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Re: 3x^2/3+2 [#permalink]
Its tricky. Any easy explanation?
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Re: 3x^2/3+2 [#permalink]
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Farina wrote:
Its tricky. Any easy explanation?


If you're comfortable factoring things where the "a" value is not 1, then you essentially just use x^1/3 as your standard x value and factor, and sort out the exponent at the end step. This is because (X^1/3)(x^1/3)=(X^2/3) (because you add the exponents like conventional numbers or fractions when multiplying the same bases raised to a power)

The trick to factoring when the "a" value is not 1 is multiplying the "a" and "c" values to determine what you are trying to make the component x values multiply to. For example, in the given problem we have a=3 b=2 c=-8
3 X -8 = -24
so we factor by finding what multiplies to -24 and adds to our "b" value of positive 2, namely 6 and -4

so then we substitute 2x (with our x in this case being x^1/3) for 6x and -4x which still add to positive 2. We can then factor by grouping and factoring out what we can.

so 3x^2 + 6x - 4x - 8 gets grouped like so:

(3x^2 + 6x) (-4x - 8)

then we factor out what we can leaving a common base inside the parenthesis (which is how you know you're on the right track)

3x(x+2)-4(x+2)

then you use what's in the parenthesis as one factor, and what's outside as the second

(x+2)(3x-4) and you're off to the races.

In the given problem, if you use x^1/3 instead of x it will factor the same but the answer will be correct. Hope this helped

Originally posted by stinkydiver on 04 Aug 2020, 01:21.
Last edited by stinkydiver on 04 Aug 2020, 10:24, edited 3 times in total.
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Re: 3x^2/3+2 [#permalink]
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Thank you sir for the amazing explanation.

Post it as text and not as images. It would be easiest to search by the students

Regards
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Re: 3x^2/3+2 [#permalink]
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Carcass wrote:
Thank you sir for the amazing explanation.

Post it as text and not as images. It would be easiest to search by the students

Regards


You're welcome!
I will post answers as text from now on.
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Re: 3x^2/3+2 [#permalink]
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Thank you Sir.

it is just a matter of practicality no more than this :wink:
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Re: 3x^2/3+2 [#permalink]
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Carcass wrote:
\(3x^{\frac{2}{3}}+2 \sqrt[3]{x}-8=0\)

Quantity A
Quantity B
\(x\)
\(\frac{64}{25}\)



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.



B is the correct answer.


Factoring the equation you get:

\((3*x^\frac{1}{3} - 4)(x^\frac{1}{3} + 2) = 0\)

Starting with the left equation:

\(3*x^\frac{1}{3} - 4 = 0\)
\(3*x^\frac{1}{3} = 4\)
\(x^\frac{1}{3} = \frac{4}{3}\)
\(x = \frac{64}{27}\)

With the second equation:

\(x^\frac{1}{3} + 2 = 0\)
\(x^\frac{1}{3} = -2\)

\(x = -8\)


\( \frac{64}{27}\) \(<\) \(\frac{64}{25}\)

Because \(\frac{64}{25}\) has the same numerator but a smaller denominator than \(\frac{64}{25}\)

\(x = -8\) \(<\) \(\frac{64}{25}\)

Is obvious.


Since both roots are less than Quantity B, Quantity B is greater, so B is the answer.
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Re: 3x^2/3+2 [#permalink]
grenico wrote:
Carcass wrote:
\(3x^{\frac{2}{3}}+2 \sqrt[3]{x}-8=0\)

Quantity A
Quantity B
\(x\)
\(\frac{64}{25}\)



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.



B is the correct answer.


