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Solving an Equation with Fractions

kuheli

New member
Oct 27, 2013
14
using properties of proportion solve

( x^3 + 3x )/341 = (3x^2 + 1)/91
 

chisigma

Well-known member
Feb 13, 2012
1,704
using properties of proportion solve

( x^3 + 3x )/341 = (3x^2 + 1)/91
If $\displaystyle f(x) = (1+x)^{3}$ then is...

$\displaystyle \text{Even} \{f(x)\} = \frac{f(x) + f(-x)}{2}$

$\displaystyle \text{Odd} \{f(x)\}= \frac{f(x)-f(-x)}{2}\ (1)$

... so that with little effort we obtain...


$\displaystyle 125\ f(x) = - 216\ f(-x) \implies 125\ (1+x)^{3}= - 216\ (1-x)^{3} \implies \frac{(1 + x)^{3}}{(1-x)^{3}} = - \frac{216}{125} \implies \frac{1+x}{1-x} = (- \frac {216}{125})^{\frac{1}{3}}\ = - \frac{6}{5}\ (2)$

Now all what we have to do is to solve a first order equation...


Kind regards


$\chi$ $\sigma$
 
Last edited:

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,709
Here is another method to approach the problem:

The existence of the terms $x^3+3x$ and $3x+1$ (as given in the problem) suggests that we might want to consider to relate the problem with $(x\pm1)^3$ and also, 341 and 91 be the sum or difference of cubes.

Observe that $341=216+125$ and $91=216-125$, hence we have

\(\displaystyle \frac{x^3+3x}{341}=\frac{3x^2+1}{91}\)

\(\displaystyle \frac{x^3+3x}{216+125}=\frac{3x^2+1}{216-125}\)

\(\displaystyle 216(x^3+3x)-125(x^3+3x)=216(3x^2+1)+125(3x^2+1)\)

\(\displaystyle 216(x^3+3x)-216(3x^2+1)=125(3x^2+1)+125(x^3+3x)\)

\(\displaystyle 216(x^3-3x^2+3x-1)=125(x^3+3x^2+3x+1)\)

\(\displaystyle 216(x-1)^3=125(x+1)^3\)

\(\displaystyle \frac{(x-1)^3}{(x+1)^3}=\frac{125}{216}=\frac{5^3}{6^3}\)

\(\displaystyle \frac{x-1}{x+1}=\frac{5}{6}\)

\(\displaystyle \therefore x=11\)
 
Last edited:

kaliprasad

Well-known member
Mar 31, 2013
1,309
I feel a little unnecessary in providing an answer when there are already 2 good answers but because my method is different I would provide it

we have

$\frac{x^3 + 3x}{341} = \frac{3x^2 + 1}{91}$

which is same as

$\frac{x^3 + 3x}{3x^2 + 1} = \frac{341}{91}$

using componendo dividendo we have

$\frac{x^3 + 3x+3x^2 + 1}{x^3 + 3x -3x^2 - 1} = \frac{341+91}{341- 91}$

rearranging the terms on LHS and simplifying RHS we get

$\frac{x^3 + 3x^2+3x + 1}{x^3 - 3x^2+ 3x - 1} = \frac{432}{250}$

or
$\frac{(x+1)^3}{(x-1)^3} = \frac{216}{125}$

or $(\frac{x+1}{x-1})^3 = (\frac{6}{5})^3$

or $\frac{x+1}{x-1} = \frac{6}{5}$

I apply componendo dividendo again

$\frac{x+1+x - 1}{(x+1)- (x-1)} = \frac{6+5 }{6- 5}$
or
$\frac{2x}{2} = \frac{11 }{1}$

or x = 11



for componendo dividendo method if some is unfamiliar

we have if

$\frac{a}{b} = \frac{c }{d}$

then

$\frac{a+b}{a-b} = \frac{c +d }{c- d}$