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.