Factoring the equation you get:

\((3*x^\frac{1}{3} - 4)(x^\frac{1}{3} + 2) = 0\)

Starting with the left equation:

\(3*x^\frac{1}{3} - 4 = 0\)
\(3*x^\frac{1}{3} = 4\)
\(x^\frac{1}{3} = \frac{4}{3}\)
\(x = \frac{64}{27}\)


With the second equation:

\(x^\frac{1}{3} + 2 = 0\)
\(x^\frac{1}{3} = -2\)

\(x = -8\)


\( \frac{64}{27}\) \(<\) \(\frac{64}{25}\)

Because \(\frac{64}{25}\) has the same numerator but a smaller denominator than \(\frac{64}{25}\)

\(x = -8\) \(<\) \(\frac{64}{25}\)

Is obvious.


Since both roots are less than Quantity B, Quantity B is greater, so B is the answer.


x^(1/3) = 4^3/3^3

means if power is in fraction, it will be the power of other numbers on opposite side?
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Re: 3x^2/3+2 [#permalink]
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we can write the eqtn like ,

3 x ^ (1 /3) x ^ (1 /3) + 2 x ^ (1 /3) - 8 = 0

let , x ^ (1 /3) = a

so the eqn is,

3 a^2 + 2 a - 8 = 0
solve for a

a = 4/3 or a = -2
i.e ,
x ^ (1 /3) = 4/3 or x ^ (1 /3) = -2
=> x = 64/ 27 or x = -8

both are less than B

so answer is B
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Re: 3x^2/3+2 [#permalink]
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Quote:
x^(1/3) = 4^3/3^3

means if power is in fraction, it will be the power of other numbers on opposite side?


x^(1/3) is the equivalent of the cube root of x.

Since we want to solve for x in the equation x^(1/3) = 4/3, we would need to cube both sides.

So you can interpret the sentence as: 4/3 is the cube root of x.
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Re: 3x^2/3+2 [#permalink]
grenico wrote:
Quote:
x^(1/3) = 4^3/3^3

means if power is in fraction, it will be the power of other numbers on opposite side?


x^(1/3) is the equivalent of the cube root of x.

Since we want to solve for x in the equation x^(1/3) = 4/3, we would need to cube both sides.

So you can interpret the sentence as: 4/3 is the cube root of x.


Cube both sides, got it. Thank you very much
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Re: 3x^2/3+2 [#permalink]
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let \(x^1/3 = p\)

we can re write the equation as \(3p^2+2p-8=0\)

after factorising we will get p=-2 and p=(4/3) as solutions

now \(x=p^3\)
x=-8 and x=64/27
so when x=-8
then quant A < quant B (since x is -ve)

when x=(64/27)
then quant A < quant B

so final answer will be B
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3x^2/3+2 [#permalink]
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\(3x^{\frac{2}{3}}+2 \sqrt[3]{x}-8=0\)

can be re-written as

\(3x^{\frac{2}{3}} + 2x^{\frac{1}{3}} - 8 = 0\)

we can see that \(x^{\frac{1}{3}}\) is common to both the quadratic (\(x^2\)) and the linear term (\(x\))

So we do a simple u-substitution

\(u = x^{1/3}\)

and rewrite the equation as

\(3u^2 + 2u - 8 = 0\)

Now, we find two numbers that multiply to \(-24\) (\(3 * -8\)) and add to \(2\). They are clearly \(-6\) and \(4\)

So we proceed

\(3u^2 - 6u + 4u - 8 = 0\)

\(3u(u + 2) - 4(u +2) = 0\)

\((3u - 4)(u + 2) = 0\)

\(3u - 4 = 0\)

\(3u = 4\)

\(u = 4/3 \)

OR

\(u + 2 = 0\)

\(u = -2\)

Now since \(u = x^{\frac{1}{3}}\)

\(x^{\frac{1}{3}} = \frac{4}{3}\)

cubing both sides

\(x = \frac{64}{27}\)

Since \(\frac{64}{27} < \frac{64}{25}\), Quantity B is greater.

and since we also have

\(u = -2\)

\(x^{\frac{1}{3}} = -2\)

cubing both sides

\( x = -8\)

And since \(-8 < \frac{64}{25}\), Quantity B is greater.

So, either way, Quantity B is greater
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