+0  
 
+8
212
6
avatar+1794 

$${\mathtt{15}} = {\frac{{\mathtt{75}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}}}}{\left({\frac{{\mathtt{60}}}{{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{10}}\right)}^{{\mathtt{2}}}}}\right)}}$$

gibsonj338  May 14, 2015

Best Answer 

 #5
avatar+26329 
+15

It's always worth drawing a graph for these sort of problems:

 

 graph

.

Alan  May 14, 2015
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6+0 Answers

 #1
avatar+91051 
+13

Hi Gibson,     

 

$$\\15=\frac{75\sqrt{x-3}}{\left(\frac{60}{(x-10)^2\right)}}\\\\
\frac{15}{75}=\frac{\sqrt{x-3}}{\left(\frac{60}{(x-10)^2\right)}}\\\\
\frac{3}{15}=(\sqrt{x-3})\div{\left(\frac{60}{(x-10)^2\right)}}\\\\
\frac{3}{15}=(\sqrt{x-3})\times{\frac{(x-10)^2}{60}}\\\\
\frac{3*60}{15}=(\sqrt{x-3})\times(x-10)^2}\\\\
12=(\sqrt{x-3})\times(x-10)^2}\\\\
144=(x-3)\times(x-10)^4}\\\\
144=(x-3)\times(x-10)^4}\\\\
144=(x-3)(x^4-4*10x^3+6*100x^2-4*1000x+10000)\\\\
$I'd keep going but i think I will just ask the calc$$$

 EDITED: Thanks Fiora for pointing out the error.

I actually put it into the web2 'correctly' but I did not put a * between the brackets and the calc misinterpreted what I was saying.  I now find that the web2 calc will not solve this which is interesting.

I guess WolframAlpha is always the fall back calc :)

 

$${\mathtt{144}} = \left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{4}}\right)}^{{\mathtt{4}}} \Rightarrow {\mathtt{0}} = {{\mathtt{x}}}^{{\mathtt{5}}}{\mathtt{\,-\,}}{\mathtt{19}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{144}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{544}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,024}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{912}} \Rightarrow {\mathtt{0}} = {{\mathtt{x}}}^{{\mathtt{5}}}{\mathtt{\,-\,}}{\mathtt{19}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{144}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{544}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,024}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{912}}$$

 

x must be greater than 3

 

You can check it with Wolfram alpha anyway :)

Melody  May 14, 2015
 #2
avatar+519 
+8

Hi,Melody.

I think you typed your last equation wrong.It,s 144=(x-3)*(x-4)^4

not 144=[(x-3)(x-4)]^4...so....

fiora  May 14, 2015
 #3
avatar+78755 
+8

Yeah.....fiora.....Melody just made a slight mistype.....WolframAlpha only shows one real solution....

 

 

  

CPhill  May 14, 2015
 #4
avatar+91051 
0

Thaks you Chris and Fiora :))

Melody  May 14, 2015
 #5
avatar+26329 
+15
Best Answer

It's always worth drawing a graph for these sort of problems:

 

 graph

.

Alan  May 14, 2015
 #6
avatar+91051 
0

Yes Alan is right.  Drawing graphs and understanding their relevance is very useful !

Thanks Alan.  

Melody  May 14, 2015

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