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Squares and roots

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by GMATGuruNY » Sun Jan 01, 2012 8:58 am
rijul007 wrote:What is the value of x?

(1) Image
(2) Image
√ means the positive root ONLY.

Statement 1: √x� = 9.
It's possible that x=3: √3� = √81 = 9.
It's possible that x=-3: √(-3)� = √81 = 9.
INSUFFICIENT.

Statement 2: √x² = -x.
It's possible that x=-3:
√(-3)² = -(-3)
√9 = 3
3=3.

It's possible that x=-1:
√(-1)² = -(-1)
√1 = 1
1=1.
INSUFFICIENT.

Statements 1 and 2 combined:
The only value that satisfies both statements is x=-3.
SUFFICIENT.

The correct answer is C.
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by neelgandham » Sun Jan 01, 2012 10:26 am
As Mitch already mentioned √ is always the positive root ONLY.
1) √x� = 9.
x� = 81 = 3^4 or (-3)^4. so the value of x can be 3 or -3
INSUFFICIENT.
2) √x² = -x.

√ is always the positive root. So -x > 0 Implies x < 0.
Let x = -2 then 2 =-(-2)(√x² = -x)
Let x = -3 then 3 =-(-3)(√x² = -x)
INSUFFICIENT.
From 1 and 2
x = -3 Sufficient Option C
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by pemdas » Sun Jan 01, 2012 4:23 pm
couple of comments here
st(1) sqrt(x^4)=|x^2|=+-x^2=+-9, Insuff
st(2) sqrt(x^2)=-|x|=+-x, hence x can be +ve or -ve, Insuff
Combined st(1&2): only with x=+9, x could be +-3 Insuff

e

rijul007 wrote:What is the value of x?

(1) Image
(2) Image
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by neelgandham » Sun Jan 01, 2012 4:41 pm
couple of comments here
st(1) sqrt(x^4)=|x^2|=+-x^2=+-9, Insuff - Square root of a number is always positive so sqrt(x^4)=x^2(which is positive and not +x^2, which leaves us with
x^2 = 9 and x = +3
st(2) sqrt(x^2)=-|x|=+-x, hence x can be +ve or -ve, Insuff - The question states sqrt(x^2) = -x and not sqrt(x^2)=-|x|
Combined st(1&2): only with x=+9, x could be +-3 Insuff
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by pemdas » Sun Jan 01, 2012 4:53 pm
@Nil :oops: -x is -1*x. You cannot disagree with sqrt(x^2)=|x| correct?

we assign |x| and further assign the sign, hence sqrt(x^2)=-|x|

in combined st(1&2): further descr. re st(1) has been provided
Combined st(1&2): only with x=+9, x could be +-3 Insuff
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by neelgandham » Sun Jan 01, 2012 4:58 pm
pemdas wrote:@Nil :oops: -x is -1*x. You cannot disagree with sqrt(x^2)=|x| correct?

we assign |x| and further assign the sign, hence sqrt(x^2)=-|x| - Is your assumption. He clearly specified that sqrt(x^2)= -x and we know that square root of any number is always >0. So for the value of -x to be greater than OR equal to 0, x should be less than or equal to 0

in combined st(1&2): further descr. re st(1) has been provided
Combined st(1&2): only with x=+9, x could be +-3 Insuff
Yes, by definition sqrt(x^2)=|x|, i.esqrt(x^2)= x if x > 0 and -x if x < 0.
Statement 2 says sqrt(x^2) = -x, Implies x is less than 0.

Comments in RED above!
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by pemdas » Sun Jan 01, 2012 5:06 pm
how did you find x<0???
was this given in this DS... or later assumed
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by neelgandham » Sun Jan 01, 2012 5:10 pm
pemdas wrote:how did you find x<0???
was this given in this DS... or later assumed
By definition
sqrt(x^2)= x if x > 0 or
sqrt(x^2)= -x if x < 0.
Statement 2 - sqrt(x^2)= -x, implies x <0 from definition of the square root. Now tell me which part of the explanation you disagree to?
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by pemdas » Sun Jan 01, 2012 5:22 pm
@Nil, you misinterpreted
i don't disagree and know beforehand what you were typing and all in red ...

what I mean is that sqrt(x^2)=|x| is true for any value of x within absolute

sqrt(9) has two roots: primary and secondary, i.e. |3|=+-3 (+ve and -ve)

when it stated sqrt(9)=-3 would be different from sqrt(x^2)=-x, as x has two signs within its modes (+ve and/or +ve), then we can switch the signs ...

that's the whole point, both statements include radicals with squares and it's Insuff for both statements together to have -3 as solution, because 3 is also solution here.

It's not right when you put sqrt(-9) but this it right when it's sqrt(9)=-3, conversely sqrt(9)=+3, since -3^2=9 and +3^2=9
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by neelgandham » Sun Jan 01, 2012 5:39 pm
pemdas wrote:@Nil, you misinterpreted
i don't disagree and know beforehand what you were typing and all in red ...

what I mean is that sqrt(x^2)=|x| is true for any value of x within absolute

sqrt(9) has two roots: primary and secondary, i.e. |3|=+-3 (+ve and -ve) - Partly agree

when it stated sqrt(9)=-3 would be different from sqrt(x^2)=-x, as x has two signs within its modes (+ve and/or +ve), then we can switch the signs ...

that's the whole point, both statements include radicals with squares and it's Insuff for both statements together to have -3 as solution, because 3 is also solution here.

It's not right when you put sqrt(-9) but this it right when it's sqrt(9)=-3, conversely sqrt(9)=+3, since -3^2=9 and +3^2=9
I hear what you are saying! My understanding of definition of square root in GMAT (from expert posts on GMAT problems/and solutions provided by Prep companies), Square root of a number is the principal square root which is the positive square root.

Every positive number x has two square roots: , which is positive, and , which is negative. Together, these two roots are denoted (see ± shorthand). Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root. - Source Wiki

So, the statement sqrt(9) has two roots: primary and secondary, i.e. |3|=+-3 (+ve and -ve) is correct BUT In general and in GMAT sqrt(9) = 3 and not +3

Check the Properties section here -> https://en.wikipedia.org/wiki/Square_root and if you agree to the definition of Square root(x^2) as provided there, then you should agree that
sqrt(9) = 3 and not +3 because if you assume
9 = -3*-3, then sqrt(9) = sqrt(-3^2) = -(-3) = 3 (Definition: sqrt(x^2)= -x if x < 0.)
and if you assume
9 = 3*3, then sqrt(9) = sqrt(3^2) = 3 (Definition: sqrt(x^2)= x if x > 0.)

p.s: I am sorry if your question is still left unanswered, but I s**k at explaining things and I am trying to improve in that region.
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by pemdas » Sun Jan 01, 2012 5:53 pm
clear now
this question tests/checks definition of the radical or that sqrt(a^2) is always positive a, primary root of the number. Hopefully such definition questions do not twist me in exam :(
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