amparvardi wrote: Determine whether there exist distinct real numbers $a, b, c, t$ for which: (i) the equation $ax^2 + btx + c = 0$ has two distinct real roots $x_1, x_2,$ (ii) the equation $bx^2 + ctx + a = 0$ has two distinct real roots $x_2, x_3,$ (iii) the equation $cx^2 + atx + b = 0$ has two distinct real roots $x_3, x_1.$ So $x_1x_2=\frac{c}{a}$, $x_2x_3=\frac{a}{b}$ and $x_3x_1=\frac{b}{c}$. So $(x_1x_2x_3)^2=1$ so $x_1x_2x_3=\pm 1$ Suppose that $x_1x_2x_3=-1$. Then $x_1=-\frac{b}{a}$, $x_2=-\frac{c}{b}$ and $x_3=-\frac{a}{c}$ Then $bt=a\left(\frac{b}{a}+\frac{c}{b}\right)$ so $t-1=\frac{ca}{b^2}$ Similarly, $t-1=\frac{ab}{c^2}=\frac{bc}{a^2}$ So $(t-1)^3=\cdot\frac{ab}{c^2}\cdot\frac{bc}{a^2}=1$ So $t=2$ So $abc=a^3=b^3=c^3$ so $a=b=c$ so $x_1=x_2=x_3$ Otherwise $x_1x_2x_3=1$. Then $x_1=\frac{b}{a}$, $x_2=\frac{c}{b}$ and $x_3=\frac{a}{c}$ Then $-bt=a\left(\frac{b}{a}+\frac{c}{b}\right)$ so $-t-1=\frac{ca}{b^2}$ Similarly, $-t-1=\frac{ab}{c^2}=\frac{bc}{a^2}$ So $(-t-1)^3=\cdot\frac{ab}{c^2}\cdot\frac{bc}{a^2}=1$ So $t=-2$ So $abc=a^3=b^3=c^3$ so $a=b=c$ so $x_1=x_2=x_3$ So $t=2$ So $abc=a^3=b^3=c^3$ so $a=b=c$ so $x_1=x_2=x_3$ Hence the answer is no.