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Find different real numbers $a,b,c,t$ for which the following conditions:

1) the equation $ax^2+btx+c=0$ has real roots $x_1,x_2$;

2) the equation $bx^2+ctx+a=0$ has real roots $x_2,x_3$;

3) the equation $cx^2+atx+b=0$ has real roots $x_1,x_3$, where $x_1,x_2,x_3 -$ different real numbers.

My work so far:

I use Vieta's_formulas:

Obviously $abc\not=0$. Then $$(x_1x_2)(x_2x_3)(x_3x_1)=\frac ca \cdot\frac ab \cdot\frac bc=1.$$ Then $$x_1x_2x_3=\pm1$$

I don't know how to solve more

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Roman83
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  • I think you are given $x_1,x_2,x_3$ to find $a,b,c,t$, not the other way around. – kennytm May 19 '16 at 07:06
  • Try to work out the following cases:
    1. $a+b+c=0$
    2. $a=b$ or any two of them is equal
    3. if all are different,

    $$ a^2-a b t^2-a c+b^2-b c+c^2 t^2=0 $$ and all symmetric versions hold, but then $a=b=c=1$.

     – asomog May 19 '16 at 07:13

1 Answers1

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By Vieta's formulas, $$x_1+x_2=-\frac{bt}{a},\quad x_1x_2=\frac ca$$ $$x_2+x_3=-\frac{ct}{b},\quad x_2x_3=\frac ab$$ $$x_3+x_1=-\frac{at}{c},\quad x_3x_1=\frac bc$$

From these, we can have $$-t=(x_1+x_2)x_2x_3=(x_2+x_3)x_3x_1=(x_3+x_1)x_1x_2,$$ i.e. $$-t=x_1x_2x_3+x_2^2x_3=x_1x_2x_3+x_3^2x_1=x_1x_2x_3+x_1^2x_2$$ from which we have $$x_2^2x_3=x_3^2x_1=x_1^2x_2$$ Since $x_1x_2x_3\not=0$, $$x_2^2=x_3x_1\quad\text{and}\quad x_3^2=x_1x_2$$ Eliminating $x_3$ gives $$\left(\frac{x_2^2}{x_1}\right)^2=x_1x_2$$ to get $$(x_1-x_2)\left(\left(x_1+\frac{x_2}{2}\right)^2+\frac{3x_2^2}{4}\right)=0$$ from which we have $x_1=x_2$ since $x_1,x_2$ are real numbers.

So there is no such $(a,b,c,t,x_1,x_2,x_3)$.

mathlove
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