How do you solve $j^ { 2} + 12j = 0$ ?

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Answer 1 · 2018-03-24T03:23:55

$j=0$ or $j=-12$

First, take $j$ common from LHS,
$rArr j(j+12)=0$
By zero product rule,
Either, $j=0$ Or, $(j+12)=0$
So, either $j=0$ or $j=-12$

That's it,
Hope this helps :)

How do you solve j^ { 2} + 12j = 0j^ { 2} + 12j = 0?