+0  
 
+5
273
6
avatar+271 

$${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{x}}}} = {\sqrt{{\frac{{\mathtt{x}}}{{\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{x}}}}}}}}$$

it solves out fine manually and all its just interesting.

TheJonyMyster  Apr 26, 2015

Best Answer 

 #4
avatar+91045 
+10

Lets just proove this TheJonyMyster

 

$$\\\sqrt[n]{x}=\sqrt[(n-1)]{\frac{x}{\sqrt[n]{x}}}\\\\
RHS=\sqrt[(n-1)]{x^{1-\frac{1}{n}}}\\\\
RHS=\sqrt[(n-1)]{x^{\frac{n-1}{n}}}\\\\
RHS=x^{\frac{n-1}{n}}\right)^{\frac{1}{n-1}}\\\\
RHS=x^{(1/n)}\\\\
RHS=\sqrt[n]{x}\\\\
RHS=LHS$$
    

Melody  Apr 28, 2015
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6+0 Answers

 #1
avatar+26329 
+10

It's true as long as x is greater than zero.

.

Alan  Apr 26, 2015
 #2
avatar+91045 
+5

Yes TheJonyMyster, it is cool how roots work :)

Melody  Apr 26, 2015
 #3
avatar+271 
+5

yeah, positive x cause of imaginary numbers. also it works for any $${\sqrt[{{\mathtt{{\mathtt{n}}}}}]{{\mathtt{x}}}} = {\sqrt[{{\mathtt{\left({\mathtt{n}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}}}]{{\frac{{\mathtt{x}}}{{\sqrt[{{\mathtt{{\mathtt{n}}}}}]{{\mathtt{x}}}}}}}}$$

TheJonyMyster  Apr 26, 2015
 #4
avatar+91045 
+10
Best Answer

Lets just proove this TheJonyMyster

 

$$\\\sqrt[n]{x}=\sqrt[(n-1)]{\frac{x}{\sqrt[n]{x}}}\\\\
RHS=\sqrt[(n-1)]{x^{1-\frac{1}{n}}}\\\\
RHS=\sqrt[(n-1)]{x^{\frac{n-1}{n}}}\\\\
RHS=x^{\frac{n-1}{n}}\right)^{\frac{1}{n-1}}\\\\
RHS=x^{(1/n)}\\\\
RHS=\sqrt[n]{x}\\\\
RHS=LHS$$
    

Melody  Apr 28, 2015
 #5
avatar+78741 
0

Nice proof,  Melody......!!!!

 

  

CPhill  Apr 28, 2015
 #6
avatar+91045 
0

thanks Chris :)

Melody  Apr 28, 2015

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