#### Question

If the quartic x^4 + 3x^3 + 11x^2 + 9x + A has roots k, l, m, and n such that kl = mn, find A.

Question

If the quartic x^4 + 3x^3 + 11x^2 + 9x + A has roots k, l, m, and n such that kl = mn, find A.

Since k, l, m, and n are the four roots of the equation x^4 + 3x^3 + 11x^2 + 9x + A=0

Applying the Vieta's formula for a quartic equation gives the system of equations below,

\begin{cases} klmn=A \\ k+l+m+n = -3 \\ kml+kmn+mnl+kln = -9 \end{cases}

Since kl=mn, the equations could be simplified by replacing kl with mn

\begin{cases} m^2n^2=A \\ k+l+m+n = -3 \\ m^2n+kmn+mnl+mn^2 = -9 \end{cases}

Collect the like terms for the third equation

mn(k+l+m+n) = -9

Substituting the second equation gives

mn = 3

Then the value of A is obtained.

A = m^2n^2 = 9

Next a short verification is given below

So the quartic equation is x^4 + 3x^3 + 11x^2 + 9x + 9 = 0

Solving the quartic equation gives 4 complex roots.

\begin{cases} x_1=−0.40474211683987−0.97201092566324i \\ x_2=−0.40474211683987+0.97201092566324i \\ x_3=−1.0952578831601−2.6303233208409i \\ x_4=−1.0952578831601+2.6303233208409i \end{cases}

Obviously, sum of the four roots gives -3 and products of four complex number results in 9.