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Determine the non-permissible values of $5a^2+80a/(50ab^2$)

I know how to determine the non-permissible values of an equation, but this is the only question that I've encounter that contains two variables; both a and b.

Would I solve for a, and disregard b or solve for them both?

Any help is greatly appreciated.

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    $\begingroup$ You mean $5a^2 + 80a / (50 ab^2)$? (BTW this is a term, not an equation) The only thing you should avoid is division by zero. Can you see what $a$ and $b$ values would cause this? $\endgroup$ – mvw Apr 26 '17 at 9:29
  • $\begingroup$ For $ab \neq 0$, $5a^2 + 80a / (50 ab^2)=5a^2 + \dfrac8{5b^2} >0$ isn't it? $\endgroup$ – scarface Apr 26 '17 at 10:48
  • $\begingroup$ As others have suggested, you want to consider $50ab^2 = 0.$ As for what to do because this contains two variables, think about why $(x-1)(x+2) = 0$ leads to $x-1=0$ or $x+2=0$ and, and then ask yourself whether the principle used can also be used in your situation. $\endgroup$ – Dave L. Renfro Apr 26 '17 at 16:08
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The only thing that makes it non-permissible is having the denominator zero, so what values of $a,b$ make $50ab^2$ zero? The trap is to avoid canceling $a$ before answering the question. $0/0$ is still unacceptable.

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