5x^2/(x^2)^(1/3)

Question

\frac{5x^2}{\sqrt {x^2}}

A.  5\sqrt {x}

B.  5x\sqrt {x}

C.  5x\sqrt {x^2}

D.  5x^2\sqrt{x^2}

E.  125x^4

shashankism · Accepted Answer

\frac{5x^2}{\sqrt {x^2} 
=  \frac{5x^2}{x^{2/3}} 
=  5 x^{2-2/3} 
=  5 x^{4/3} 
=  5x^{1+1/3} 
=  5x\sqrt {x}

Answer B

rocko911 · Answer

What is wrong with cubing numerator and denominator?

It will be
 125 * (x)power 6 / (x)power 2
Ans will be

125 * (x)power 4

shashankism · Answer

Hey Rocko911,

Why do you want to take cube of the given expression.. Its no where written to find cube.

shashankism · Answer

If you want to solve by cubing the expression then you will have to take cube root of the same..

\frac{5x^2}{\sqrt {x^2} 
= \sqrt {(\frac{5x^2}{\sqrt {x^2})^3} 
= \sqrt {(\frac{125x^6}{x^2})} 
= \sqrt {125x^4} 
=  5x\sqrt {x}

rocko911 · Answer

If you want to solve by cubing the expression then you will have to take cube root of the same..

\frac{5x^2}{\sqrt {x^2} 
= \sqrt {(\frac{5x^2}{\sqrt {x^2})^3} 
= \sqrt {(\frac{125x^6}{x^2})} 
= \sqrt {125x^4} 
=  5x\sqrt {x}

Why do we need to take cube root???? Please help

broall · Answer

In case that you want to get rid of cube root of $\sqrt {x}$

rocko911 · Answer

In case that you want to get rid of cube root of $\sqrt {x}$

yeah to get rid of $\sqrt {x}$ we are cubing the whole expression.... but as told by Shashankism, He cubed the whole problem AND took CUBE ROOT ALSO.......SO WHY BOTH? Why not only cubing the whole expression and get rid of $\sqrt {x}$ ??????????

broall · Answer

He did so to make the expression more simple for calculate. You could make cube only of $\sqrt {x^2}$, however, you still have to go around to come to the answer.

rocko911 · Answer

No, i dont get it

See how I did this..

5x^2 / (x^2)^(1/3)

5x(x^2) / x^(2/3)

CUBING NUMERATOR AND DENOMINATOR

125 x^(6) / x^(2)

125 x^4

what is wrong with this ?

broall · Answer

Your way is wrong as I explained above

rocko911 · Answer

5×x2x23=5×x2x23 ?

CUBING NUMERATOR AND DENOMINATOR

125 x^(6) / x^(2) .....Hey, this step is wrong. You can't left the cube root sign like this. You have to keep it like this:
=(5×x2x23)3−−−−−−−−⎷3=125×x6x2−−−−−−√3=(5×x2x23)33=125×x6x23

125 x^4 .....As this step led to incorrect value. This step must be =125x4−−−−−√3

cube root was in denominator and when i already did x^(2/3) from where did the cube and a cube root came from?????????? i dont get it,, can u explain by basic and step by step approach?

broall · Answer

I don't get your question.

Let make another example. For example, simplify this $\sqrt {8x^3}$

The correct way is $\sqrt {8x^3}=\sqrt {(2x)^3}=2x$

Your way is like this $\sqrt {8x^3}=8x^3$. You left out cube root and get wrong value.

However, you could left out cube when solving equation.
$\sqrt {8x^3}=1 \implies 8x^3=1$

Hope this clears your problem

kandarlin · Answer

In this case you are not necessarily multiplying the expression by 1, so you have created a different term by cubing both the numerator and denominator and then simplifying. So taking the cube root of your solution 125x^4 should get you back to to the answer.

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5x^2/(x^2)^(1/3)

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