Question 1073256
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<pre>
This assignment is for those who firmly knows the Vieta's theorem and Vieta's formulas.

    For the second degree polynomial (quadratic)  P(x)= {{{ax^2 +bx + c}}}  roots {{{x[1]}}} and {{{x[2]}}}  of the equation  P(x)=0 satisfy

    {{{x[1]}}} + {{{x[2]}}} = {{{-b/a}}}   and   {{{x[1]*x[2]}}} = {{{c/a}}}.     (1)     (see Wikipedia, <A HREF=https://en.wikipedia.org/wiki/Vieta's_formulas>this article</A>).  


Concretely, for the given equation  {{{2x^2-x-4}}} = 0  its roots p and q satisfy 
     p + q = {{{1/2}}} and p*q = -2.     (2)

We can state it based in Vieta's theorem even without making explicit calculations of the roots.


Now, they want we construct the quadratic equation/polynomial with the roots {{{p - q/p}}} and {{{q - p/q}}}.


1.  Then the coefficient at x of this polynomial, based on the Vieta's theorem,  must be the opposite number to

        {{{p-q/p}}} + {{{q - p/q}}} = {{{p+q}}} - {{{q/p+p/q}}} = {{{1/2}}} - {{{(q^2 + p^2)/(pq)}}}.       (3)

        From (2), you have {{{q^2 + p^2}}} = {{{(q+p)^2 - 2qp}}} = {{{(1/2)^2 -2*(-2)}}} = {{{4}}}{{{1/4}}} = {{{17/4}}}.

        Therefore, we can continue and complete (3) in this way

        {{{p-q/p}}} + {{{q - p/q}}} = {{{1/2}}} - {{{(q^2 + p^2)/(pq)}}} = {{{1/2 - (17/(4*(-2)))}}} = {{{1/2 + 17/8}}} = {{{4/8 + 17/8}}} = {{{21/8}}}.

        So, the opposite number to it, {{{-21/8}}} is the coefficient at x in the polynomial under the question.


2.  Next step we should calculate the product {{{(p-q/p)}}}.{{{(q - p/q)}}} to determine the constant term of our polynomial.  The product is 

        {{{(p-q/p)}}}.{{{(q - p/q)}}} = {{{pq}}} - {{{q^2/p}}} - {{{p^2/q}}} + {{{1}}} = {{{(pq+1)}}} - {{{(p^3 + q^3)/pq}}}.

        Now, replace pq by -2 and replace {{{p^3 + q^3}}}  by  {{{(p+q)*(p^2 - pq + q^2)}}} = {{{(1/2)*((p+q)^2-3pq)}}} = {{{(1/2)*((1/2)^2-3*(-2))}}} = {{{(1/2)*(1/4+6)}}} = {{{(1/2)*(25/4)}}} = {{{25/8}}}.

        You will get  {{{(p-q/p)}}}.{{{(q - p/q)}}} = ((-2) + 1) - {{{((25/8))/(-2)}}} = {{{-1 + 25/16}}} = {{{9/16}}}.


Thus your equation under the question is  {{{x^2 - (21/8)x + 9/16}}} = 0.

Or, if you want to have it with integer coefficients, multiply everything by 16, and you will get

        {{{16x^2 - 42x + 9}}} = 0.
</pre>

Solved.


Using this approach, you have a priviledge of making calculations with rational numbers.


By doing it using another approach with explicit irrationalities, you will be forced to work with radicals.


It is the advantage of using Vieta's theorem.



--------------------
To extend your horizon:


<pre>
     The way I showed here, is <U>THE ONLY WAY</U> to solve the problems like this one.

     The problem is especially designed and intended to teach you this method.
     It is not my fantasy. It is <U>the way you should know</U> and <U>the approach you must follow</U> when solving such problems.

     It may happen you will have a desire to find the roots of the original equations and then manipulate with them.

     It is <U>NOT THE WAY</U> they want you follow in your solution.
     Saying "they", I mean those who assigned you this problem.

     They <U>DEFINITELY</U> want to know, whether you know the method I showed you, and
                                   whether you are able to use it.
</pre>


To see similar solved problems, look into 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=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>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-functions-of-roots-of-a-cubic-and-quartic-equation.lesson>HOW TO evaluate functions of roots of a cubic and quartic 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>".



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Good luck and &nbsp;&nbsp;&nbsp;&nbsp;H&nbsp;A&nbsp;P&nbsp;P&nbsp;Y &nbsp;&nbsp;&nbsp;&nbsp;L&nbsp;E&nbsp;A&nbsp;R&nbsp;N&nbsp;I&nbsp;N&nbsp;G &nbsp;!&nbsp;!