2021/077) Let $a>b>c$ be the real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$. Find $b(a+c)$.

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 positive 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)