## GRE 0568 question 11

Forum for the GRE subject test in mathematics.
jijicavalerul
Posts: 1
Joined: Thu Oct 30, 2008 8:34 pm

### GRE 0568 question 11

Hey guys,

Can anyone help me with this problem please:

What is the best approximation of 1.5^{1/2}*266^{3/2}?

I've tried using a Taylor approximation on f(x)=x*266^x, with f(1)=266, but don't get the right answer 5,300. Does anyone have any ideas?

Thanks!

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
You have to use f(x,y)=(x)^(1/2)*(y)^(3/2)
then choose x=1, y=256

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am
Try 1.5^(1/2) * 266 ^ (3/2) = [ (1.5 * 266) ^ (3/2) ] / 1.5 = (3 / 2 * 266) ^ 3/2 / (3/2) = (399)^(3/2) / (3/2)

Your answer will be slightly higher if you use 400 instead of 399, so try (400)^(3/2) / 1.5 = 8000 / 1.5 = 16000 / 3 = 5333.3333

ralphhumacho
Posts: 8
Joined: Tue Mar 04, 2008 1:02 am
Here's how I did it:

266^(3/2) = sqrt(266)sqrt(266)sqrt(266) = 16 * 16 * 16 = 4096 = 4100, since 16^2 = 256, which is definitely close enough in this case considering the range of answer choices.

To find sqrt(1.5) = sqrt(3)/sqrt(2) = 1.7/1.4 (rough approx). Thus, the final answer will be significantly higher than 4100, so 5300 makes the most sense.

Note that since the answers are each like 1000 units apart, you have HUGE leeway in approximating.

nate
Posts: 1
Joined: Wed Nov 05, 2008 7:23 pm
This is what I did:
266^(3/2) is 266*266^(1/2)
so we can rewrite as 266*(1.5*266)^(1/2)
= 266*sqrt(399)
which is approximately 266*sqrt(400) or 266*20 = 5320