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x= \sqrt[4]{(x^3 + 6x^2)}
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Updated on: 16 Sep 2026, 10:09
Updated on: 16 Sep 2026, 10:09
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\(x = \sqrt[4]{(x^3 + 6x^2)}\)
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
sum of all possible values of x |
1 |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Originally posted by GreenlightTestPrep on 02 Feb 2019, 07:07.
Last edited by Carcass on 16 Sep 2026, 10:09, edited 3 times in total.
Last edited by Carcass on 16 Sep 2026, 10:09, edited 3 times in total.
Edited by Carcass
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: x= \sqrt[4]{(x^3 + 6x^2)}
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03 Dec 2021, 06:19
03 Dec 2021, 06:19
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JKGrowling wrote:
This is incorrect the answer should be C.
(16)^1/2 has 2 solutions, 4 and -4
For a better representation, check the graphical representation of y = x^2. It is a parabola, with 2 x-intercepts(roots) centered around the y axis. The graph for y = x^4 is similar, except that is steeper because y rises faster than the run of x since it is a higher power.
(16)^1/2 has 2 solutions, 4 and -4
For a better representation, check the graphical representation of y = x^2. It is a parabola, with 2 x-intercepts(roots) centered around the y axis. The graph for y = x^4 is similar, except that is steeper because y rises faster than the run of x since it is a higher power.
Be careful, although there are two x-values (2 and -2) that satisfy the equation x^4 = 16, the fourth root notation tells us to take the positive root only.
So, ∜(16) = 2 (not -2)
If you want a graphical solution, it's better to graph the equation y = ∜x
If you do, you'll get something like this:
As you can see, there's only one value of y corresponding to x = 16
General Discussion
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: x= \sqrt[4]{(x^3 + 6x^2)}
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02 Feb 2019, 07:09
02 Feb 2019, 07:09
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GreenlightTestPrep wrote:
x = ∜(x³ + 6x²)
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
sum of all possible values of x |
1 |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Given: x = ∜(x³ + 6x²)
Raise both sides to the power of 4 to get: x⁴ = x³ + 6x²
Rewrite equation as follows: x⁴ - x³ - 6x² = 0
Factor to get: x²(x² - x - 6) = 0
Factor more to get: x²(x - 3)(x + 2) = 0
This means x² = 0, (x - 3) = 0, or (x + 2) = 0
So, the possible solutions are x = 0, x = 3 and x = -2
IMPORTANT: At this point, we should plug each answer choice back into the equation to identify any possible extraneous roots.
Test x = 0. We get: 0 = ∜(0³ + 6*0²) = ∜(0). WORKS
Test x = 3. We get: 3 = ∜(3³ + 6*3²) = ∜(81). WORKS
Test x = -2. We get: -2 = ∜(-2³ + 6*-2²) = ∜(16). DOESN’T work.
So, the only VALID solutions are x = 0 and x = 3
So, the SUM of all possible values of x = 0 + 3 = 3
We get:
QUANTITY A: 3
QUANTITY B: 1
Answer: A
Cheers,
Brent
