Question

Solve the following equation:
$\dfrac{{{\text{2}}{{\text{x}}^3} - 3{{\text{x}}^2} + {\text{x}} + 1}}{{{\text{2}}{{\text{x}}^3} - 3{{\text{x}}^2}{\text{ - x - }}1}} = \dfrac{{{\text{3}}{{\text{x}}^3} - {{\text{x}}^2} + 5{\text{x - }}13}}{{{\text{3}}{{\text{x}}^3} - {{\text{x}}^2} - 5{\text{x + }}13}}$

Answer 1

Hint: In order to solve this type of equations we use the dividendo and componendo rule of fractions, which states that if the ratio of the numbers a to b is equal to c to d then the ratio of a + b to a – b is equal to the ratio of c + d to c – d. We apply this to the given equation respectively.

Complete step by step solution:
Given data,
 $\dfrac{{{\text{2}}{{\text{x}}^3} - 3{{\text{x}}^2} + {\text{x}} + 1}}{{{\text{2}}{{\text{x}}^3} - 3{{\text{x}}^2}{\text{ - x - }}1}} = \dfrac{{{\text{3}}{{\text{x}}^3} - {{\text{x}}^2} + 5{\text{x - }}13}}{{{\text{3}}{{\text{x}}^3} - {{\text{x}}^2} - 5{\text{x + }}13}}$

$\dfrac{{{\text{2}}{{\text{x}}^3} - 3{{\text{x}}^2} + {\text{x}} + 1}}{{{\text{2}}{{\text{x}}^3} - 3{{\text{x}}^2}{\text{ - x - }}1}} = \dfrac{{{\text{3}}{{\text{x}}^3} - {{\text{x}}^2} + 5{\text{x - }}13}}{{{\text{3}}{{\text{x}}^3} - {{\text{x}}^2} - 5{\text{x + }}13}}$
Using the dividendo and componendo rule we get,
 $
   \Rightarrow \dfrac{{{\text{2}}{{\text{x}}^3} - 3{{\text{x}}^2} + {\text{x}} + 1{\text{ + }}\left( {{\text{2}}{{\text{x}}^3} - 3{{\text{x}}^2}{\text{ - x - }}1} \right)}}{{{\text{2}}{{\text{x}}^3} - 3{{\text{x}}^2} + {\text{x}} + 1{\text{ - }}\left( {{\text{2}}{{\text{x}}^3} - 3{{\text{x}}^2}{\text{ - x - }}1} \right)}} = \dfrac{{{\text{3}}{{\text{x}}^3} - {{\text{x}}^2} + 5{\text{x - }}13{\text{ + }}\left( {{\text{3}}{{\text{x}}^3} - {{\text{x}}^2} - 5{\text{x + }}13} \right)}}{{{\text{3}}{{\text{x}}^3} - {{\text{x}}^2} - 5{\text{x + }}13{\text{ - }}\left( {{\text{3}}{{\text{x}}^3} - {{\text{x}}^2} - 5{\text{x + }}13} \right)}} \\
   \Rightarrow \dfrac{{{\text{4}}{{\text{x}}^3} - 6{{\text{x}}^2}}}{{{\text{2x + 2}}}} = \dfrac{{{\text{6}}{{\text{x}}^3} - 2{{\text{x}}^2}}}{{{\text{10x - 26}}}} \\
  {\text{Now we take }}{{\text{x}}^2}{\text{ common from the numerator and 2 from the denominator}} \\
   \Rightarrow \dfrac{{{\text{2x}} - 3}}{{{\text{x + 1}}}} = \dfrac{{{\text{3x}} - 1}}{{{\text{5x - 13}}}} \\
   \Rightarrow {\text{10}}{{\text{x}}^2} - 15{\text{x - 26x + 39 = 3}}{{\text{x}}^2} - {\text{x + 3x - 1}} \\
   \Rightarrow {\text{7}}{{\text{x}}^2} - 43{\text{x + 40 = 0}} \\
   \Rightarrow {\text{7}}{{\text{x}}^2} - 35{\text{x - 8x + 40 = 0}} \\
   \Rightarrow \left( {{\text{7x - 8}}} \right)\left( {{\text{x - 5}}} \right) = 0 \\
   \Rightarrow {\text{x = 5, }}\dfrac{8}{7},{\text{ 0 }}\left( {\because {\text{Since we eliminated }}{{\text{x}}^2},{{\text{x}}^2} = 0{\text{ is also a solution}}} \right) \\
$

Therefore the solutions of the given equation are${\text{x = 5, }}\dfrac{8}{7},{\text{ 0}}$.

Note: In order to solve this type of problems we need to have an insight on how to go about solving this type of equation by looking at it. We can achieve this by solving more equations of this type. Using the dividendo and componendo rule simplifies the equation to an extent. To further simplify the equation we took the term ${{\text{x}}^2}$common from both sides of the equation and eliminated it, which is why x = 0 became a solution. It is important to watch out for x = 0 also as a solution, we tend to forget this. We further solved the quadratic equation to get the answers.