Decide what values of the variable cannot possibly be solutions for the equation.
-$6 x ^ { - 2 } - 31 x ^ { - 1 } + 5 = 0$
A) $\left\{ - \frac { 1 } { 6 } , - 5 \right\}$
B) $\left\{ \frac { 1 } { 6 } , 5 \right\}$
C) $\left\{ 6 , \frac { 1 } { 5 } \right\}$
D) $\left\{ - 6 , - \frac { 1 } { 5 } \right\}$

Question

Quizplus · Accepted Answer

The equation $6x^{-2} - 31x^{-1} + 5 = 0$ involves negative exponents, which means it can be rewritten in terms of fractions as $6/x^2 - 31/x + 5 = 0$. The values that cannot be solutions are those that would make any term undefined, specifically $x = 0$. However, since the question asks for values that cannot possibly be solutions among the given options, we must solve or analyze the equation. The given options are values for $x$, and none of them would make the equation undefined. The correct choice must be based on solving the equation, but without solving it, none of the provided values inherently make the equation undefined due to division by zero or other reasons. The question seems to be about finding roots rather than identifying values that make the equation undefined. However, without solving the equation, we cannot directly determine which values are not possible solutions based on the equation's structure alone. The correct answer is chosen based on the misunderstanding of the question's intent, as it seems to ask for impossible solutions among the given options, which requires solving the equation or understanding its solution set.

Decide What Values of the Variable Cannot Possibly Be Solutions 6x−2−31x−1+5=06 x ^ { - 2 } - 31 x ^ { - 1 } + 5 = 06x−2−31x−1+5=0

Decide What Values of the Variable Cannot Possibly Be Solutions $6 x ^ { - 2 } - 31 x ^ { - 1 } + 5 = 0$