Indian Statistical Institute B.Math & B.Stat : Polynomials

If the roots of the equation $x^4 + ax^3 + bx^2 + cx + d = 0$ are in geometric progression then show that $c^2 = a^2d$. Let the roots of the equation be $x_1,x_2,x_3,x_4$. Since the roots are in geometric progression we have $x_1x_4 = x_2x_3$. Also using Vieta's Formulas ( relation between roots and coefficients ) we have $$x_1+x_2+x_3+x_4 = - a$$ $$(x_1+x_4)(x_2+x_3)+x_1x_4+x_2x_3 = b$$ $$x_1x_4(x_2+x_3)+x_2x_3(x_1+x_4) = -c$$ $$x_1x_2x_3x_4 = d$$ Since $x_1x_4 = x_2x_3$ and $x_1x_4(x_2+x_3)+x_2x_3(x_1+x_4) = -c$ we have $x_1x_4(x_2+x_3+x_1+x_4) = -c$. Now using $x_1+x_2+x_3+x_4 = - a$ we have $x_1x_4 \times -a = -c \implies x_1x_4 = \frac{c}{a}$. Thus $x_1x_4 = x_2x_3 = \frac{c}{a}$. Again since $x_1x_2x_3x_4 = d \implies \frac{c}{a} \times \frac{c}{a} = d \implies c^2 = da^2$