How does this not equal?

Question

i got 1 and 6. but mgmat is saying 1 not valid. how? i plugged in 1 and still saw it was valid. it's an equation. both sides are equal. x = y. what am i missing?

mvictor · Answer

mate, take a close look what the question is asking:

suppose you have x=1.

1+3 is 4, and 4 under radical is 2.

now look at the second expression: x-3. if x=1, then the result is -2. as we know, something under radical cannot be negative. 
since we get 2=-2, this is an extraneous root, aka root that does not yield a valid result when testing.

you have to be very careful when answering questions with roots, since you have to check the solutions as well for extraneous roots.

Ekland · Answer

Thanks Mvictor. I get that. But i don't want to pretend that your explanation has cleared it for me. but it's not just 4 under any radical but specifically under the square root radical. square root of 4 is +-(plus or minus) 2. Am i wrong with the result of square root of 4? if i'm right then that means the other side of the equation is - or + whatever. then why did 6 fit in but 1 didnt? 6 yielded a positive result on the right hand side. 1 yielded a negative result. So why was 1 taken out as wrong, since the radical should result in negative or positive?
Is my fundamentals wrong?
or are there other rules I'm missing. It didnt even say that x signifies "number of kids buying icecream in a shop". Why the "radical" expulsion of negative?

mvictor · Answer

In GMAT world, everything that is under square root will result in a positive number. 
yes it is true that -2 squared is 4, and thus it should work, but only when the condition is x^2 = 4. but in this case, you have sqrt(4), which must always be positive.

for just in case, try with your calculator if such things are possible:
1. square root of a negative number;
2. result of a square root of any positive number.

ENGRTOMBA2018 · Answer

I think  mvictor  has very ably answered the question but let me chime in.

When you have  \sqrt{x} = 1 , the only possible value is x=1 and NOT x= \pm  1

But, when you have  x^2 =1  ---> you get x =  \pm  1. This is because,  x^2=1  is a quadratic equation and by the very definition of 'roots of an equation', the number of roots of an equation = power of the polynomial. In this case the power of the polynomial = 2 (for a quadratic) while  \sqrt{x} = abc  is linear (power =1) in x.

Coming back to your question, when you substitute x=1 , you get a negative number (1-3=-2) on the RHS and as for GMAT math,  \sqrt{x}  can only be defined for x  \geq  0

Thus x=1 is not a valid option.

If you are looking the question as  \sqrt{x+3} = x-3  and then you end up squaring both sides to get  x+3=x^2+9-6x  ---> you get 2 solutions x=1 or 6. But as you squared the ORIGINAL equation and thus doubled the number of TRUE Solutions, you must go back and figure out a way to eliminate 1 of the 2 solution to arrive at the only correct solution, which in this case is x=6.

Hope this helps.

EMPOWERgmatRichC · Answer

Hi Nez,

Both mvictor and Engr2012 have properly explained the concepts involved, so I won't rehash any of that here. You WILL end up being tested on these concepts on the GMAT (although only a couple of times at most), so it's important that you memorize them and have them 'at the ready' when needed. At higher scoring levels, the GMAT will test you on concepts that you know, but in ways that you're not necessarily used to thinking about (eg. in this prompt, we had a variable under the radical; that's a bit more difficult than just having a number under the radical, even though the rules are still the same).

GMAT assassins aren't born, they're made,
Rich

Ekland · Answer

something just got into place as I read your replies. You guys are awesome. I was missing a tiny crucial fundamental. I get it now. @Engr2012's added value to @mvictor's.  EMPOWERgmatRichC  dusted it up clean.
Moreover, i just read the posting rules Engr2012 linked. I didn't quite follow procedures in posting. I'll make it up to gmatclub.

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