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What are the two solutions to (4x-1)^2=11

 Jul 16, 2015

Best Answer 

 #1
avatar+128406 
+5

(4x -1)^2 = 11    expand the left side....

 

16x^2 - 8x + 1   = 11      subtract 11 from both sides

 

16x^2 - 8x - 10   = 0      divide through by 2

 

8x^2 - 4x - 5   =  0      this does not factor...using the on-site solver, we get.....

 

$${\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{11}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.579\: \!156\: \!197\: \!588\: \!85}}\\
{\mathtt{x}} = {\mathtt{1.079\: \!156\: \!197\: \!588\: \!85}}\\
\end{array} \right\}$$

 

Note......we could have also used the square root property to solve this, if we wished.....!!!!!

 

 

  

 Jul 16, 2015
 #1
avatar+128406 
+5
Best Answer

(4x -1)^2 = 11    expand the left side....

 

16x^2 - 8x + 1   = 11      subtract 11 from both sides

 

16x^2 - 8x - 10   = 0      divide through by 2

 

8x^2 - 4x - 5   =  0      this does not factor...using the on-site solver, we get.....

 

$${\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{11}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.579\: \!156\: \!197\: \!588\: \!85}}\\
{\mathtt{x}} = {\mathtt{1.079\: \!156\: \!197\: \!588\: \!85}}\\
\end{array} \right\}$$

 

Note......we could have also used the square root property to solve this, if we wished.....!!!!!

 

 

  

CPhill Jul 16, 2015

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