Roots of a Quartic (Vieta's Formulas) 

Question: The quartic polynomial $x^4 −8x^3 + 19x^2 +kx+ 2$ has four distinct real roots denoted
    $a, b, c,d$ in order from smallest to largest. If $a + d = b + c$ then
(a) Show that $a + d = b + c = 4$.
(b) Show that $abcd = 2$ and $ad + bc = 3$.
(c) Find $ad$ and $bc.$
(d) Find $a, b, c, d$ and $k$.



My attempt:
$$ x^4 −8x^3 + 19x^2 +kx+ 2 $$
With Vieta's formulas;
$$ a+b+c+d = 8 $$
$$ab+ac+ad+bc+bd+cd = 19$$
$$abc + abd + acd + bcd = -k$$
$$ abcd = 2 $$ 


(a) Show that $a + d = b + c = 4$

As $$ a+b+c+d = 8 $$
but $b+c = a+d $
$$ 2a+2d = 8 $$
$$ a+d = 4 $$
Hence $$ a+d = b+c = 4 $$


(b) Show that $abcd = 2$ and $ad + bc = 3$

As $$ abcd = 2 $$ 
and 
$$ab+ac+ad+bc+bd+cd = 19$$
$$ ad+bc + a(b+c) + d(b+c) = 19 $$
$$ ad+bc + (a+d)(b+c) = 19 $$
$$ ad+bc + (4)(4) = 19 $$
$$ ad+bc + 16 = 19 $$
$$  ad +bc = 3 $$


(c) Find $ad$ and $bc.$

Given $ad +bc = 3$ and $ abcd = 2 $
Hence $$ bc = \frac {2}{ad}$$
$$ad +bc = 3$$
$$ ad + \frac {2}{ad} = 3 $$
Let $ad = z $
$$ z^2 - 3z + 2 = 0 $$
$$ ad = 2 $$
$$ bc = 1 $$


(d) Find $a, b, c, d$ and $k$.

Given 
$$abc + abd + acd + bcd = -k$$
$$ ad(b+c)+ bc(a+d) = - k $$
But $b+c=a+d=4$
$$ 4ad+ 4bc = - k $$
$$ 4(ad+bc) = -k $$
$$ 4(3) = -k $$
$$ k = -12 $$
Now this is the part which I am stuck on.. How do I find $a,b,c,d$?

Hint If you know that $ad = 2$ and $a + d = 4$, then $$(x - a)(x - d) = x^2 - (a + d) x + ad = x^2 - 4 x + 2 ,$$ so finding $a, d$ is just finding the roots of that quadratic. Of course, finding $b, c$ is analogous.
