+0  
 
0
770
2
avatar

x(x+2)=55

what is x?

 May 7, 2014

Best Answer 

 #2
avatar+130511 
+5

Simplifying, we have  x^2 +2x - 55 = 0

I can tell this one is going to be "nasty."   Let's use the on-site calculator to solve.

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{55}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{14}}}}{\mathtt{\,-\,}}{\mathtt{1}}\\
{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{14}}}}{\mathtt{\,-\,}}{\mathtt{1}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{8.483\: \!314\: \!773\: \!547\: \!882\: \!8}}\\
{\mathtt{x}} = {\mathtt{6.483\: \!314\: \!773\: \!547\: \!882\: \!8}}\\
\end{array} \right\}$$

Anonymous was not technically incorrect. One answer was, indeed, around 6.5.

Since it's a quadratic, we'll also (usually) have a second different solution, as well.

 May 7, 2014
 #1
avatar
0

it will be around 6.4 to 6.5

 May 7, 2014
 #2
avatar+130511 
+5
Best Answer

Simplifying, we have  x^2 +2x - 55 = 0

I can tell this one is going to be "nasty."   Let's use the on-site calculator to solve.

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{55}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{14}}}}{\mathtt{\,-\,}}{\mathtt{1}}\\
{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{14}}}}{\mathtt{\,-\,}}{\mathtt{1}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{8.483\: \!314\: \!773\: \!547\: \!882\: \!8}}\\
{\mathtt{x}} = {\mathtt{6.483\: \!314\: \!773\: \!547\: \!882\: \!8}}\\
\end{array} \right\}$$

Anonymous was not technically incorrect. One answer was, indeed, around 6.5.

Since it's a quadratic, we'll also (usually) have a second different solution, as well.

CPhill May 7, 2014

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