2015/009) find b(a+c) given they are roots of equation $\sqrt{2015}x^3-4031x^2+2=0$ and a > b > c

To avoid surds let $t=\sqrt{2015}$

so we get $tx^3 - (2t^2+1)x^2 + 2 = 0$
or $tx^3 - x^2 - 2t^2x^2+2=0$
or $x^2(xt-1) - 2(x^2t^2-1) = 0$
or $(x^2(xt-1) - 2 (xt-1)(xt+1) = 0$

so xt = 1 or $x^2-2xt -2 = 0$
2nd equation gives

$(x-t)^2 = t^2+ 2 = 2017$

so 3 roots are $x=\dfrac{1}{\sqrt{2015}}$ or $x= \sqrt{2015} \pm \sqrt{2017}$

clearly $ \sqrt{2015} +\sqrt{2017} \gt \dfrac{1}{\sqrt{2015}} \gt \sqrt{2015} - \sqrt{2017}$

so $a= \sqrt{2015} +\sqrt{2017}$
$b = \dfrac{1}{\sqrt{2015}}$
$c = \sqrt{2015} -\sqrt{2017}$

so b(a+c) = -2

Note:

I  have  taken the problem from  site http://mathhelpboards.com/challenge-questions-puzzles-28/find-b-c-14115.html#post66990.  I personally feel that my solution is simpler