\(A=(\sqrt{x}+ \sqrt{x} )^2=\sqrt{x} ^2+\sqrt{x} ^2+ 2*\sqrt{x}* \sqrt{x}=x+x+2x=4x=x+3x\)
\(B=(x + x \sqrt{x})\)

so we are comparing \(x+3x\) and \((x+x\sqrt{x})\).....
since x is >1 we compare \(3x\) and \((x\sqrt{x})\) or \(3=\sqrt{9}\) and \(\sqrt{x}\)
since x is less than 9 but >1, \(\sqrt{9}\) > \(\sqrt{x}.......A>B\)

A

Please fix quantity B in solution. It is x+x√x

Please fix quantity B in solution. It is x+x√x[/quote]

Good catch - thanks!
I have edited my solution accordingly.

Cheers,
Brent

Answer: A

1 < x < 9 and x is an integer
A: (√x + √x)2 = (2√x)^2 = 4x
B: x+x√x = x(1+√x)

We now should compare 4 and 1+√x
x is between 2 and 8 so the maximum value for √x equals √8 which is less than 3 so 1+√x < 1+3 =4 So A is bigger than B

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.