Question 1199157
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Show that the real solutions of the equation ax^2 + bx + c = 0 are the reciprocation 
the real solutions of the equation cx^2 + bx + a = 0. 
Assume that b^2 - 4ac is greater than or equal to 0.
Can someone get me started here?
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            As given,  the problem's formulation is not  100%  accurate.
            To be accurate,  it must assume that the coefficient  "c"  is not zero,
            because otherwise an equation  cx^2 + bx + a = 0  IS  NOT a  QUADRATIC.


            This assumption is equivalent to say that no one of the routes of 
            the original equation is  0  (zero).


            It is a  NECESSARY  condition to that the reciprocal to the roots do exist.


            So,  in my solution below  I  will assume that  a=/=0,  c=/=0.



<pre>
Let x be a root of the equation ax^2 + bx + c = 0.
Since c=/= 0, it implies that x=/=0, so the reciprocal  {{{1/x}}}  does exist.


Divide equation  {{{ax^2 + bx + c}}} = 0 by  {{{x^2}}}.  You will get

    {{{a + b*(1/x) + c*(1/x)^2}}} = 0.


It means that  {{{1/x}}}  is the root of the equation  {{{a + bx + cx^2}}} = 0.
</pre>

Q.E.D.


Solved.



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By the way, the statement of the problem is true for complex roots, too,

as it is true for real roots.



The same formal proof works for the complex root.



So, the requirement of the problem to the roots to be real is not necessary: it is EXCESSIVE.