BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

3 prep questions on square root equations

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 48
Joined: Sat Nov 28, 2009 1:01 pm
Thanked: 1 times
Followed by:1 members

3 prep questions on square root equations

by Mr_T » Tue Jan 19, 2010 2:24 pm
Hi everyone,

I seem to have some problems with some equations containing the root of something. It would be great if someone took a look.

1) if M = 4^1/2 + 4^1/3 + 4^1/4, then the value of M is...
a) Less than 3
b) Equal to 3
c) between 3 and 4
d) equal to 4
e) greater than 4 -> Ans, but how? by approx.?

2) If x and y are positive, which of the following must be greater than 1 / (x+y)^1/2

I. (x+y)^1/2 / 2x
II. (x^1/2 + y^1/2) / x+y
III. (x^1/2 - y^1/2) / x+y

3) Loss the question but I think it was: Solve ( ((9+80^1/2)^1/2 + (9-80^1/2)^1/2)^2


Thanks,

mr T
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 98
Joined: Mon Nov 23, 2009 2:30 pm
Thanked: 26 times
Followed by:1 members

by ace_gre » Tue Jan 19, 2010 4:04 pm
Hi, Here is my approach..

1. M = 4^1/2 + 4^1/3 + 4^1/4
M = 4 ^6/12 + 4^ 4/12 + 4^3/12
M = 4^(3/12) *( 4^2+ 4^1 + 1)
M = 4^1/4*(21)
M=2^1/2 * 21
M ~1.414*21 and is greater than 4.

2. Lets say x=1,y=1,
1/(x+y)^1/2 = 1/ 2^1/2

I. 2^1/2 / 2 ==> 1/ 2^1/2 which is equal. Hence not true.

II. x=1,y=1
1/(x+y)^1/2 = 1/ 2^1/2 ==>(2)^1/2
(x^1/2 + y^1/2) / x+y ==> 1. true.
Check for fractions. lets say x= 1/4 and y=1/4.
exp II is greater

III. (x^1/2 - y^1/2) / x+y
when x=1, y=1, II becomes 0. Hence not greater.

IMO II is true.

3. Based on the question, I say let a = 9, b= 80^1/2
exp = ((a+b) ^1/2 + (a-b)^ 1/2 )^2
==> a+b +a-b +2 (a+b)^1/2 (a-b) ^1/2
==>2a +2(a^2 - b^2)^1/2
==>2(9) + 2( 81 - 80) ^1/2
==>20.

IMO 20
Top
Senior | Next Rank: 100 Posts
Posts: 48
Joined: Sat Nov 28, 2009 1:01 pm
Thanked: 1 times
Followed by:1 members

by Mr_T » Tue Jan 19, 2010 7:08 pm
Thanks!

Actually, you made a mistake in the first problem when you tried to factor out 4^3/12. It's ok I figured out how to approximate the answer. For question 2, I wanted to see if there was a way to simplify the equations instead of plugging numbers. Your answer for the third one is dead on. I should have seen that the exponents were the same so I could multiply their base...

Thanks again.

mr T
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Tue Jan 05, 2010 11:02 am
Thanked: 1 times

by nicolas » Sun Jan 31, 2010 1:01 am
ace_gre wrote:Hi, Here is my approach..

1. M = 4^1/2 + 4^1/3 + 4^1/4
M = 4 ^6/12 + 4^ 4/12 + 4^3/12
M = 4^(3/12) *( 4^2+ 4^1 + 1)
M = 4^1/4*(21)
M=2^1/2 * 21
M ~1.414*21 and is greater than 4.
Hi ace_gre,

actually you got a flaw in your calculation:
M = 4 ^6/12 + 4^ 4/12 + 4^3/12 is not equal to M = 4^(3/12) *( 4^2+ 4^1 + 1), as you multiplied the exponents, although you have to add them.
If you want to factor out the right equation would be:
M = 4 ^6/12 + 4^ 4/12 + 4^3/12
M = 4 ^3/12 *(4 ^3/12 + 4^1/12 + 1)
..
I would find another method beneficial, namely:
Take a look:
4^1/2 + 4^1/3 + 4^1/4

--> The first term 4^1/2 = 2
--> The second term is 4^1/3, we don't know what that is, but it states that x*x*x = 4, i.e. at least x>1 has to be true!
--> The third term is 4^1/4 = (2^2)^1/4 = 2^1/2, i.e. the square root of 2, which is something about 1.4, i.e. >1

So we have 2 + >1 + >1 and that is at least > 4!

Thus E
Top
Post new topic Post Reply
• Page 1 of 1