# If $a+b+c=0$ then the roots of $ax^2+bx+c=0$ are rational?

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 : $(1-\sqrt{2})x^2-2x+(1+\sqrt{2})=0$

• If you take $x^2-4x+4=0$, this has rational roots, but it is not true that $a+b+c=0$. The condition for having rational roots is that $b^2-4ac$ is the square of a rational number. – Crostul Aug 28 '16 at 8:52
• You also could choose two zeros, say $x_0, x_1$ of the equation is $(x-x_0)(x-x_1)= x^2 -(x_0+x_1) x + x_0x_1$. Then you are asking for $1-x_0-x_1 + x_0x_1=0$ which is not true for any two rationals $x_0, x_1$. – user60589 Aug 28 '16 at 9:01
• The condition needs $\frac ca or \frac ba \in \mathbb Q$. – Takahiro Waki Aug 28 '16 at 11:38

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".

If $$c = -a-b$$, then the discriminant is $$b^2-4ac = b^2 +4a(a+b) = b^2 +4ab +4a^2 = (b+2a)^2.$$

Since the discriminant is a perfect square, then the roots are always rational.

Their values are

$$\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} =\dfrac{-b \pm (b+2a)}{2a} \in \left\{ 1, -\dfrac{a+b}{a} \right\}$$

Of course, now that you know this, you can say that

$$(x-1)(ax+a+b) = ax^2 + bx -(a+b) = ax^2+ bx + c$$