Vieta’s Formulas

Let $x_1$ and $x_2$ be the solutions of the quadratic equation

$$ ax^2+bx+c$$

Vieta’s Formulas state that

$$ x_1+x_2 = – \frac{b}{a} $$

$$ x_1x_2 = \frac{c}{a} $$

Problem (Russia, 1996)

Find all real numbers $a$, $b$, $c$, $d$, such that the equation $x^2+ax+b$ has solutions $x=c$ and $x=d$, and the equation $x^2+cx+d$ has solutions $x=a$ and $x=b$.

 

Solution

Answer: $(a,b,c,d)$ is $(a,0,-a,0)$ or $(1,-2,1,-2)$.

By Vieta’s Formulas we have

$$ a=-(c+d) $$

$$ c=-(a+b) $$

$$ b=cd $$

$$ d= ab $$

 

From the first two equalities we have that $a+c+d=a+b+c=0$, which implies that $b=d$. Therefore from the last two equalities we have $ab=cd$.

Case 1: $b=0$. Therefore $d=0$ and $c=-a$.

Case 2: $b \neq 0$. Therefore $a=c$, which implies that $a=c=1$ and $b=d=-2$.

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