Solving an Equation with Fractions

[kuheli]

using properties of proportion solve

( x^3 + 3x )/341 = (3x^2 + 1)/91

 

[chisigma]

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 $\chi$ $\sigma$

 

[anemone]

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

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

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

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

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

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

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

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

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

\(\displaystyle \therefore x=11\)

 

[kaliprasad]

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

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

 

[CellCoree]

I would approach this problem by first understanding the properties of proportion. Proportion is a relationship between two ratios or fractions that are equal, and it can be represented as a/b = c/d. In this case, we have two fractions on either side of the equation, and we need to use properties of proportion to solve for the variable x.

The first step would be to cross-multiply the fractions, which means multiplying the numerator of one fraction by the denominator of the other fraction, and vice versa. This would give us the equation (x^3 + 3x) * 91 = (3x^2 + 1) * 341.

Next, we would distribute the numbers outside the parentheses to simplify the equation. This would give us 91x^3 + 273x = 1023x^2 + 341.

We can now move all the terms to one side of the equation and set it equal to zero by subtracting 1023x^2 and 341 from both sides. This would give us 91x^3 - 1023x^2 + 273x - 341 = 0.

Using the principles of algebra, we can factor this equation to find the values of x. After factoring, we would get (7x - 19)(13x^2 - 19x + 19) = 0.

This gives us two possible solutions for x: x = 19/7 or x = (19 ± √(-691))/26. However, since we cannot have a negative value under the square root, the only valid solution is x = 19/7.

In conclusion, by using the properties of proportion and algebraic principles, we have solved the given equation with fractions and found the value of x to be 19/7. This solution can be verified by substituting x = 19/7 back into the original equation.