amanvermagmat wrote:

Could you please explain me how should I deal with the absolute value "b+2" after squaring both sides of the equation of option 2? I was confused by this part...

Really appreciate

Hi

Your question is not very clear. If you are talking about statement 1, there the right hand side is \sqrt{b + 2}[/m]

When squared, it will become (b + 2). Where does 'absolute value' come from?

If you are talking about statement 2, then it just tells us that x > 0. So [m]9^x + 9^{-x} can also be written as:

9^x + 1/9^x

But since we are not told anything about 'b' in statement 2, there is no way this statement is sufficient to answer this question.

Can you please explain what you meant by absolute value in this case?[/quote]

Hi there, thanks for coming back.

Sorry that it should be statement 1 instead of 2.

I think more or less got myself the answer.

My question is previously (1) 3x+3−x=b+2‾‾‾‾‾√3x+3−x=b+2 --> square both sides --> 9x+2∗3x∗13x+9−x=b+29x+2∗3x∗13x+9−x=

|b+2|, b+2 must be absolute |b+2|, which means the right side of the equition could be

b+2 or

-b-2....I suppose.

But since the first part " 9x+2∗3x∗13x+9−x" are all positive numbers, there is no way that |b+2| will be negative. I hope my logic here makes sense to you...

Anyway, thank you for ur answer

[/quote]

Hi

In statement 1, when you square the RHS (right hand side), then the square of √(b+2) will be just (b+2) and NOT |b+2|.

Whenever we square √x, we get 'x', not |x|. Well its a different thing though that if we are taking √x (square root of x), then x must be non-negative, because square root of a negative quantity is not defined.

So in this question, statement 1, RHS is √(b+2) and its understood that (b+2) cannot be negative here, so when we take square of √(b+2) we will just write b+2.[/quote]

Ah! I see.... Thank you very much for clarifying! I can finally realise what has been going wrong in my logic...