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sqrt(x)*sqrt(x+1) equals ?

 Jul 8, 2015

Best Answer 

 #1
avatar+423 
+5

$${\sqrt{{\mathtt{x}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}$$

Let's combine the surds. To do this, square one of them (so put it in terms of it's square root)

$${\sqrt{{{\sqrt{{\mathtt{x}}}}}^{\,{\mathtt{2}}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

Now simplify since obviously the square of a square root is...

$${\sqrt{{\mathtt{x}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

Expand out the equation inside the surd and you get:

$${\sqrt{{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}}}$$

.
 Jul 9, 2015
 #1
avatar+423 
+5
Best Answer

$${\sqrt{{\mathtt{x}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}$$

Let's combine the surds. To do this, square one of them (so put it in terms of it's square root)

$${\sqrt{{{\sqrt{{\mathtt{x}}}}}^{\,{\mathtt{2}}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

Now simplify since obviously the square of a square root is...

$${\sqrt{{\mathtt{x}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

Expand out the equation inside the surd and you get:

$${\sqrt{{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}}}$$

Sir-Emo-Chappington Jul 9, 2015

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