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

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.

Thank you Sir.

it is just a matter of practicality no more than this

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?

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

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

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

\(3x^{\frac{2}{3}}+2 \sqrt[3]{x}-8=0\)

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