of theory named after him (he died at the age of 20 in 1832 from a gunshot wound suffered in a pistol duel).

Wolfram's Mathworld Links

Cubic Formula

http://mathworld.wolfram.com/CubicFormula.html

Quartic Equation

http://mathworld.wolfram.com/QuarticEquation.html


--Elucidus

 

[mobilln2]

How about using a conjugate? I once saw a poster for the roots of a 5th (quintic) degree. That guy had a lot of extra time.

 

[srijithju]

Thank You very much Elucidus for the PDF link ... the steps involved seem very simple and clear (for the cubic equation ) ... but who would have thought to do those steps in that way ... not me anyway !