If $a+b+c=0$ then the roots of $ax^2+bx+c=0$ are rational ?
Is it a "If and only if " statement or "only if " statement ?
For $a,b,c \in \mathbb Q$ , I think it is a "if and only if" statement . Am I correct ?
I can prove that if $a+b+c=0$ and $a,b,c \in \mathbb Q$ , then roots are rational.
But I can not prove that if roots are rational and $a,b,c \in \mathbb Q$ , then $a+b+c=0$.
Any help ? or any conditions that $ax^2 +bx+c =0 $
should satisfy in order to have rational roots ?
My clarifications :
1) What is a rational number ? It is a number which can be written in the form $\frac{p}{q} $ , where $q \neq 0$ and $p , q \in \mathbb Z$
2)
In this case : Let's take the quadratic equation $ax^2+bx+c=0$ where $a \neq 0$.
We all know that the roots are given by , $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
Case 1 : Suppose that $a,b,c \in \mathbb Q$.
Then $x$ is rational if and only if $b^2-4ac$ is a perfect square or zero $(0,1,4,9,16,…)$.
Now we need to make $b^2-4ac$ a perfect square !
Now observe that if $a+b+c=0$ , then $b^2-4ac = (-a-c)^2-4ac=(a-c)^2$
That is if $a,b,c \in \mathbb Q$ and $a+b+c=0$ then the solutions are rational.
Case 2 : Suppose that $a,b,c \in \mathbb Q$.
If $c=0$ , then all the roots are rational. (This is easy if all $a,b,c$ are rationals)
Case 3 : Suppose that $b,c \in \mathbb R- \mathbb Q$.(If $a$ is irrational we can divide by $a$ )
$a+b+c=0$ condition does not satisfy.
Ex : $(1-\sqrt{2})x^2-2x+(1+\sqrt{2})=0$
Best Answer
If $a+b+c= 0 \implies x = 1$ is a root and is a rational number, and the other root is $x = \dfrac{c}{a}$ also a rational number since $a, c \in \mathbb{Q}$. To see a counter example for the other part, take $a = c = 3, b = 10$, then $a+b+c = 16 \neq 0$, yet the equation $3x^2+10x+3 = 0$ has rational roots $x = -3, -\dfrac{1}{3}$. Thus it is an "IF" statement and not "IF AND ONLY IF".