dhlee922 wrote:hi, i have a question regarding what is the correct way of doing this problem

given that 5 - (y^2)^(1/2) = -3, what is x?


the y term is actually squared and underneath a radical sign. i thought you could solve it as:
5 - y = -3
y = 8

i thought that the y^2^(1/2) would cancel each other out, but the book says that you should do it this way:

5+3 = (y^2)^(1/2)
64 = y^2
y = +8, -8

i actually also think there is a third way of doing it:

25 - y^2 = 9
y^2 = 16
y = +4, -4

but when plugging 4 back into the original, 4 doesnt work, but the math sure works when solving for 4, could someone elaborate?

thank you

Your third way is having a flaw.


you are squaring both side of the equation, so your equation should be like this:-


[5 -(y^2)^(1/2)]^2 = (-3)^2
25 + y^2 -2*5*(y^2)^(1/2) = 9.

imagine a = 5 ,b = (y^2)^(1/2) and c =-3

a-b = c
so when you are squaring both side it should be:-
(a-b)^2 = c^2 not a^2 - b^2 = c^2.

Also the book choice is more appropriate because when you are removing the square, you are eliminating one possible answer, which is not advisable.

Hope this helps.

Cheers,
Hemant

Great stuff, Hemant - and let me just add on this one...methods 1 and 3 are WRONG and if you were to see this on any above-average questions on the GMAT they'll almost certainly hinge on that.

The question appears here but not answer choices - my hunch is that this is either Data Sufficiency (in which case that first method says "sufficient", that x must equal 8 and that's wrong) or it's a Problem Solving question in which you need to determine something that must be true about x:

Given that 5 - (y^2)^(1/2) = -3, which of the following must be true:

I. x > 7
II. x < 9
III. x^2 = 64

(A) I and II only
(B) I and III only
(C) II and III only
(D) all of the above
(E) none of the above

(and in this case, since x could be -8, it's II and III)

The GMAT is a reasoning test and it loves to bait you with things like this - algebra where it's "convenient" but not correct to forget all about negative values, nonintegers, 0, etc. So the lesson on this one isn't just the method for doing the algebra - it's that this kind of thinking is crucial on this test, particularly when it's either Data Sufficiency or a Problem Solving question that asks about "must be" or "could be".

thanks hemant and brian. i do understand why 3 is wrong sort of.

if a-b=c

what's the algebraic rule that a-b has to be treated as 1 term. if i'm squaring everything, why does it matter?

also, if method 1 is wrong, how come the y squared raised to the 1/2, why doesnt the square cancel out the root?

thanks