JEE Biquadratic

Let the roots be:

$z_1 = x_1+y_1i\quad \quad z_2 = x_2+y_2i\quad \quad z_3 = x_3+y_3i \quad \quad z_4 = x_4+y_4i$

$z_1+z_2 = 3 + 4i \quad \quad z_3z_4 = 13 + i$

Using Vieta Formulas, for we have:

$z_1+z_2+z_3+z_4 = -a +0i \quad \Rightarrow y_3+y_4 = -(y_1+y_2) = -4$

$z_1z_2z_3z_4 = d + 0i \quad \Rightarrow z_1z_2 = \overline {z_3z_4} = 13-i$

$z_1z_2z_3 + z_1z_2z_4 + z_1z_3z_4 + z_2z_3z_4 = - c + 0i$ $\quad \Rightarrow z_1z_2(z_3+z_4) + z_3z_4(z_1+z_2) = - c +0i$ $\quad \Rightarrow (13-i)[(x_3+x_4)+(y_3+y_4)i] + (13+i)(3+4i) = - c +0i$ $\quad \Rightarrow -(x_3+x_4)+13(-4) + 52+3 = 0\quad \Rightarrow x_3+x_4 = 3$

$\Rightarrow z_3+z_4 = (x_3+x_4)+(y_3+y_4)i = 3 -4i$

Now,

$b = z_1z_2 + z_1z_3 +z_1z_4 +z_2z_3 +z_2z_4 +z_3z_4$ $\quad = z_1z_2 + (z_1+z_2)(z_3 +z_4) +z_3z_4$ $\quad = 13-i + (3+4i)(3-4i) +13+i = 26 + 9 + 16 = \boxed{51}$

We have the result that non real roots of a polynomial with purely real coefficients always occur as conjugate pairs (see: Complex conjugate root theorem).

Thus, 3-4i is the sum of the other two roots. Similarly, 13-i is the product of the other two roots.

If e,f,g,h are the roots of the given bi-quadratic equation, then b is the sum of the roots taken 2 at a time.

Grouping terms, we get:

b= (e+f) * (g+h) + e * f+ g * h

Where e*f=13+i and g+h=3+4i (given conditions)

All terms are related as conjugates(from the aforementioned result) and the answer follows immediately.

Let A,B,C,D be the complex roots of our bi-quad,(A,B & C,D are pair-wise conjugate) Then we know the following:-

A+D=3+4i

B+C=3-4i

BC=13+i

AD=13-i

Using vieta's ,b=

(AB+BC+CD+DA+BD+AC)

b=(B+C)(A+D)+BC+AD

b=(9+16)+13+i+13-i

b=25+ 26

b=51

HINT1:

Write the roots as :

root1

root2

conjugate of root1

conjugate of root2

HINT 2:

Write the polynomial as the product of two Quadratic equations.