Question

Solve the equations:
$\frac{3x^4+x^2-2x-3}{3x^4-x^2+2x+3}=\frac{5x^4+2x^2-7x+3}{5x^4-2x^2+7x-3}$.

Answer 1

Using $\frac{a}{b}=\frac{a+b}{a-b}$

Apply this property both sides, we get

$\frac{3x^4+x^2-2x-3}{3x^4-x^2+2x+3}=\frac{5x^4+2x^2-7x+3}{5x^4-2x^2+7x-3}$

$\Rightarrow \frac{6x^4}{2x^2-4x-6}=\frac{10x^4}{4x^2-14x+6}$

When $x=0$, the equation is satisfied. So, one solution is zero.

Now, when $x$ is not equal to zero, cancel $x^4$ on both sides and cancel factor of $2$ in the denominator and numerator

$\Rightarrow \frac{3}{x^2-2x-3}=\frac{5}{2x^2-7x+3}$

Now after cross multiplying and simplifying the equation, we get

$\Rightarrow x^2-11x+24=0$

$\Rightarrow (x-8)(x-3)=0$

So, $3$ and $8$ are also the solution of equation.

So, $x=0,3,8$ are the solutions.