Answer
$3$ $or$ $-5$
Work Step by Step
$15{{x}^{-2}}-2{{x}^{-1}}-1=0$
It can be written as,
$\frac{15}{{{x}^{^{2}}}}-\frac{2}{x}-1=0$
Multiply with ${{x}^{2}}$ on both the sides,
$\begin{align}
& 15-2x-{{x}^{2}}=0 \\
& {{x}^{2}}+2x-15=0 \\
\end{align}$
Compare the above equation with $a{{x}^{2}}+bx+c=0$
$a=1$,$b=2$ and $c=-15$
Apply the quadratic formula, $x=\frac{-b\pm \sqrt{{{b}^{^{2}}}-4ac}}{2a}$
Substitute values of a, b and c in the above equation,
$\begin{align}
& x=\frac{-2\pm \sqrt{{{2}^{2}}-4\left( 1 \right)\left( -15 \right)}}{2\left( 1 \right)} \\
& =\frac{-2\pm \sqrt{4+60}}{2} \\
& =\frac{-2\pm \sqrt{64}}{2} \\
& =\frac{-2\pm 8}{2}
\end{align}$
Further,
$\begin{align}
& x=\frac{-2+8}{2} \\
& =\frac{6}{2} \\
& =3
\end{align}$
$\begin{align}
& x=\frac{-2-8}{2} \\
& =\frac{-10}{2} \\
& =-5
\end{align}$
Thus, the values of x for the equation $15{{x}^{-2}}-2{{x}^{-1}}-1=0$ are 3 or -5.
