"

13 Radical Expressions

The nth radical is defined by :    \sqrt[n]{x}=x^{1/n}


\displaystyle{ \sqrt[n]{ a b}=\sqrt[n]{a} \sqrt[n]{b} \qquad \qquad \sqrt[n]{\sqrt[m]{x}}=\sqrt[mn]{x} \qquad \qquad \sqrt[n]{ \frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} }

A conjugate expression for  a + b  is the expression a- b.
For rationalizing an expression, we multiply  it by its  conjugate.  

 

Watch a Video


View on YouTube

 

Show Examples

 

Exercise 1

 

Show/Hide Solution.

We have

\displaystyle{\frac{ x- {16}}{ \sqrt{x} - {4} } = \frac{( x- {16})(\sqrt{x} + {4} ) }{ ( \sqrt{x} - {4} ) (\sqrt{x} + {4} ) } = \frac{( x- {16})(\sqrt{x} + {4} ) }{ x- {16} } =\sqrt{x} + {4} }.

 

 

Exercise 2

 

Show/Hide Solution.

We have

\displaystyle{ \sqrt {x^2+ {17}} - x = \frac{ (\sqrt {x^2 + {17}} - x) ( \sqrt {x^2 + {17}} + x) }{ \sqrt {x^2 + {17}} + x } =\frac{(\sqrt {x^2 + {17}})^2 - x^2}{ \sqrt {x^2 + {17}} + x }= \frac{{17}}{ \sqrt {x^2 + {17}} + x }}.

 

Exercise 3

 

Show/Hide Solution.

We have

\displaystyle{ \frac{ x^2 - {16} y^2 }{ \sqrt{x}- {2}\sqrt{y} } = \frac{( x- \sqrt{16} y ) ( x +\sqrt{16} y ) }{ \sqrt{x}- {2}\sqrt{{y}} } = ( \sqrt{x}+2\sqrt{{y}} ) ( x + 4 y ) }.

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Guide to Precalculus Review Copyright © 2025 by Samia CHALLAL is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Share This Book