If the equation z4+a1z3+a2z2+a3z+a4=0, where a1, a2, a3, a4 are real coefficients different from zero, has a purely imaginary root, then the expression a3a1a2+a1a4a2a3 has the value equal to

Let xi be the root where x ≠ 0 and x ∈ Rx4−a1x3i−a2x2+a3xi+a4=0⇒ x4−a2x2+a4=0         (1)and a1x3−a3x=0                 (2)From Eq. (2),a1x2−a3=0⇒ x2=a3a1 as x≠0)Putting the value of x2 in Eq. (1), we geta32a12−a2a3a1+a4=0⇒ a32+a4a12=a1a2a3⇒ a3a1a2+a1a4a2a3=1 dividing by a1a2a3