the answer given in gregmat is d for the first q and c for the second q which acc to me are wrong
You canât just say something is wrong and provide no justification lol. Depending on your justification, you could /could not be right.
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the first is right but in the sec q the ans could be + or -6
the squared root of 36 is +6, which is greater that -6. What do you mean it can be either + or -?
-6*-6=36 
I think you are overcomplicating yourself doing all of that. Additionally, there is no true square root of a negative number.
The \sqrt{.} by convention refers to the principal square root (only the positive root ). Your definition is also fine, but then youâre admitting to not treating square root as a function (but rather as a multi-valued function), which is clearly not what weâre going for here. Different definitions leads to different answers, but the definition here is clear and honestly you generally donât allow square root to be multivalued until you learn about order relations and stuff because itâs impractical.
Tldr; \sqrt{36} = 6 alone because sqrt() is a function and a function can have at most one output for each input. If not, then itâd be equivalent to you saying 6 = -6, which is obviously false.
okay so x^2=36 will give ±6 but sqrt of 36 gives 6 by convention
x^2 = 36 still doesnât contradict the fact that \sqrt{36} = 6.
x^2 = 36 \implies \sqrt{x^2} = |x| = \sqrt{36} = 6 and you know that |x| = 6 means x = \pm 6.
i think you are confused I had agreed with you
lol, iâm not confused. I was just adding onto that as an âappendixâ because the way you stated it made it seem like these two occurrences rely on different definitions, which it doesnât (in both cases sqrt(36) is indeed 6).
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