Let $f = 2x^4 + 2(a - 1)x^3 + (a^2 + 3)x^2 + bx + c.$ ,Find out $a, b, c ∈ R$ and its roots knowing that all roots are real. 

Let $f = 2x^4 + 2(a - 1)x^3 + (a^2 + 3)x^2 + bx + c.$
Find out $a, b, c ∈ R$  and its roots knowing that all roots are real.



The first thing that came into my mind was to use vieta's formulas so let roots be $\alpha , \beta , \gamma , \delta$
$$ \alpha + \beta +\gamma+\delta=-\frac{b}{a}$$
$$ \Rightarrow \alpha + \beta +\gamma+\delta = 1-a$$
$$ \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma+\beta\delta+\gamma\delta=\frac{c}{a}$$
$$ \Rightarrow \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma+\beta\delta+\gamma\delta =\frac{a^2+3}{2}$$
$$ \alpha\beta\gamma + \alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=\frac{-d}{a}$$
$$ \Rightarrow  \alpha\beta\gamma + \alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta= \frac{-b}{2}$$
$$ \alpha\beta\delta\gamma = \frac{e}{a} $$
$$ \Rightarrow \alpha\beta\delta\gamma = \frac{c}{2} $$
But that did not get me anywhere...
Also I took the second derivative and set it equal to zero but that led me to complex roots... any ideas?
 A: Between any two real roots of a polynomial there is a root of its derivative.  Between any two real roots of the derivative there is a root of the second derivative.  So your second derivative should have all real roots.  Take another look at it.
A: As you said, $f''$ has complex roots which shows that double derivative is always positive. So this implies that $f'$ has only one real root because it's always a increasing function. From this you can conclude that $f$ will have either 2 coincident roots ($p,p,q,q$) or all 4 coincident roots ($p,p,p,p$) or no real roots. So we will neglect the third case as we want f to have real roots.
Now
$ Case-1 \ \ $ When all the roots are real and coincident -:
Let that root be $\alpha$
From here you will get 4 equations 
$$4\alpha=-(a-1)$$
$$6\alpha^2=\frac {a^2+3}{2}$$
$$4\alpha^3=-\frac{b}{2}$$
$$\alpha^4=\frac{c}{2}$$
Frome here you can get the value's of $[a,b,c,\alpha]$ very easily. These values are $[-3,-8,2,1]$ respectively.
$ Case-2 \ \ $ When two of the roots are coincident and real -:
Let the roots be $\alpha,\beta$
Now again write 4 equations using these roots
$$2(\alpha+\beta)=-(a-1)$$
$$\alpha^2+\beta^2+4\alpha\beta=\frac {a^2+3}{2}$$
$$2(\alpha^2\beta+\alpha\beta^2)=-\frac{b}{2}$$
$$\alpha^2\beta^2=\frac{c}{2}$$
From first two equations, you can get the vlaue of $\alpha\beta=\frac{a^2+2a+5}{8}$ and from first equation $\alpha+\beta=-\frac{(a-1)}{2}$
So consider a equation $\delta^2+\frac{(a-1)}{2}\delta+
\frac{a^2+2a+5}{8}=0$. Clearly, this equation has roots $\alpha,\beta$
Now Disciminant of this equation is $D=-\frac{(a+3)^2}{4}<0$ for all values of $a$. So $\alpha,\beta$ can not be real for any value of $a$. So we have to reject this case.
So finally $$[a,b,c]=[-3,-8,2]$$ for roots to be real.
Hope this will help !
A: Hint: Since $f(x)$ has $4$ roots, the equation $f''(x)=0$ has at least one solution. Note that 
$$
f''(x)=24x^2+12(a-1)x+2(a^2+3)
$$
The discriminate is 
$$
\Delta=12^2(a-1)^2-4\cdot 24\cdot 2(a^2+3)=-48(a+3)^2
$$
In order to have a root we must have that $a=-3$. 
