Solve Rational equations (Quadratic)

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Note that $x \neq-1$ and $x \neq-3$

**Answer**

$\frac{5}{x+1}-\frac{57}{x+3}-\frac{4}{\left(x+1\right)\left(x+3\right)}=-16$

Or, $5\left(x+3\right)-57\left(x+1\right)-4=-16\left(x+1\right)\left(x+3\right)$

Or, $5x+15-57x-57-4=-16\left(x^{2}+4x+3\right)$

Or, $-52x-46=-16x^{2}-64x-48$

Or, $-52x-46+16x^{2}+64x+48=0$Or, $16x^{2}+12x+2=0$The solutions of a *quadratic equation* $ax^2+bx + c = 0$ are given by$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

In this *quadratic equation*, $a=16$, $b=12$, and $c=2$

Substituting these values of $a$, $b$ and $c$ in the above formula, we get

$x=\frac{-12\pm \sqrt{144-128}}{32}$

or, $x = \frac{-12\pm \sqrt{16}}{32}$

or, $x = \frac{-12\pm 4}{32}$

or, $x = \frac{-8}{32}$ and $x = \frac{-16}{32}$

or, $x=-\frac{1}{4}$ and $x =-\frac{1}{2}$

There are two distinct solutions given by $x = -\frac{1}{4}$ and $x = -\frac{1}{2}$