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# Assigning var

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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

#3
+92256
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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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#1
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I believe it is possible....but I might be wrong...

happy7  Dec 12, 2014
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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
+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
#4
+92256
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I have added this to the Sticky Topic - "Great Answers to Learn From"

Melody  Dec 14, 2014

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