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What is the value of x?

Expert replies
by suzeemunkee » Thu Jul 29, 2010 7:37 am
What is the value of x?

(1) sqrt(x^4) = 9
(2) sqrt(x^2) = -x

The issue I can't figure out with this problem is: Why can't we assume that Statement 1 will always yield a positive number? Here's my logic: Since sqrt(x^2) equals the absolute value of x, why can't we assume that sqrt(x^4), which is the same as sqrt(x^2*x^2), will also always be positive?

The OA is C. The question is from beat the gmat's SmartGMAT (which I love!) practice.

Thanks guys for any help you can provide :)
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Source: — Data Sufficiency |

by selango » Thu Jul 29, 2010 7:56 am
stmt1,

sqrt(x^4) = 9

x can be sqrt(3) or -sqrt(3)

Insuff

stmt2,

sqrt(x^2) = -x

No info abt X or sqrt(x^2)

Insuff

Combining 1 and 2,

sqrt(x^4) = 9

x^2=9

sqrt(x^2)=3

x=sqrt(3) or x=-sqrt(3)

we know that sqrt(x^2) = -x

-->x=-sqrt(3)

Pick C
--Anand--
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by selango » Thu Jul 29, 2010 8:06 am
The issue I can't figure out with this problem is: Why can't we assume that Statement 1 will always yield a positive number? Here's my logic: Since sqrt(x^2) equals the absolute value of x, why can't we assume that sqrt(x^4), which is the same as sqrt(x^2*x^2), will also always be positive?
sqrt(x^4) = 9

sqrt(x^2*x^2)=9

x^2 is positive.However x can be positive or negative.

For ex sqrt(x^4)=4

sqrt(x^2*x^2)=4

x^2 is 4.Here x^2 is positive.

However x^2=4.x can be 2 or -2

Hence we cant determine value of x.

Hope this clarify
--Anand--
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by suzeemunkee » Thu Jul 29, 2010 8:23 am
selango wrote:
The issue I can't figure out with this problem is: Why can't we assume that Statement 1 will always yield a positive number? Here's my logic: Since sqrt(x^2) equals the absolute value of x, why can't we assume that sqrt(x^4), which is the same as sqrt(x^2*x^2), will also always be positive?
sqrt(x^4) = 9

sqrt(x^2*x^2)=9

x^2 is positive.However x can be positive or negative.

For ex sqrt(x^4)=4

sqrt(x^2*x^2)=4

x^2 is 4.Here x^2 is positive.

However x^2=4.x can be 2 or -2

Hence we cant determine value of x.

Hope this clarify

A beam of light was just shed upon my noggin :) Thank you selango.
Join the discussion