Finding $b(a+c)$ for Real Roots of $\sqrt{2014}x^3-4029x^2+2=0$

[anemone]

Let $a>b>c$ be the real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$. Find $b(a+c)$.

 

[kaliprasad]

Spoiler: my Solution

To avoid radicals let $\sqrt{2014}=p$
So we get $px^3-(2p^2+1)x^2 +2 = 0$
Or factoring we get $(px-1)(x^2-2px-2)$ = 0
So one root is $x= \frac{1}{p}$ and other two roots are roots of $x^2-2px-2=0$
For the equation $x^2-2px-2=0$ sum of the roots is 2p and product is -2. so one root has to be -ve and
the postiive root shall be above 2p
So $b=\frac{1}{p}\cdots(1)$
And c is the -ve root and $a> 2p$
a,c are roots of $x^2-2px-2=0$ so $a+c = 2p\cdots(2)$
Hence $b(a+c) = 2$ using (1) and (2)