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Roots

Expert replies
by yellowho » Wed Mar 16, 2011 9:24 pm
Sqroot(x)>x

Can you square both sides? (We know nothing about the signs. I guess x cannot be negative because sqroot of x would then be imaginary).
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Source: — Problem Solving |

by 6983manish » Wed Mar 16, 2011 9:56 pm
yellowho wrote:Sqroot(x)>x

Can you square both sides? (We know nothing about the signs. I guess x cannot be negative because sqroot of x would then be imaginary).

Better, we can multiply both sides by sqrt(x)

which becomes x > x^2 , which is possible in case when x < 1.

As per GMAT guidelines , all the numbers used in problems there must be real. ( Please correct if I am wrong ).
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by Anurag@Gurome » Wed Mar 16, 2011 10:04 pm
yellowho wrote:Sqroot(x)>x

Can you square both sides? (We know nothing about the signs. I guess x cannot be negative because sqroot of x would then be imaginary).
Also, note that in GMAT, sqrt(x) always means the positive square root of x.
For example, sqrt(4) is always 2 and not -2.
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by yellowho » Wed Mar 16, 2011 11:15 pm
I'm sorry, I don't understand. So can we square both sides?

Also, if we multiply both sides by srt(x) dont we get X>xsqrt(x)? How did you get squares?



[quote="Anurag@Gurome"][quote="yellowho"]Sqroot(x)>x

Can you square both sides? (We know nothing about the signs. I guess x cannot be negative because sqroot of x would then be imaginary).[/quote]

Also, note that in GMAT, sqrt(x) always means the positive square root of x.
For example, sqrt(4) is always 2 and not -2.[/quote]
Join the discussion

by yellowho » Wed Mar 16, 2011 11:19 pm
I'm sorry, I don't understand. So can we square both sides?

Also, if we multiply both sides by srt(x) dont we get X>xsqrt(x)? How did you get squares?



[quote="Anurag@Gurome"][quote="yellowho"]Sqroot(x)>x

Can you square both sides? (We know nothing about the signs. I guess x cannot be negative because sqroot of x would then be imaginary).[/quote]

Also, note that in GMAT, sqrt(x) always means the positive square root of x.
For example, sqrt(4) is always 2 and not -2.[/quote]
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by hoji » Thu Mar 17, 2011 2:05 pm
Sqroot(x)>x in this case first thing we do is to find determinant for this inequality;
if Sqroot(x), then it must be x>0 , otherwise inequality will be meaningless,
so x is positive we can freely square both sides.
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