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x=(x^3+6x^2)
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22 Aug 2019, 02:57
22 Aug 2019, 02:57
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Question Stats:
30% (01:44) correct
69% (01:13) wrong based on 113 sessions
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\(x=(x^3+6x^2)^\frac{1}{4}\)
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
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
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Re: x=(x^3+6x^2)
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22 Aug 2019, 13:17
22 Aug 2019, 13:17
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smartchin77 wrote:
\(x=(x^3+6x^2)^\frac{1}{4}\)
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
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
General Discussion
Re: x=(x^3+6x^2)
[#permalink]
22 Aug 2019, 04:42
22 Aug 2019, 04:42
Expert Reply
\(x= x^2(x+6)^{1/4}\)
\((x^2)^{1/4} \times (x+6)^{1/4}\)
Just only the first root x is \(x^{1/2}\)
AND the second one \((x+6)^{1/4}\) must be \(> 1\)
Answer is A
\((x^2)^{1/4} \times (x+6)^{1/4}\)
Just only the first root x is \(x^{1/2}\)
AND the second one \((x+6)^{1/4}\) must be \(> 1\)
Answer is A
