Question

Solve the following equation :$\dfrac{{3{x^4} + {x^2} - 2x - 3}}{{3{x^4} - {x^2} + 2x + 3}} = \dfrac{{5{x^4} + 2{x^2} - 7x + 3}}{{5{x^4} - 2{x^2} + 7x - 3}}$.

Answer 1

Hint: Here, we go through by rearranging the equation so that we eliminate it easily and factorise the final equation to find the value of x.

Complete step-by-step answer:
We will rewrite question as
$\dfrac{{3{x^4} + 3{x^4} - 3{x^4} + {x^2} - 2x - 3}}{{3{x^4} - {x^2} + 2x + 3}} = \dfrac{{5{x^4} + 5{x^4} - 5{x^4} + 2{x^2} - 7x + 3}}{{5{x^4} - 2{x^2} + 7x - 3}}$
$ \Rightarrow \dfrac{{3{x^4} + 3{x^4}}}{{3{x^4} - {x^2} + 2x + 3}} - \dfrac{{3{x^4} - {x^2} + 2x + 3}}{{3{x^4} - {x^2} + 2x + 3}} = \dfrac{{5{x^4} + 5{x^4}}}{{5{x^4} - 2{x^2} + 7x - 3}} - \dfrac{{5{x^4} - 2{x^2} + 7x - 3}}{{5{x^4} - 2{x^2} + 7x - 3}}$
$ \Rightarrow \dfrac{{6{x^4}}}{{3{x^4} - {x^2} + 2x + 3}} - 1 = \dfrac{{10{x^4}}}{{5{x^4} - 2{x^2} + 7x - 3}} - 1$
$ \Rightarrow \dfrac{{6{x^4}}}{{3{x^4} - {x^2} + 2x + 3}} = \dfrac{{10{x^4}}}{{5{x^4} - 2{x^2} + 7x - 3}}$
And we apply cross multiplication,
$ \Rightarrow 3 \times \left( {5{x^4} - 2{x^2} + 7x - 3} \right) = 5 \times \left( {3{x^4} - {x^2} + 2x + 3} \right)$
$
   \Rightarrow {\text{15}}{{\text{x}}^4} - 6{x^2} + 21x - 9 = 15{x^4} - 5{x^2} + 10x + 15 \\
   \Rightarrow {x^2} - 11x + 24 = 0 \\
   \Rightarrow \left( {x - 8} \right)\left( {x - 3} \right) = 0 \\
 $
$\therefore x = 8,x = 3$

Note: Whenever we face such types of questions you have to simplify the equation as possible and rearrange equations to get the desired answer.