Question 1105837
.
<U>Solution 1</U>


<pre>
If  {{{x}}} + {{{1/x)}}} = -3,  then (square both sides) 

{{{(x+1/x)^2}}} = 9  ====>

{{{x^2}}} + {{{2}}} + {{{1/x^2}}} = 9  ====>

{{{x^2}}} + {{{1/x^2}}} = 9-2 = 7  ====>

{{{x^2}}} - {{{2}}} + {{{1/x^2}}} = 7 - 2 = 5  ====>

{{{(x-1/x)^2}}} = 5  ====>  {{{x}}} - {{{1/x}}} = +/- {{{sqrt(5)}}}.


And both values  {{{sqrt(5)}}} and {{{-sqrt(5)}}} are achievable.
</pre>


<U>Solution 2</U>


<pre>
If  {{{x}}} + {{{1/x)}}} = -3,  then 

{{{x^2 + 3x + 1}}} = 0  ====>

{{{x[1,2]}}} = {{{(-3 +- sqrt(9-4))/2}}} = {{{(-3 +- sqrt(5))/2}}}.



1.  If  x = {{{(-3 + sqrt(5))/2}}},  then

    {{{1/x}}} = {{{1/(-3 + sqrt(5))/2}}} = {{{2/(-3 + sqrt(5))}}} = {{{2*(-3-sqrt(5))/(9-5)}}} = {{{(-3+sqrt(5))/2}}}.


    Then  {{{x}}} - {{{1/x}}} = {{{(-3 + sqrt(5))/2}}} - {{{(-3 - sqrt(5))/2}}} = {{{sqrt(5)}}}.



2.  If  x = {{{(-3 - sqrt(5))/2}}},  then similarly

    {{{1/x}}} = {{{1/(-3 - sqrt(5))/2}}} = {{{2/(-3 - sqrt(5))}}} = {{{2*(-3+sqrt(5))/(9-5)}}} = {{{(-3+sqrt(5))/2}}}.


    Then  {{{x}}} - {{{1/x}}} = {{{(-3 - sqrt(5))/2}}} - {{{(-3 + sqrt(5))/2}}} = {{{-sqrt(5)}}}.
</pre>

Solved (two times with the identical answers !).
</pre>


The answer by @josgarithmetic is &nbsp;&nbsp;<U>I N C O R R E C T</U>.


-----------------
For many similar solved problems see the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-expressions-involving-x%2Binv%28x%29-x2%2Binv%28x2%29-and-x%5E3%2Binv%28x%5E3%29.lesson>HOW TO evaluate expressions involving &nbsp;{{{(x + 1/x)}}}, &nbsp;{{{(x^2+1/x^2)}}}, &nbsp;{{{(x^3 + 1/x^3)}}}  &nbsp;and &nbsp;{{{(x^5+1/x^5)}}}</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Advanced-lesson-on-evaluating-expressions.lesson>Advanced lesson on evaluating expressions</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-functions-of-roots-of-a-square-equation.lesson>HOW TO evaluate functions of roots of a quadratic equation</A>

in this site.



Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic "<U>Evaluation, substitution</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.