Question

Using the properties of proportion, solve the following equation for x.

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

• A. 12
• B. 24
• C. 26
• D. 11

Answer 1

The correct option is D

11

Applying componendo and dividendo, we get;

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

$\Rightarrow \frac{(x+1{)}^3}{(x-1{)}^3}=\frac{432}{250}$ $\left[\right(a+b{)}^3=a^3+b^3+3ab(a+b)]$

$\Rightarrow \frac{(x+1{)}^3}{(x-1{)}^3}=\frac{216}{125}={\left(\frac{6}{5}\right)}^3$

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

Again, appplying componendo and dividendo we get:

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

$\Rightarrow \frac{2x}{2}=\frac{11}{1}\Rightarrow x=11$