Question 1117462
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Part of my solution to the problem is basically the same as the other tutor's; the other part is far different.<br>
As he says, put the equation in pure quadratic form:<br>
{{{(abc^2)x^2+(3a^2c+b^2c)x+(-6a^2-ab+2b^2) = 0}}}<br>
In a quadratic equation<br>
{{{px^2+qx+r=0}}}<br>
the sum of the roots is {{{-q/p}}} and the product of the roots is {{{r/p}}}.<br>
In this equation, the product of the roots is<br>
{{{-6a^2-ab+2b^2)/(abc^2)}}}<br>
or<br>
{{{(-(3a+2b)(2a-b))/(abc^2)}}}
With abc^2 as the denominator in the product of the two roots, the denominators of the two roots are most likely ac and bc.  There are then four possible pairs of roots:<br>
(1) {{{-(3a+2b)/(ac)}}} and {{{(2a-b)/(bc)}}}
(2) {{{-(3a+2b)/(bc)}}} and {{{(2a-b)/(ac)}}}
(3) {{{(3a+2b)/(ac)}}} and {{{-(2a-b)/(bc)}}}
(4) {{{(3a+2b)/(bc)}}} and {{{-(2a-b)/(ac)}}}<br>
To find which is the correct pair of roots, we use the fact that the sum of the roots has to be<br>
{{{(-3a^2c-b^2c)/(abc^2) = (-3a^2-b^2)/(abc)}}}<br>
The second pair of roots gives us this sum:<br>
{{{-(3a+2b)/(bc)+(2a-b)/(ac)}}}
{{{(-3a^2-2ab+2ab-b^2)/(abc)}}}
{{{(-3a^2-b^2)/(abc)}}}<br>
So the two roots are<br>
{{{-(3a+2b)/(bc)}}} and {{{(2a-b)/(ac)}}}<br>
And as long as none of a, b, or c is 0, those roots are rational.