Hello ,

In this question why can we never have y as (-)ve value? (As explained in the video solution)

If we write the expression as y = [(-2)^2 - x^2]^1/2

In this case , if we substitute x = 0 , then we can get

“y = -2”

So we can consider the full circle area right?

Can anyone please clarify this doubt?

Thank you

Suggestion: please take a screenshot or link to the problem as this is very hard to read. But no, it’s not a full circle.

Hello , I have uploaded a clearer screenshot .

Also , can you please explain why y = (-2) is not possible ?

Classic GRE rule: the root (on the RHS) is always positive. Also use Latex wherever needed.

But , a root of 4 can be either +2 or -2 no?

The value of y in y = \sqrt{x} is always nonnegative.

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So , in a typical GRE question , I can never consider the case 2 ?

Can you please confirm ( Sorry for repeated questions)

Thanks!

Not in general. The problem is that \sqrt{x} must be nonnegative, which means that -a will fail.

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Ok thank you for the help