This is what I've done so far:
Using the formula, we get
$x = \frac{-c \pm \sqrt{c^2 - 24}}{2}$
For solutions to exist, $c \geq 5$. For solutions to be rational, $c^2 - 24$ has to be a perfect square.
I'm kind of stuck now. I'd appreciate if someone can give me a hint or two on how to solve this question. Thanks!
On
Observe that if $x \in \mathbb{Q}$ which is a solution of the equation $x^2 + cx + 6 = 0\text{ } (1)$, then solve for $c$ in terms of $x$ we obtain: $c = -\dfrac{x^2+6}{x}$. Thus if $c = - \dfrac{r^2+6}{r}$ with $r \in \mathbb{Q}\setminus \{0\}$, we prove the solutions of $(1)$ are rational numbers. We have: $x^2+ cx + 6 = 0 \implies x^2 - \left(\dfrac{r^2+6}{r}\right)x + 6 = 0\implies x^2r-r^2x - 6(x-r) = 0\implies (xr-6)(x-r)=0\implies x = r, \dfrac{6}{r}$ which are both rationals.
On
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 the solutions of $x^2+cx+6$ are rational, then $$ x^2+cx+6=(x-a)(x-b)=x^2-(a+b)x+ab $$ for rational $a,b$. Then we have $ab=6$, and $c$ is given by $c=-a-b$. We can choose any rational number $a\neq 0$, then $b=6/a$ and $c=-\frac{a^2+6}{a}$.