Last visit was: 17 May 2024, 16:28 It is currently 17 May 2024, 16:28

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
avatar
Intern
Intern
Joined: 07 Feb 2019
Posts: 36
Own Kudos [?]: 30 [5]
Given Kudos: 0
Send PM
Verbal Expert
Joined: 18 Apr 2015
Posts: 28756
Own Kudos [?]: 33335 [0]
Given Kudos: 25245
Send PM
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11734 [0]
Given Kudos: 136
Send PM
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11734 [4]
Given Kudos: 136
Send PM
Re: x=(x^3+6x^2) [#permalink]
3
1
Bookmarks
smartchin77 wrote:
\(x=(x^3+6x^2)^\frac{1}{4}\)

Quantity A
Quantity B
The sum of all possible roots of x
1



A)Quantity A is greater.
B)Quantity B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given


TAKE: \(x=(x^3+6x^2)^\frac{1}{4}\)

Raise both sides to the power of 4 to get: \(x^4=x^3+6x^2\)

Set equation equal to zero: \(x^4-x^3-6x^2=0\)

Factor to get: \(x^2(x^2-x-6)=0\)

Factor part in brackets: \(x^2(x-3)(x+2)=0\)

Solutions: x = 0, x = 3 and x = -2
HOWEVER, we should check these solutions for extraneous roots by plugging the values into the original equation.

Test \(x = 0\)
We get: \(0=(0^3+6(0)^2)^\frac{1}{4}\)
Simplify: \(0=(0)^\frac{1}{4}\)
WORKS!

Test \(x = 3\)
We get: \(3=(3^3+6(3)^2)^\frac{1}{4}\)
Simplify: \(3=(81)^\frac{1}{4}\)
WORKS!

Test \(x = -2\)
We get: \(-2=((-2)^3+6(-2)^2)^\frac{1}{4}\)
Simplify: \(-2=(16)^\frac{1}{4}\)
NO GOOD!

So, the solutions are \(x = 0\) and \(x = 3\)

We get:
Quantity A: \(0+3=3\)
Quantity B: \(1\)

Answer: A

Cheers,
Brent
Senior Manager
Senior Manager
Joined: 23 Jan 2021
Posts: 294
Own Kudos [?]: 153 [0]
Given Kudos: 81
Concentration: , International Business
Send PM
x=(x^3+6x^2) [#permalink]
GreenlightTestPrep wrote:
smartchin77 wrote:
\(x=(x^3+6x^2)^\frac{1}{4}\)

Quantity A
Quantity B
The sum of all possible roots of x
1



A)Quantity A is greater.
B)Quantity B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given


TAKE: \(x=(x^3+6x^2)^\frac{1}{4}\)

Raise both sides to the power of 4 to get: \(x^4=x^3+6x^2\)

Set equation equal to zero: \(x^4-x^3-6x^2=0\)





Factor to get: \(x^2(x^2-x-6)=0\)

Factor part in brackets: \(x^2(x-3)(x+2)=0\)

Solutions: x = 0, x = 3 and x = -2
HOWEVER, we should check these solutions for extraneous roots by plugging the values into the original equation.

Test \(x = 0\)
We get: \(0=(0^3+6(0)^2)^\frac{1}{4}\)
Simplify: \(0=(0)^\frac{1}{4}\)
WORKS!

Test \(x = 3\)
We get: \(3=(3^3+6(3)^2)^\frac{1}{4}\)
Simplify: \(3=(81)^\frac{1}{4}\)
WORKS!

Test \(x = -2\)
We get: \(-2=((-2)^3+6(-2)^2)^\frac{1}{4}\)
Simplify: \(-2=(16)^\frac{1}{4}\)
NO GOOD!

So, the solutions are \(x = 0\) and \(x = 3\)

We get:
Quantity A: \(0+3=3\)
Quantity B: \(1\)

Answer: A

Cheers,
Brent



Sir, do i need to check in original equation? If i put x=-2 in \(x^4=x^3+6x^2\), I get the equal no in both side, Is this ambiguous question?
Intern
Intern
Joined: 09 Sep 2021
Posts: 1
Own Kudos [?]: 1 [1]
Given Kudos: 12
GRE 1: Q169 V153
Send PM
x=(x^3+6x^2) [#permalink]
1
GreenlightTestPrep wrote:
Test \(x = -2\)
We get: \(-2=((-2)^3+6(-2)^2)^\frac{1}{4}\)
Simplify: \(-2=(16)^\frac{1}{4}\)
NO GOOD!


Doesn't make sense. Coz,

\((16)^\frac{1}{4} = 2\) or \(-2\)

So, \(-2\) checks out.

The summation of the possible values of \(x\) is \(0 + 3 - 2 = 1\).

The answer should be C.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11734 [1]
Given Kudos: 136
Send PM
Re: x=(x^3+6x^2) [#permalink]
1
mushaf wrote:
GreenlightTestPrep wrote:
Test \(x = -2\)
We get: \(-2=((-2)^3+6(-2)^2)^\frac{1}{4}\)
Simplify: \(-2=(16)^\frac{1}{4}\)
NO GOOD!


Doesn't make sense. Coz,

\((16)^\frac{1}{4} = 2\) or \(-2\)

So, \(-2\) checks out.

The summation of the possible values of \(x\) is \(0 + 3 - 2 = 1\).

The answer should be C.


\((16)^\frac{1}{4} = \sqrt[4]{16}\), and the fourth root notation tells us to take only the POSITIVE fourth root of 16.

This is very similar to the square root notation that tells us to take the positive root of a number.
For example, \(\sqrt{9}=3\) (and not -3)

If, however, we're given a similar equation but WITHOUT the root notation, there will be a positive and a negative solution.
For example, if \(x^4 = 81\), then there are two solutions: \(x = 3\) and \(x = -3\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11734 [0]
Given Kudos: 136
Send PM
Re: x=(x^3+6x^2) [#permalink]
COolguy101 wrote:
Sir, do i need to check in original equation? If i put x=-2 in \(x^4=x^3+6x^2\), I get the equal no in both side, Is this ambiguous question?


When testing for extraneous roots, we must plug the potential solution into the original equation

In this case, the original equation is \(x=(x^3+6x^2)^\frac{1}{4}\).
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 4501
Own Kudos [?]: 68 [0]
Given Kudos: 0
Send PM
Re: x=(x^3+6x^2) [#permalink]
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.
Prep Club for GRE Bot
[#permalink]
Moderators:
Moderator
1085 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne