Question 1104382
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I will solve (iii) only.


<pre>
{{{m - 1/m}}} = 5  ====>

m^2 - 5m - 1 = 0  ====>  (apply the quadratic formula)

{{{m[1,2]}}} = {{{(5 +- sqrt(5^2 - 4*(-1)))/2}}} = {{{(5 +- sqrt(29))/2}}}.


In other words,   {{{m[1]}}} = {{{(5 + sqrt(29))/2}}},  {{{m[2]}}} = {{{(5 - sqrt(29))/2}}}.


Notice that   {{{m[1]*m[2]}}} = -1.


You can check it by making direct calculations

   {{{m[1]*m[2]}}} = {{{((5 + sqrt(29))/2)*((5 - sqrt(29))/2)}}} = {{{(25-29)/(2*2)}}} = {{{(-4)/4}}} = -1,

or derive it from the Vieta's theorem.


Thus  {{{m[1]^(-1)}}} = {{{1/m[1]}}} = {{{-m[2]}}}   and  {{{m[2]^(-1)}}} = {{{1/m[2]}}} = {{{-m[1]}}}.


It implies   {{{m[1]^(-2)}}} = {{{m[2]^2}}} = {{{((5 - sqrt(29))/2)^2}}} = {{{(25-10*sqrt(29) + 29)/4}}} = {{{(54-10*sqrt(29))/4}}} = {{{(27-5*sqrt(29))/2}}},   and

             {{{m[2]^(-2)}}} = {{{m[1]^2}}} = {{{((5 + sqrt(29))/2)^2}}} = {{{(25+10*sqrt(29) + 29)/4}}} = {{{(54+10*sqrt(29))/4}}} = {{{(27+5*sqrt(29))/2}}}.


Thus you have

            {{{m[1]^2}}} = {{{(27+5*sqrt(29))/2}}},  {{{1/m[1]^2}}} = {{{(27-5*sqrt(29))/2}}},  which implies  {{{m[1]^2}}} - {{{1/m[1]^2}}} = {{{5*sqrt(29)}}},   and


            {{{m[2]^2}}} = {{{(27-5*sqrt(29))/2}}},  {{{1/m[2]^2}}} = {{{(27+5*sqrt(29))/2}}},  which implies  {{{m[2]^2}}} - {{{1/m[2]^2}}} = {{{-5*sqrt(29)}}},
</pre>

Answer</U>.  &nbsp;&nbsp;If  &nbsp;{{{m}}} - {{{1/m}}} = 5,  &nbsp;then this equation has &nbsp;<U>TWO solutions</U>; &nbsp;correspondingly, &nbsp;"m"&nbsp;  has &nbsp;<U>TWO values</U>; 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;correspondingly,  &nbsp;{{{m^2}}} &nbsp;has &nbsp;<U>TWO values</U>,  &nbsp;&nbsp;and, &nbsp;correspondingly, &nbsp;{{{m^2}}} - {{{1/m^2}}} &nbsp;has two values &nbsp;&nbsp;+/- {{{5*sqrt(29)}}}.


Solved.