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  4. Let r, s, t, u be the roots of the equation x4 + Ax3 + Bx2 + Cx + D = 0. If rs = tu, then

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Q. Let $r, s, t, u$ be the roots of the equation $x^4 + Ax^3 + Bx^2 + Cx + D = 0$. If $rs = tu$, then

Complex Numbers and Quadratic Equations

Solution:

Let $rs = tu = K $
Let $r + s = P$ ; $t + u = Q$
$\therefore \, x^4 + Ax^3 + Bx^2 + Cx + D$
= $(x^2 - Px + K) (x^2 - Qx + K) $
$ \therefore A = - (P + Q)$ ; $B = PQ + 2K$
$C = -(P + Q)K$ ; $D = K^2$
$ \therefore \, \frac{C}{A} = K$
$\Rightarrow \, \frac{C^2}{A^2} = K^2 = D$
$ \therefore \, A^2D = C^2$.