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Is there a way to define a variable to use an equation? E.x.: 3x^2-2x+5=y, and I want to assign x a value without changing it out. Is it possible?

CatBoy13  Dec 12, 2014

Best Answer 

 #3
avatar+92256 
+10

Welcome to web2.0calc forum Catboy13  

Umm - Interesting question.  

 

$$\begin{array}{rll}
3x^2-2x+5&=&y\\\\
3x^2-2x&=&y-5\\\\
x^2-\frac{2}{3}x&=&\frac{y-5}{3}\\\\
x^2-\frac{2}{3}x+\left(\frac{2}{6}\right)^2&=&\frac{y-5}{3}+\left(\frac{2}{6}\right)^2\\\\
x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{y-5}{3}+\left(\frac{1}{3}\right)^2\\\\
x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{3(y-5)}{9}+\frac{1}{9}\\\\
\left(x-\frac{1}{3}\right)^2&=&\frac{3y-15+1}{9}\\\\
\left(x-\frac{1}{3}\right)^2&=&\frac{3y-14}{9}\\\\
x-\frac{1}{3}&=&\pm\sqrt{\frac{3y-14}{9}}\\\\
x&=&\frac{1}{3}\pm\sqrt{\frac{3y-14}{9}}\\\\
x&=&\frac{1\pm\sqrt{3y-14}}{3}}\\














\end{array}$$

Melody  Dec 13, 2014
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4+0 Answers

 #1
avatar+7188 
0

I believe it is possible....but I might be wrong...

happy7  Dec 12, 2014
 #2
avatar
0

If you first define a function f as

$${f}{\left({\mathtt{x}}\right)} = {\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}$$

 

then for any particular x the function f(x) has a value and you can write things such as

f(2) + f(0) - 4 = 14

Guest Dec 12, 2014
 #3
avatar+92256 
+10
Best Answer

Welcome to web2.0calc forum Catboy13  

Umm - Interesting question.  

 

$$\begin{array}{rll}
3x^2-2x+5&=&y\\\\
3x^2-2x&=&y-5\\\\
x^2-\frac{2}{3}x&=&\frac{y-5}{3}\\\\
x^2-\frac{2}{3}x+\left(\frac{2}{6}\right)^2&=&\frac{y-5}{3}+\left(\frac{2}{6}\right)^2\\\\
x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{y-5}{3}+\left(\frac{1}{3}\right)^2\\\\
x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{3(y-5)}{9}+\frac{1}{9}\\\\
\left(x-\frac{1}{3}\right)^2&=&\frac{3y-15+1}{9}\\\\
\left(x-\frac{1}{3}\right)^2&=&\frac{3y-14}{9}\\\\
x-\frac{1}{3}&=&\pm\sqrt{\frac{3y-14}{9}}\\\\
x&=&\frac{1}{3}\pm\sqrt{\frac{3y-14}{9}}\\\\
x&=&\frac{1\pm\sqrt{3y-14}}{3}}\\














\end{array}$$

Melody  Dec 13, 2014
 #4
avatar+92256 
0

I have added this to the Sticky Topic - "Great Answers to Learn From"

Melody  Dec 14, 2014

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