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Clarification for Square Roots

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by Ryandmitri » Sun Apr 24, 2011 11:43 pm
Dear All,

From what I understand, if one comes across the equation(x^2) = 4, then x = +2 or -2. However √(4) = +2 only (not -2).

I came across a question which required me to find a value for the following equation : √(x^2 - 6x + 9) + √(2 - x) + x - 3. To be precise, the question was as follows :

If each expression under the square root is greater than or equal to 0, what is √(x^2 - 6x + 9) + √(2 - x) + x - 3?

a. √(2-x)
b. 2x - 6 + √(2-x)
c. √(2-x) + x - 3
d. 2x - 6 + √(x-2)
e. x + √(x-2)



Now if the first square root is taken into consideration, √(x^2 - 6x + 9) = √(x-3)^2. So shouldnt the solution for this be +(x-3) as well as - (x-3). But while solving this problem only the positive root was taken into consideration. I am confused as to when we need to consider both roots and when only the +ve root is to be considered.
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by Stuart@KaplanGMAT » Mon Apr 25, 2011 12:36 am
Hi,

you answered your own question early on, when you said:
However √(4) = +2 only (not -2).
By mathematical convention, the "√x" should be read as "the non-negative root of x" (so √0 = 0, but in all other cases √x is the positive root of x). Like all math rules, this applies regardless of how complicated the expression under the root sign.

In fact, that's a very important thing to remember for all GMAT math: no matter how complicated something looks, the same basic rules apply. The GMAT loves to "dress up" questions to see if you can identify the basic rules under wild and wacky circumstances.

For this particular question, I'd never even attempt an algebraic solution: picking numbers is going to be far less confusing. Let's give it a go!

We certainly don't want any negative numbers under the roots, so since each "rooted" term has a -x component, let's pick a negative value for x to be safe. If x = -3 (a nice simple number), we get:

√(9 + 18 + 9) + √(2 + 3) - 3 - 3
= √36 + √5 - 6
= 6 + √5 - 6
= √5

Plugging -3 into the choices:

a. √(2-x) = √5
b. 2x - 6 + √(2-x) = -6 -6 + √5 = -12 + √5
c. √(2-x) + x - 3 = √5 - 3 - 3 = √5 - 6
d. 2x - 6 + √(x-2) = -6 - 6 + √-5 = imaginary
e. x + √(x-2) = -3 + √-5 = imaginary

Only (A) matches our target of √5... choose (A)!
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by Ryandmitri » Mon Apr 25, 2011 1:32 am
Thanks Stuart.

So can I safely say that if I get anything under a square root I should consider only the +ve root and if I need to find the solution for anything under a square sign then I consider both roots?

Thank You.
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by Ryandmitri » Mon Apr 25, 2011 1:37 am
Thanks Stuart.

So can I safely say that if I get anything under a square root I should consider only the +ve root and if I need to find the solution for anything under a square sign then I consider both roots?

Thank You.
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by Stuart@KaplanGMAT » Mon Apr 25, 2011 6:07 am
Ryandmitri wrote:Thanks Stuart.

So can I safely say that if I get anything under a square root I should consider only the +ve root and if I need to find the solution for anything under a square sign then I consider both roots?

Thank You.
Correct (if by square sign you mean anything raised to the power of two, like in your original example)!
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by Ryandmitri » Tue Apr 26, 2011 1:02 am
If √(x+4)^2= 3, which of the following could be the value of x - 4?
1) -11
2) -7
3) -4
4) -3
5) 5

For this question, shouldnt we be cnsidering only (x+4)= 3 ??
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by pankajks2010 » Tue Apr 26, 2011 9:58 am
Hi,

I too faced a similar problem with regard to square roots. You might want to refer to the following link:
https://www.beatthegmat.com/squares-root ... tml#358944

As for your recent-most question: Going by the reasoning that the experts have provided, we should consider only the positive root, which is (x+4) in this case. However, for this, the value of x becomes -1 and thus, (x-4)=-5 which does not matches with any of the options.

However, if we square both LHS & RHS of the equation, we get a quadratic equation: x^2+8x+7=0
Solving we get x=-7 and x=-1. From these options, we can get the required answer as -11
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